Heights & Distances

Let AB (h) be the height of the pole. Then,

BC = 15 m , ∠ACB = 30°

In triangle ABC,

We know that,

$$\begin{aligned}\Rightarrow \tan \theta = \frac{\text{Perpendicular}}{\text{base}} \\[1em] \Rightarrow \tan 30° = \frac{h}{15} \\[1em] \Rightarrow \frac{1}{\sqrt3} = \frac{h}{15} \\[1em] \Rightarrow h = \frac{15}{\sqrt3} \\[1em] \Rightarrow h = \frac{15}{1.732}\\[1em] \Rightarrow h = 8.66 \text{ m} .\end{aligned}$$

Hence, height of the pole is 8.66 m.

Let AB be the cliff.

Let the distance of the buoy from the foot of the cliff be d (BC). Then,

AB (h) = 50 m

From figure,

Angle of Elevation ∠ACB = Angle of Depression ∠CAD = 30° [Alternate interior angles]

In triangle ABC,

We know that,

$$\begin{aligned}\Rightarrow \tan \theta = \frac{\text{perpendicular}}{\text{base}} \\[1em] \Rightarrow \tan 30° = \frac{AB}{BC} \\[1em] \Rightarrow \frac{1}{\sqrt3} = \frac{50}{d} \\[1em] \Rightarrow d = 50 \times 1.732 \\[1em] \Rightarrow d = 86.6 \text{ m} .\end{aligned}$$

Hence, the distance of the buoy from the foot of the cliff 86.6 m.

Let AB be the height of pole = 12 m.

Length of shadow BC = $12\sqrt{3}$ m

In triangle ABC,

$$\begin{aligned}\Rightarrow \tan \theta = \frac{\text{perpendicular}}{\text{base}} \\[1em] \Rightarrow \tan \theta = \frac{12}{12\sqrt3} \\[1em] \Rightarrow \tan \theta = \frac{1}{\sqrt3} \\[1em] \Rightarrow \tan \theta = \tan 30° \\[1em] \Rightarrow \theta = 30°.\end{aligned}$$

Hence, angle of elevation of the sun is 30°.

Let AC be the length of string = 80 m and height of kite above ground be AB (h).

Angle of elevation = 60°

In triangle ABC,

$$\begin{aligned}\Rightarrow \sin \theta = \frac{\text{perpendicular}}{\text{hypotenuse}} \\[1em] \Rightarrow \sin 60° = \frac{h}{80} \\[1em] \Rightarrow \frac{\sqrt3}{2} = \frac{h}{80} \\[1em] \Rightarrow h = 80 \times \frac{\sqrt3}{2} \\[1em] \Rightarrow h = 40 \times 1.732 \\[1em] \Rightarrow h = 69.28 \text{ m}\end{aligned}$$

Hence, height of the kite above the ground is 69.28 m.

Let the required height AB = h metres and length of a string AC = 85 m.

Given,

tan θ = $\Big(\frac{15}{8}\Big) = \frac{\text{perpendicular}}{\text{base}}$

Let the Height be 15x and Base be 8x.

Hypotenuse = $\sqrt{(15x)^2 + (8x)^2}$

Hypotenuse = $\sqrt{289x^2}$ = 17x

In triangle ABC,

$$\begin{aligned}\Rightarrow \sin \theta = \frac{\text{perpendicular}}{\text{hypotenuse}} = \frac{h}{85} \\[1em] \Rightarrow \frac{15x}{17x} = \frac{h}{85} \\[1em] \Rightarrow \frac{15}{17} \times 85 = h \\[1em] \Rightarrow h = 5 \times 15 \\[1em] \Rightarrow h = 75 \text{ m}.\end{aligned}$$

Hence, height of kite from ground = 75 m.

Given,

cos θ = 0.53

From table of cosines, we have,

θ = 58°

Let AB be the height of tower = 20 m and distance of man form foot of tower be CB = x

In right angled triangle ABC,

$$\begin{aligned}\Rightarrow \tan \theta = \frac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \tan \theta = \Big(\frac{AB}{CB}\Big) \\[1em] \Rightarrow \tan 58° = \frac{20}{x} \\[1em] \Rightarrow 1.6 = \frac{20}{x} \\[1em] \Rightarrow x = \frac{20}{1.6} \\[1em] \Rightarrow x = 12.5 \text{ m}.\end{aligned}$$

Hence, distance of man form foot of tower is 12.5 m.

Let h be the height of the tower (AB), x be the distance from foot of tower to second observation D,

Since the man walked 64 m towards the tower, the distance from C to the tower is (x + 64) m.

In right angled triangle ABD,

$$\begin{aligned}\Rightarrow \tan \phi = \frac{\text{perpendicular}}{\text{base}} = \frac{h}{x} \\[1em] \Rightarrow \frac{3}{4} = \frac{h}{x} \\[1em] \Rightarrow x = \frac{4h}{3}.\end{aligned}$$

In right angled triangle ABC,

$$\begin{aligned}\Rightarrow \tan \theta = \frac{\text{perpendicular}}{\text{base}} = \frac{h}{x + 64} \\[1em] \Rightarrow \frac{7}{12} = \frac{h}{x + 64} \\[1em] \Rightarrow \frac{7}{12} = \frac{h}{\Big(\frac{4h}{3}\Big) + 64} \\[1em] \Rightarrow 7 \times \Big(\frac{4h}{3} + 64\Big) = 12h \\[1em] \Rightarrow \Big(\frac{28h}{3} + 448\Big) = 12h \\[1em] \Rightarrow 28h+ 1344 = 36h \\[1em] \Rightarrow 1344 = 36h - 28h \\[1em] \Rightarrow 1344 = 8h \\[1em] \Rightarrow h = \frac{1344}{8} \\[1em] \Rightarrow h = 168 \text{ m}.\end{aligned}$$

Hence, height of the tower is 168 m.

Let the height of the building be CD = 10 m,

Let the distance from the base of the building to point B be x and point A be y.

In right angled triangle CBD,

$$\begin{aligned}\Rightarrow \tan \theta = \frac{\text{perpendicular}}{\text{base}} \\[1em] \Rightarrow \tan 60° = \frac{CD}{BD} \\[1em] \Rightarrow \sqrt{3} = \frac{10}{x} \\[1em] \Rightarrow x = \frac{10}{\sqrt3}.\end{aligned}$$

In right angled triangle CBD,

$$\begin{aligned}\Rightarrow \tan 30° = \frac{CD}{AD} \\[1em] \Rightarrow \frac{1}{\sqrt{3}} = \frac{10}{y} \\[1em] \Rightarrow y = 10\sqrt{3}.\end{aligned}$$

The distance between the two points is the difference between their distances from the tower (y - x):

$$\begin{aligned}\Rightarrow AB = y - x \\[1em] \Rightarrow AB = 10\sqrt3 - \frac{10}{\sqrt3} \\[1em] \Rightarrow AB = \frac{30 - 10}{\sqrt3} \\[1em] \Rightarrow AB = \frac{20}{\sqrt3} \\[1em] \Rightarrow AB = \frac{20}{1.732} \\[1em] \Rightarrow AB = 11.547 \approx 11.55 \text{ m}.\end{aligned}$$

Hence, the distance between A and B is 11.55 m.

Given,

The height of the tower AB = h metres and the length of its shadow = x metres when the sun's altitude is 45°. When the sun's altitude is 30°, then the length of shadow of tower is 10 m longer, i.e., BA = h meters, AD = x meters and CD = 10 metres.

From right angled △ABD, we get

$$\begin{aligned}\Rightarrow \tan 45° = \frac{BA}{AD} \\[1em] \Rightarrow 1 = \frac{h}{x} \\[1em] \Rightarrow h = x\end{aligned}$$

From right angled △BCA, we get

$$\begin{aligned}\Rightarrow \tan 30° = \frac{BA}{CA} \\[1em] \Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{CD + AD} \\[1em] \Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{10 + x} \\[1em] \Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{10 + h} [\because h = x] \\[1em] \Rightarrow 10 + h = \sqrt3h \\[1em] \Rightarrow 10 = \sqrt3h - h \\[1em] \Rightarrow 10 = h(1.732 - 1) \\[1em] \Rightarrow 10 = 0.732h \\[1em] \Rightarrow h = \frac{10}{0.732} \\[1em] \Rightarrow h = 13.66 \approx 13.7 \text{ m}.\end{aligned}$$

Hence, the height of the tower is 13.7 m.

Given,

AB is the tower of height h meters, BC = a meters and BD = b meters.

From figure,

In △ABD,

$$\begin{aligned}\Rightarrow \tan (90° - \theta) = \frac{AB}{BD} \\[1em] \Rightarrow \tan (90° - \theta) = \frac{h}{b} \\[1em] \Rightarrow \cot \theta = \frac{h}{b}....(1) \\[1em] \text{ In △ABC,} \\[1em] \Rightarrow \tan \theta = \frac{\text{perpendicular}}{\text{base}} \\[1em] \Rightarrow \tan \theta = \frac{AB}{BC} \\[1em] \Rightarrow \tan \theta = \frac{h}{a}.....(2)\end{aligned}$$

Multiplying (1) by (2) we get,

$$\begin{aligned}\Rightarrow \tan \theta \cot \theta = \frac{h}{a} \times \frac{h}{b} \\[1em] \Rightarrow \frac{\sin \theta}{\cos \theta} \times \frac{\cos \theta}{\sin \theta} = \frac{h^2}{ab} \\[1em] \Rightarrow 1 = \frac{h^2}{ab} \\[1em] \Rightarrow h^2 = ab \\[1em] \Rightarrow h = \sqrt{ab}.\end{aligned}$$

Hence, proved that h = $\sqrt{ab}$ meters.

Let man in the boat be originally at point D and after 1.5 minutes it reaches the point C and AB be the lighthouse.

AB = 150 meters.

In △ABD,

$$\begin{aligned}\Rightarrow \tan 60° = \frac{\text{BA}}{AD} \\[1em] \Rightarrow \sqrt3 = \frac{150}{a} \\[1em] \Rightarrow a = \frac{150}{\sqrt3} \times \frac{\sqrt3}{\sqrt3} \\[1em] \Rightarrow a = \frac{150\sqrt3}{3} \\[1em] \Rightarrow a = 50\sqrt3 \text{ meters.}\end{aligned}$$

In △ABC,

$$\begin{aligned}\Rightarrow \tan 45° = \frac{BA}{AC} \\[1em] \Rightarrow 1 = \frac{150}{a + x} \\[1em] \Rightarrow a + x = 150 \text{ meters.}\end{aligned}$$

⇒ x = 150 - a

= 150 - 50 $\sqrt3$

= 150 - 86.6

= 63.4 meters.

In 1.5 minutes boat covers 63.4 meters or boat covers 63.4 meters in 90 seconds.

$\text{ speed} = \frac{\text{Distance}}{\text{time}} = \frac{63.4}{90}$ = 0.70 m/sec.

Hence, the speed of boat = 0.70 m/sec.

Given,

AB and CD are the two towers of height h meters. E is a point in the roadway BD such that BD = 120 m, ∠AEB = 60° and ∠CED = 30°.

In ∆ABE,

$$\begin{aligned}\Rightarrow \tan 60° = \frac{AB}{BE} \\[1em] \Rightarrow \sqrt3 = \frac{h}{BE} \\[1em] \Rightarrow BE = \frac{h}{\sqrt3} \\[1em] \Rightarrow BE = \frac{h}{\sqrt3} \text{ .....(1)}\end{aligned}$$

In △CDE,

$$\begin{aligned}\Rightarrow \tan 30° = \frac{CD}{ED} \\[1em] \Rightarrow \frac{1}{\sqrt3} = \frac{CD}{ED} \\[1em] \Rightarrow ED = \sqrt3CD \\[1em] \Rightarrow ED = \sqrt3h \text{ ....(2)}\end{aligned}$$

We know that,

⇒ BD = 120 m

⇒ BE + ED = 120 m

From (1) and (2), we get :

$$\begin{aligned}\Rightarrow \frac{h}{\sqrt3} + \sqrt{3}h = 120 \\[1em] \Rightarrow \frac{h + 3h}{\sqrt3} = 120 \\[1em] \Rightarrow \frac{4h}{\sqrt3} = 120 \\[1em] \Rightarrow 4h = 120\sqrt3 \\[1em] \Rightarrow h = \frac{120 \times 1.732}{4} \\[1em] \Rightarrow h = 51.96 \text{ meters}\end{aligned}$$

From equation (1),

$BE = \frac{h}{\sqrt3} = \frac{51.96}{1.732}$ = 30 meters.

Hence, height of each pillar is 51.96 m and the point E is 30 m from the pillar AB.

Given,

XY = 40 m

⇒ PQ = h (let)

⇒ ∠PXQ = 60° and ∠RYQ = 45°

⇒ RP = XY = 40 m and RQ = PQ - RP = h - 40

In △RYQ,

$$\begin{aligned}\Rightarrow \tan 45° = \frac{RQ}{RY} \\[1em] \Rightarrow 1 = \frac{h - 40}{RY} \\[1em] \Rightarrow RY = h - 40.\end{aligned}$$

From figure,

PX = RY = h - 40

In △PXQ,

$$\begin{aligned}\Rightarrow \tan 60° = \frac{PQ}{PX} \\[1em] \Rightarrow \sqrt3 = \frac{h}{h - 40} \\[1em] \Rightarrow h\sqrt3 - 40\sqrt3 = h \\[1em] \Rightarrow h\sqrt3 - h = 40\sqrt3 \\[1em] \Rightarrow h = \frac{40\sqrt3}{(\sqrt3 - 1)} \\[1em] \Rightarrow h = \frac{40\sqrt3}{(\sqrt3 - 1)} \times \frac{(\sqrt3 + 1)}{(\sqrt3 + 1)} \\[1em] \Rightarrow h = \frac{40\sqrt3(\sqrt3 + 1)}{(3 - 1)} \\[1em] \Rightarrow h = \frac{40\sqrt3(\sqrt3 + 1)}{2} \\[1em] \Rightarrow h = 20\sqrt3(\sqrt3 + 1) \\[1em] \Rightarrow h = 60 + 20 \times \sqrt3 \\[1em] \Rightarrow h = 60 + 20 \times 1.73 \\[1em] \Rightarrow h = 60 + 34.6 \\[1em] \Rightarrow h = 94.6 \text{ m}\end{aligned}$$

In △QPX,

$$\begin{aligned}\Rightarrow \sin 60° = \frac{PQ}{XQ} \\[1em] \Rightarrow \frac{\sqrt3}{2} = \frac{PQ}{XQ} \\[1em] \Rightarrow XQ = \frac{2PQ}{\sqrt{3}} \\[1em] \Rightarrow XQ = \frac{2h}{\sqrt{3}} \\[1em] \Rightarrow XQ = \frac{2 \times 94.64}{\sqrt3}\\[1em] \Rightarrow XQ = \frac{189.28}{1.732}\\[1em] \Rightarrow XQ = 109.28 \text{ m}.\end{aligned}$$

Hence, height of tower PQ is 94.64 and XQ = 109.28 m.

Given,

AB is the lamp post and CD is the height of man and CE is the shadow of the man.

CE || DB.

Take AB = x and CD = 1.8 m

FA = CD = 1.8 m

DF = CA = 3.6 m

BF = AB - FA = (x - 1.8) m

Shadow (EC) = 5.4 m

Considering right angled △BDF, we get :

$$\begin{aligned}\Rightarrow \tan \theta = \frac{BF}{DF} \\[1em] \Rightarrow \tan \theta = \frac{x - 1.8}{3.6} \text{ ....(1)}\end{aligned}$$

Considering right angled △DCE, we get :

$$\begin{aligned}\Rightarrow \tan \theta = \frac{CD}{EC} \\[1em] \Rightarrow \tan \theta = \frac{1.8}{5.4} = \frac{1}{3} \text{ ....(2)}\end{aligned}$$

Comparing Eq 1 and Eq 2 we get,

$$\begin{aligned}\Rightarrow \frac{x - 1.8}{3.6} = \frac{1}{3} \\[1em] \Rightarrow 3x - 5.4 = 3.6 \\[1em] \Rightarrow 3x = 5.4 + 3.6 \\[1em] \Rightarrow 3x = 9 \\[1em] \Rightarrow x = 3.\end{aligned}$$

Hence, the height of the lamp post is 3 meters.

Let's take AD to be the height of the man, AD = 2 m.

From figure, BE = AD = 2 m.

Also,

CE = BC - BE = (10 - 2) = 8 m.

In ΔCED,

$$\begin{aligned}\Rightarrow \tan \theta = \frac{perpendicular}{\text{base}} \\[1em] \Rightarrow \tan x^{\circ} = \frac{CE}{DE} \\[1em] \Rightarrow \frac{2}{5} = \frac{8}{DE} \\[1em] \Rightarrow DE = \frac{8 \times 5}{2} \\[1em] \Rightarrow DE = \frac{40}{2} \\[1em] \Rightarrow DE = 20 \text{ m.}\end{aligned}$$

From figure,

AB = DE = 20 m.

Hence, AB = 20 m.

⇒ AB = DE = 10 m.

In ∆AED,

$$\begin{aligned}\Rightarrow \tan y^{\circ} = \frac{\text{perpendicular}}{\text{base}} \\[1em] \Rightarrow \tan y^{\circ} = \frac{DE}{AE} \\[1em] \Rightarrow \frac{1}{4} = \frac{DE}{AE} \\[1em] \Rightarrow AE = 4DE \\[1em] \Rightarrow AE = 4 \times 10 \\[1em] \Rightarrow AE = 40 \text{ m}.\end{aligned}$$

In ∆AEC,

$$\begin{aligned}\Rightarrow \tan x^{\circ} = \frac{\text{perpendicular}}{\text{base}} \\[1em] \Rightarrow \tan x^{\circ} = \frac{CE}{AE} \\[1em] \Rightarrow \frac{5}{2} = \frac{CE}{AE} \\[1em] \Rightarrow CE = \frac{5}{2}AE \\[1em] \Rightarrow CE = \frac{5}{2} \times 40 \\[1em] \Rightarrow CE = 100 \text{ m.}\end{aligned}$$

From figure,

CD = DE + CE = 10 + 100 = 110 m.

Hence, height of tower (CD) = 110 m.

Let A be the position of the aeroplane and B be the point on the sea surface vertically below the plane.

Let C and D be the positions of the two ships.

Let AB be the altitude of the aeroplane = 900 m

BC = y meters and CD = x meters

In ∆ABC,

$$\begin{aligned}\Rightarrow \tan 60^{\circ} = \frac{\text{perpendicular}}{\text{base}} = \frac{AB}{BC} \\[1em] \Rightarrow \sqrt3 = \frac{900}{y} \\[1em] \Rightarrow y = \frac{900}{\sqrt3} \\[1em] \Rightarrow y = \frac{900}{\sqrt3} \times \frac{\sqrt3}{\sqrt3} \\[1em] \Rightarrow y = 300\sqrt3 \text{ m .......(1)}\end{aligned}$$

In ∆ABD,

$$\begin{aligned}\Rightarrow \tan 30^{\circ} = \frac{\text{perpendicular}}{\text{base}} = \frac{AB}{BD} \\[1em] \Rightarrow \frac{1}{\sqrt3} = \frac{900}{x + y} \\[1em] \Rightarrow x + y = 900\sqrt3 \text{ m .............(2)}\end{aligned}$$

Subtracting equation (1) from (2), we get :

(x + y) - y = $900\sqrt3 - 300\sqrt3$

x = $600\sqrt3$

x = 600(1.732)

x = 1039.20 m.

Hence, the distance between the ships = 1039.20 m

Considering right angled △ABC, we get

$$\begin{aligned}\Rightarrow \tan 45^{\circ} = \frac{\text{Perpendicular}}{\text{Base}} = \frac{AB}{AC} \\[1em] \Rightarrow 1 = \frac{180}{AC} \\[1em] \Rightarrow AC = 180 \text{ m}\end{aligned}$$

Considering right angled △ADB, we get

$$\begin{aligned}\Rightarrow \tan 60^{\circ} = \frac{\text{Perpendicular}}{\text{Base}} = \frac{AB}{AD} \\[1em] \Rightarrow \sqrt{3} = \frac{180}{AD} \\[1em] \Rightarrow AD = \frac{180}{\sqrt{3}} \\[1em] \Rightarrow AD = 103.92 \text{ m.}\end{aligned}$$

Distance between two objects (CD) = CA - DA = 180 - 103.92 = 76.08 m

Hence,the distance between two objects = 76.08 meters.

Let AB be the height of church = 96 m.

Hence, the angle of elevation of the first vehicle at position D to the top of the church is x° and that of the second vehicle at position C is y°.

Considering right angled △ABD, we get

$$\begin{aligned}\Rightarrow \tan x^{\circ} = \frac{\text{perpendicular}}{\text{base}} = \frac{AB}{BD} \\[1em] \Rightarrow \frac{1}{4} = \frac{96}{BD} \\[1em] \Rightarrow BD = 96 \times 4 \\[1em] \Rightarrow BD = 384 \text{ m.}\end{aligned}$$

Considering right angled △ABC, we get

$$\begin{aligned}\Rightarrow \tan y^{\circ} = \frac{\text{perpendicular}}{\text{base}} = \frac{AB}{BC} \\[1em] \Rightarrow \frac{1}{7} = \frac{96}{BC} \\[1em] \Rightarrow BC = 96 \times 7 \\[1em] \Rightarrow BC = 672 \text{ m.}\end{aligned}$$

The distance between the vehicles = 672 m - 384 m = 288 m.

Hence, distance between the vehicles is 288 m.

From figure,

CD is the distance between two persons.

In △ABC, we get

$$\begin{aligned}\Rightarrow \tan 50^{\circ} = \frac{\text{perpendicular}}{\text{base}} = \frac{AB}{BC} \\[1em] \Rightarrow \tan 50^{\circ} = \frac{70}{x} \\[1em] \Rightarrow x = \frac{70}{\tan 50^{\circ}} \\[1em] \Rightarrow x = \frac{70}{1.1917} \\[1em] \Rightarrow x = 58.74 \text{ m.}\end{aligned}$$

In △ABD, we get

$$\begin{aligned}\Rightarrow \tan 25^{\circ} = \frac{\text{Perpendicular}}{\text{base}} = \frac{AB}{BD} \\[1em] \Rightarrow \tan 25^{\circ} = \frac{70}{x + y} \\[1em] \Rightarrow x + y = \frac{70}{0.4663} \\[1em] \Rightarrow x + y = 150.12 \text{ m.}\end{aligned}$$

CD = BD - BC = (x + y) - x = 150.12 - 58.74 = 91.38 m.

Hence, distance between the two men is 91.38 m.

(i) Given,

Height of the tower CD = 60 m

Angle of depression to the foot of pole (A) = 60°

Angle of depression to the top of pole (B) = 30°

Let the horizontal distance between the foot of the pole and the foot of the tower be x.

Considering right angled △ACD, we get :

$$\begin{aligned}\Rightarrow \tan 60^{\circ} = \frac{\text{perpendicular}}{\text{base}} = \frac{CD}{AC} \\[1em] \Rightarrow \sqrt3 = \frac{60}{AC} \\[1em] \Rightarrow AC = \frac{60}{\sqrt3}\\[1em] \Rightarrow AC = \frac{60\times \sqrt3}{\sqrt3 \times \sqrt3}\\[1em] \Rightarrow AC = 20\sqrt3 \\[1em] \Rightarrow AC = 34.64 \text{ m.}\end{aligned}$$

Hence, horizontal distance between pole and tower = 34.64 m.

(ii) In △DEB,

$$\begin{aligned}\Rightarrow \tan 30^{\circ} = \frac{\text{Perpendicular}}{\text{Base}} = \frac{DE}{BE} \\[1em] \Rightarrow \frac{1}{\sqrt3} = \frac{x}{20\sqrt3} \\[1em] \Rightarrow 1 = \frac{x}{20}\\[1em] \Rightarrow x = 20 \text{ m.}\end{aligned}$$

Height of the pole AB = CE = 60 - DE

= 60 - x

= 60 - 20

= 40 m.

Hence, height of the pole = 40 m.

(i) Let AB be the cliff, BC be the pillar and O be the point of observation.

Then, ∠AOB = 34°, ∠AOC = 36° and OA = 200 m

In △AOB,

$$\begin{aligned}\Rightarrow \tan 34° = \frac{\text{Perpendicular}}{\text{Base}} = \frac{AB}{OA} \\[1em] \Rightarrow 0.6745 = \frac{AB}{200} \\[1em] \Rightarrow AB = (0.6745 \times 200) \\[1em] \Rightarrow AB = 134.9 \text{ m.}\end{aligned}$$

Hence, height of the cliff 134.9 m.

(ii) In △AOC,

$$\begin{aligned}\Rightarrow \tan 36° = \frac{\text{Perpendicular}}{\text{Base}} = \frac{AC}{OA} \\[1em] \Rightarrow 0.726 = \frac{AC}{200} \\[1em] \Rightarrow AC = (0.726 \times 200) \\[1em] \Rightarrow AC = 145.2 \text{ m.}\end{aligned}$$

Height of pillar = AC - AB = 145.2 - 134.9 = 10.3 m

Hence, height of the pillar 10.3 m.

Let the height of the tower AB = h.

Let the speed of the car be x m/s.

Let the time taken to travel from D to A be t seconds.

CD (distance) = speed × time = x m/s × 600 (s) = 600x meters.

DA = (tx) meters.

In △ABD,

$$\begin{aligned}\Rightarrow \tan 45° = \frac{\text{Perpendicular}}{\text{Base}} = \frac{AB}{AD} \\[1em] \Rightarrow 1 = \frac{h}{tx} \\[1em] \Rightarrow h = tx ....(1)\end{aligned}$$

In △ABC,

$$\begin{aligned}\Rightarrow \tan 30° = \frac{\text{Perpendicular}}{\text{Base}} = \frac{AB}{CA} \\[1em] \Rightarrow \frac{1}{\sqrt3} = \frac{h}{600x + tx} \\[1em] \Rightarrow h = \frac{600x + tx}{\sqrt3} ....(2)\end{aligned}$$

From (1) and (2), we get :

$$\begin{aligned}\Rightarrow tx = \frac{600x + tx}{\sqrt3} \\[1em] \Rightarrow \sqrt3tx = 600x + tx \\[1em] \Rightarrow \sqrt3t = 600 + t \\[1em] \Rightarrow \sqrt3t - t = 600 \\[1em] \Rightarrow t(\sqrt3 - 1) = 600 \\[1em] \Rightarrow t = \frac{600}{(\sqrt3 - 1)} \\[1em] \Rightarrow t = \frac{600}{(\sqrt3 - 1)} \times \frac{(\sqrt3 + 1)}{(\sqrt3 + 1)} \\[1em] \Rightarrow t = \frac{600(\sqrt3 + 1)}{(3 - 1)} \\[1em] \Rightarrow t = \frac{600(\sqrt3 + 1)}{2} \\[1em] \Rightarrow t = 300(\sqrt3 + 1) = 819.6 \text{ sec.}\end{aligned}$$

Time taken to reach = $\frac{819.6}{60}$ = 13.66 mins i.e. 13 min 40 seconds.

Hence, time taken to reach tower is 13 min 40 sec.

Let C be the position of the cloud, l be the surface of the lake and D be reflection of the cloud.

Let AB be x and CB be h,

In △ABC,

$$\begin{aligned}\Rightarrow \tan 30° = \frac{\text{Perpendicular}}{\text{Base}} = \frac{BC}{AB} \\[1em] \Rightarrow \frac{1}{\sqrt3} = \frac{h}{x} \\[1em] \Rightarrow \sqrt3h = x ....(1)\end{aligned}$$

In △ABD,

$$\begin{aligned}\Rightarrow \tan 60° = \frac{\text{Perpendicular}}{\text{Base}} = \frac{BD}{AB} \\[1em] \Rightarrow \sqrt3 = \frac{25 + 25 + h}{x} \\[1em] \Rightarrow \sqrt3x = 50 + h ....(2)\end{aligned}$$

Substituting value of x from equation (1) in (2), we get :

$\Rightarrow \sqrt3(\sqrt3h) = 50 + h$

⇒ 3h = 50 + h

⇒ 2h = 50

⇒ h = 25 m.

Height of cloud above lake-level = OC = 25 + h = 25 + 25 = 50 m.

Hence, height of the cloud above lake-level is 50 m.

From figure,

Let height of the building CD be H meters and height of tower AB be h meters.

Given,

BD = 100 m

E is the midpoint of CD

∴ CE = $\frac{CD}{2} = \frac{H}{2}$

From figure,

⇒ ∠CAP = ∠XCA = 35° [Alternate angles are equal]

⇒ ∠EAP = ∠YEA = 14° [Alternate angles are equal]

From figure,

PD = AB = h

AP = BD = 100 m

In ΔACP,

$$\begin{aligned}\Rightarrow \text{tan 35°} = \frac{\text{Perpendicular}}{\text{Base}}\\[1em] \Rightarrow \text{tan 35°} = \frac{CP}{AP} \\[1em] \Rightarrow \text{tan 35°} = \frac{CD - PD}{AP} \\[1em] \Rightarrow 0.70 = \frac{H - h}{100} \\[1em] \Rightarrow 0.70 × 100 = H - h \\[1em] \Rightarrow H - h = 70 .....(1)\end{aligned}$$

In ΔAEP,

$$\begin{aligned}\Rightarrow \text{tan 14°} = \frac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \text{tan 14°} = \frac{EP}{AP}\\[1em] \Rightarrow \text{tan 14°} = \frac{ED - PD}{AP} \\[1em] \Rightarrow 0.25 = \frac{\frac{H}{2} - h}{100} \\[1em] \Rightarrow \frac{H}{2} - h = 0.25 × 100 \\[1em] \Rightarrow \frac{H}{2} - h = 25 .....(2)\end{aligned}$$

Subtracting equation (2) from (1), we get:

$$\begin{aligned}\Rightarrow H - h - \Big({\frac{H}{2} - h}\Big) = 70 - 25 \\[1em] \Rightarrow H - h - {\frac{H}{2} + h} = 45 \\[1em] \Rightarrow H - \frac{H}{2} = 45 \\[1em] \Rightarrow \frac{H}{2} = 45 \\[1em] \Rightarrow H = 45 \times 2 = 90 \text{ m}.\end{aligned}$$

⇒ H - h = 70

⇒ 90 - h = 70

⇒ h = 90 - 70

⇒ h = 20 m.

Hence, height of the tower AB = 20 m and height of the building CD = 90 m.

Let height of tower PT be h meters.

Considering right angled ΔPQR, we get

$$\begin{aligned}\Rightarrow \tan \theta = \frac{\text{perpendicular}}{\text{base}} \\[1em] \Rightarrow \tan 60^{\circ} = \frac{QR}{PQ} \\[1em] \Rightarrow \sqrt{3} = \frac{50}{PQ} \\[1em] \Rightarrow PQ = \frac{50}{\sqrt{3}} \text{ m.}\end{aligned}$$

Now considering right angled ΔPQT, we get

$$\begin{aligned}\Rightarrow \tan \theta = \frac{\text{perpendicular}}{\text{base}} \\[1em] \Rightarrow \tan 30^{\circ} = \frac{h}{PQ} \\[1em] \Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{\frac{50}{\sqrt{3}}} \\[1em] \Rightarrow \frac{1}{\sqrt{3}} = \frac{\sqrt{3} \times h}{50} \\[1em] \Rightarrow h = \frac{50}{\sqrt{3} \times \sqrt{3}} \\[1em] \Rightarrow h = \frac{50}{3} \\[1em] \Rightarrow h = 16.7 \text{ m.}\end{aligned}$$

On correcting to nearest meter, h = 17 m.

Hence, the height of the tower PT = 17 m.

Let tower be QR and initial position of man be P, since the initial angle of elevation is 45°, considering right angled △PQR we get,

$$\begin{aligned}\Rightarrow \tan 45° = \frac{\text{perpendicular}}{\text{base}} \\[1em] \Rightarrow 1 = \frac{QR}{PQ} \\[1em] \Rightarrow PQ = QR.\end{aligned}$$

After covering 20 m let the man be at point S, so PS = 20 m and SQ = PQ - PS = PQ - 20 = QR - 20.

Now considering right angled △SQR we get,

$$\begin{aligned}\Rightarrow \tan 60^{\circ} = \frac{QR}{SQ} \\[1em] \Rightarrow \sqrt3 = \frac{QR}{QR-20} \\[1em] \Rightarrow \sqrt3(QR - 20) = QR \\[1em] \Rightarrow QR\sqrt3 - 20\sqrt3 = QR \\[1em] \Rightarrow QR\sqrt3 - QR = 20(1.732) \\[1em] \Rightarrow QR(\sqrt3 - 1) = 34.64 \\[1em] \Rightarrow QR(1.732 - 1) = 34.64 \\[1em] \Rightarrow 0.732QR = 34.64 \\[1em] \Rightarrow QR = \frac{34.64}{0.732} \\[1em] \Rightarrow QR = 47.32.\end{aligned}$$

On correcting to 2 significant figures QR = 47.

Hence, the height of the tower is 47 m.

(i) Let R be the top of the cliff and Q be the foot of the cliff such that RQ = 60 m.

Let P and T be the positions of the two boats such that the angles of depression from R are 30° and 60° respectively.

From figure,

∠RPQ = 60° and ∠RTQ = 30°

Boats on the same side of the cliff

From right angled ΔTQR, we get

$$\begin{aligned}\Rightarrow \tan \theta = \frac{\text{perpendicular}}{\text{base}} \\[1em] \Rightarrow \tan 30^{\circ} = \frac{RQ}{TQ} \\[1em] \Rightarrow \frac{1}{\sqrt{3}} = \frac{60}{TQ} \\[1em] \Rightarrow TQ = 60\sqrt{3}\end{aligned}$$

From right angled ΔPQR, we get

$$\begin{aligned}\Rightarrow \tan \theta = \frac{\text{perpendicular}}{\text{base}} \\[1em] \Rightarrow \tan 60^{\circ} = \frac{RQ}{PQ} \\[1em] \Rightarrow \sqrt{3} = \frac{60}{PQ} \\[1em] \Rightarrow PQ = \frac{60}{\sqrt{3}}\end{aligned}$$

Distance between the boats,

$$\begin{aligned}\Rightarrow PT = TQ - PQ \\[1em] \Rightarrow PT = 60\sqrt{3} - \frac{60}{\sqrt{3}} \\[1em] \Rightarrow PT = \frac{180 - 60}{\sqrt{3}} \\[1em] \Rightarrow PT = \frac{120}{\sqrt{3}} \\[1em] \Rightarrow PT = 40\sqrt{3} \\[1em] \Rightarrow PT = 69.28 \text{ m.}\end{aligned}$$

Hence, the distance between the boats is 69.28 m when they are on the same side of the cliff

(ii) Let R be the top of the cliff and Q be the foot of the cliff such that RQ = 60 m.

Let P and T be the positions of the two boats on opposite sides of cliff, such that the angles of depression from R are 30° and 60° respectively.

From figure,

∠RPQ = 60° and ∠RTQ = 30°

The distance between the boats, when they are on opposite sides of cliff,

$$\begin{aligned}\Rightarrow PT = TQ + PQ \\[1em] \Rightarrow PT = 60\sqrt{3} + \frac{60}{\sqrt{3}} \\[1em] \Rightarrow PT = \frac{180 + 60}{\sqrt{3}} \\[1em] \Rightarrow PT = \frac{240}{\sqrt{3}} \\[1em] \Rightarrow PT = 80\sqrt{3} \\[1em] \Rightarrow PT = 138.56 \text{ m.}\end{aligned}$$

Hence, boats are 138.56 m when they are on the opposite sides of the cliff.

Given,

A is the top of the tower and B the foot. C and D be two consecutive kilometer stones with depression angles 30° and 45° respectively.

Since stones are consecutive kilometer stones hence distance between them = 1 km.

From figure,

∠XAD = ∠ADB = 30° [Alternate angles are equal]

∠XAC = ∠ACB = 45° [Alternate angles are equal]

CD = 1 km

DB = x + 1

From right angled ΔABC, we get

$$\begin{aligned}\Rightarrow \tan \theta = \frac{\text{perpendicular}}{\text{base}} \\[1em] \Rightarrow \tan 45^{\circ} = \frac{AB}{CB} \\[1em] \Rightarrow 1 = \frac{h}{x} \\[1em] \Rightarrow h = x .\end{aligned}$$

From right angled ΔADB, we get

$$\begin{aligned}\Rightarrow \tan \theta = \frac{\text{perpendicular}}{\text{base}} \\[1em] \Rightarrow \tan 30^{\circ} = \frac{AB}{DB} \\[1em] \Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{x + 1} \\[1em] \Rightarrow \frac{1}{\sqrt{3}} = \frac{x}{x + 1} \\[1em] \Rightarrow x + 1 = \sqrt3 x \\[1em] \Rightarrow \sqrt{3}x - x = 1 \\[1em] \Rightarrow x(\sqrt{3} - 1) = 1 \\[1em] \Rightarrow x = \frac{1}{\sqrt3 - 1} \\[1em] \Rightarrow x = \frac{1}{\sqrt3 - 1} \times \frac{\sqrt3 + 1}{\sqrt3 + 1} \\[1em] \Rightarrow x = \frac{\sqrt3 + 1}{(\sqrt{3})^2 - (1)^2} \\[1em] \Rightarrow x = \frac{2.73}{2} = 1.36 \text{ km}\end{aligned}$$

DB = x + 1 = 1.36 + 1 = 2.36 km.

Hence, the distance of two stones from hill are 1.36 km and 2.36 km.

From figure,

O is the position of aeroplane and P and Q are the position of ships.

OA = 1500 m

Let,

AQ = y

QP = x

In right angled ΔOAQ,

$$\begin{aligned}\Rightarrow \tan \theta = \frac{\text{perpendicular}}{\text{base}} \\[1em] \Rightarrow \tan 45^{\circ} = \frac{OA}{AQ} \\[1em] \Rightarrow 1 = \frac{1500}{y} \\[1em] \Rightarrow y = 1500 \text{ m.}\end{aligned}$$

In right angled ΔOAP,

$$\begin{aligned}\Rightarrow \tan \theta = \frac{\text{perpendicular}}{\text{base}} \\[1em] \Rightarrow \tan 30^{\circ} = \frac{OA}{AP} \\[1em] \Rightarrow \frac{1}{\sqrt{3}} = \frac{1500}{AQ + QP} \\[1em] \Rightarrow \frac{1}{\sqrt{3}} = \frac{1500}{y + x} \\[1em] \Rightarrow y + x = 1500\sqrt{3} \\[1em] \Rightarrow 1500 + x = 1500\sqrt{3} \\[1em] \Rightarrow x = 1500\sqrt{3} - 1500 \\[1em] \Rightarrow x = 1500(\sqrt{3} - 1) \\[1em] \Rightarrow x = 1500( 1.73 - 1) \\[1em] \Rightarrow x = 1500( 0.73) \\[1em] \Rightarrow x = 1095 \text{ m.}\end{aligned}$$

AQ = y = 1500 m

PQ = x = 1095 m

Hence, the distance between the two ships = 1095 m.

Given,

Aeroplane is at point A and boats are at point B and C. Since, aeroplane is at an altitude of 250 m ,

∴ AD = 250 m.

Considering right angled ΔACD, we get

$$\begin{aligned}\Rightarrow \tan \theta = \frac{\text{perpendicular}}{\text{base}} \\[1em] \Rightarrow \tan 60^{\circ} = \frac{AD}{DC} \\[1em] \Rightarrow \sqrt{3} = \frac{250}{y} \\[1em] \Rightarrow y = \frac{250}{\sqrt{3}} \\[1em] \Rightarrow y = \frac{250}{1.732} \\[1em] \Rightarrow y = 144.34 \text{ m.}\end{aligned}$$

Considering right angled ΔABD, we get

$$\begin{aligned}\Rightarrow \tan \theta = \frac{\text{perpendicular}}{\text{base}} \\[1em] \Rightarrow \tan 45^{\circ} = \frac{AD}{BD} \\[1em] \Rightarrow 1 = \frac{250}{x} \\[1em] \Rightarrow x = 250 \text{ m.}\end{aligned}$$

Width of the river (BC) = x + y = 144.34 + 250 = 394.34 meters.

Rounding off to nearest meter BC = 394 meters.

Hence, the width of the river is 394 meters.

Let CD be the tower.

From figure,

⇒ ∠A = ∠EDA = 45° (Alternate angles are equal)

⇒ ∠B = ∠FDB = 38° (Alternate angles are equal)

In ΔACD,

$$\begin{aligned}\Rightarrow \tan \theta = \frac{\text{perpendicular}}{\text{base}} \\[1em] \Rightarrow \tan 45^{\circ} = \frac{CD}{AC} \\[1em] \Rightarrow 1 = \frac{100}{AC} \\[1em] \Rightarrow AC = 100 \text{ m.}\end{aligned}$$

In ΔBCD,

$$\begin{aligned}\Rightarrow \tan \theta = \frac{\text{perpendicular}}{\text{base}} \\[1em] \Rightarrow \tan 38^{\circ} = \frac{CD}{BC} \\[1em] \Rightarrow 0.78 = \frac{100}{BC} \\[1em] \Rightarrow BC = \frac{100}{0.78} \\[1em] \Rightarrow BC = 128.20 \text{ m.}\end{aligned}$$

From figure,

The distance between ships A and B = AC + BC

= 100 + 128.20

= 228.20 m.

Hence, the distance between the two ships A and B = 228.20 m.

From figure,

AB is the first tower and CD is the second tower.

From figure,

AC = ED = 120 m.

In ΔBED,

$$\begin{aligned}\Rightarrow \tan \theta = \frac{\text{perpendicular}}{\text{base}} \\[1em] \Rightarrow \tan 30^{\circ} = \frac{BE}{ED} \\[1em] \Rightarrow \frac{1}{\sqrt{3}} = \frac{BE}{120} \\[1em] \Rightarrow BE = \frac{120}{\sqrt{3}} \\[1em] \Rightarrow BE = \frac{120}{1.732} \\[1em] \Rightarrow BE = 69.3 \text{ m.}\end{aligned}$$

In ΔEDA,

$$\begin{aligned}\Rightarrow \tan \theta = \frac{\text{perpendicular}}{\text{base}} \\[1em] \Rightarrow \tan 24^{\circ} = \frac{EA}{ED} \\[1em] \Rightarrow 0.445 = \frac{EA}{120} \\[1em] \Rightarrow EA = 120 \times 0.445 \\[1em] \Rightarrow EA = 53.4 \text{ m.}\end{aligned}$$

⇒ AB = AE + EB = 53.4 + 69.3 = 122.7 meters.

⇒ CD = EA = 53.4 meters.

Hence, height of two towers = 122.7 meters and 53.4 meters.

AB is the cliff and CD is the tower.

From figure,

∠ACE = ∠FAC = 45° (Alternate angles are equal)

∠ADB = ∠FAD = 60° (Alternate angles are equal)

Let BD = x.

From figure,

EC = BD = x meters.

EB = CD = 20 meters.

In ΔAEC,

⇒ tan 45° = $\frac{AE}{EC}$

⇒ 1 = $\frac{AE}{x}$

⇒ AE = x meters

In ΔABD,

$$\begin{aligned}\Rightarrow \tan 60^{\circ} = \frac{AB}{BD} \\[1em] \Rightarrow \sqrt{3} = \frac{AE + EB}{x} \\[1em] \Rightarrow \sqrt{3} = \frac{x + 20}{x} \\[1em] \Rightarrow \sqrt{3}x = x + 20 \\[1em] \Rightarrow \sqrt{3}x - x = 20 \\[1em] \Rightarrow x(\sqrt{3} - 1) = 20 \\[1em] \Rightarrow x = \frac{20}{\sqrt{3} - 1} \\[1em] \Rightarrow x = \frac{20(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} \\[1em] \Rightarrow x = \frac{20(\sqrt{3} + 1)}{3 - 1} \\[1em] \Rightarrow x = \frac{20(\sqrt{3} + 1)}{2}\\[1em] \Rightarrow x = 10(\sqrt{3} + 1) = 27.32 \text{ meters}.\end{aligned}$$

From Figure,

Height of cliff (AB) = AE + EB

= x + 20

= 27.32 + 20 = 47.32 meters.

Hence, the height of cliff = 47.32 meters.

(ii) From figure,

Distance between cliff and tower (BD) = x meters = 27.32 meters.

Hence, distance between cliff and tower = 27.32 meters.

Given,

AB = 100 m

CD = 100 m

∠APB = 40°

∠CPD = 60°

Let PB = x and PD = y.

In triangle ABP,

In triangle ABP,

$$\begin{aligned}\Rightarrow \tan \theta = \frac{\text{perpendicular}}{\text{base}} \\[1em] \Rightarrow \tan 40° = \frac{AB}{BP} \\[1em] \Rightarrow \tan 40° = \frac{100}{x} \\[1em] \Rightarrow 0.8391 = \frac{100}{x} \\[1em] \Rightarrow x = \frac{100}{0.8391} \\[1em] \Rightarrow x = 119.17 \text{ m.}\end{aligned}$$

PB = x = 119.17 m

In triangle CDP,

We know that,

$$\begin{aligned}\Rightarrow \tan \theta = \frac{\text{perpendicular}}{\text{base}} \\[1em] \Rightarrow \tan 60° = \frac{CD}{PD} \\[1em] \Rightarrow \sqrt{3} = \frac{100}{y} \\[1em] \Rightarrow y = \frac{100}{\sqrt{3}} \\[1em] \Rightarrow y = \frac{100}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} \\[1em] \Rightarrow y = \frac{100\sqrt{3}}{3} \text{ m.}\end{aligned}$$

PD = y = $\frac{100\sqrt{3}}{3}$ m

Hence, the distances are PB = 119.17 m and PD = $\frac{100\sqrt{3}}{3}$ m.

Let the height of the rod AB be h and length of shadow BC be s.

In triangle ABC,

$$\begin{aligned}\Rightarrow \tan \theta = \frac{\text{perpendicular}}{\text{base}} \\[1em] \Rightarrow \tan \theta = \frac{h}{s} \\[1em] \Rightarrow \tan \theta = \frac{1}{\sqrt3} \\[1em] \Rightarrow \tan \theta = \tan 30^{\circ} \\[1em] \Rightarrow \theta = 30^{\circ}.\end{aligned}$$

Hence, option 1 is the correct option.

Let the height of the pole AB be h and length of shadow BC be s.

Given,

h = s

In triangle ABC,

$$\begin{aligned}\Rightarrow \tan \theta = \frac{\text{perpendicular}}{\text{base}} \\[1em] \Rightarrow \tan \theta = \frac{h}{s} \\[1em] \Rightarrow \tan \theta = \frac{h}{h} \\[1em] \Rightarrow \tan \theta = 1 \\[1em] \Rightarrow \tan \theta = \tan 45^{\circ} \\[1em] \Rightarrow \theta = 45^{\circ}.\end{aligned}$$

Hence, option 2 is the correct option.

Let the height of kite (AC) be h.

Length of string (AB) = 2h

In triangle ABC,

$$\begin{aligned}\Rightarrow \sin \theta = \frac{\text{Perpendicular}}{\text{hypotenuse}} \\[1em] \Rightarrow \sin \theta = \frac{h}{2h} \\[1em] \Rightarrow \sin \theta = \frac{1}{2} \\[1em] \Rightarrow \sin \theta = \sin 30^{\circ} \\[1em] \Rightarrow \theta = 30^{\circ}.\end{aligned}$$

Hence, option 1 is the correct option.

Let the height of tower (AB) be h meters.

Angle of elevation = 30°

$$\begin{aligned}\Rightarrow \tan \theta = \frac{\text{perpendicular}}{\text{base}} \\[1em] \Rightarrow \tan 30^{\circ} = \frac{h}{100} \\[1em] \Rightarrow \frac{1}{\sqrt3} = \frac{h}{100} \\[1em] \Rightarrow h = \frac{100}{\sqrt3} \text{ m}.\end{aligned}$$

Hence, option 1 is the correct option.

Let the length of ladder (AB) be L.

Distance (CB) from wall (AC) to the foot of ladder = 2 m

The angle the ladder makes with the ground is 60°.

In triangle ABC,

$$\begin{aligned}\Rightarrow \cos \theta = \frac{\text{Base}}{\text{hypotenuse}} \\[1em] \Rightarrow \cos 60^{\circ} = \frac{2}{L} \\[1em] \Rightarrow \frac{1}{2} = \frac{2}{L} \\[1em] \Rightarrow L = 4 \text{ m}.\end{aligned}$$

Hence, option 4 is the correct option.

Length of string (AC) = 90 m

Let the height of the kite (AB) = h

The angle the string makes with the ground is 60°.

$$\begin{aligned}\Rightarrow \sin \theta = \frac{\text{Perpendicular}}{\text{hypotenuse}} \\[1em] \Rightarrow \sin 60^{\circ} = \frac{h}{90} \\[1em] \Rightarrow \frac{\sqrt3}{2} = \frac{h}{90} \\[1em] \Rightarrow \frac{\sqrt3}{2} \times 90 = h \\[1em] \Rightarrow h = 45\sqrt3 \text{ m}.\end{aligned}$$

Hence, option 1 is the correct option.

Height of pole AB = 15 m

Let D be the position where angle of elevation is 30° and C be the point where angle of elevation is 60°.

In triangle ABC,

We know that,

$$\begin{aligned}\Rightarrow \tan 60^{\circ} = \frac{\text{Perpendicular}}{\text{Base}} = \frac{AB}{BC} \\[1em] \Rightarrow \sqrt3 = \frac{15}{BC} \\[1em] \Rightarrow BC = \frac{15}{\sqrt3} \\[1em] \Rightarrow BC = \frac{15}{\sqrt{3}} \times \frac{\sqrt3}{\sqrt3} \\[1em] \Rightarrow BC = \frac{15\sqrt3}{3} \\[1em] \Rightarrow BC = 5\sqrt3 \text{ m.}\end{aligned}$$

In triangle ABD,

We know that,

$$\begin{aligned}\Rightarrow \tan 30^{\circ} = \frac{\text{Perpendiuclar}}{\text{Base}} = \frac{AB}{BD} \\[1em] \Rightarrow \frac{1}{\sqrt3} = \frac{15}{BD} \\[1em] \Rightarrow BD = 15\sqrt3 \text{ m.}\end{aligned}$$

From figure,

CD = BD - BC

CD = $15\sqrt3 - 5\sqrt3$

CD = $10\sqrt3 \text{ m.}$

Hence, option 4 is the correct option.

Let the base of rectangle be AB and breadth be BC,

Diagonal AC = 6 cm

In triangle ABC,

$$\begin{aligned}\Rightarrow \cos 30^{\circ} = \frac{\text{base}}{\text{hypotenuse}} = \frac{AB}{AC} \\[1em] \Rightarrow \frac{\sqrt3}{2} = \frac{AB}{6} \\[1em] \Rightarrow \frac{\sqrt3}{2} \times 6 = AB \\[1em] \Rightarrow AB = 3\sqrt3 \text{ cm.}\end{aligned}$$

Also,

$$\begin{aligned}\Rightarrow \sin 30^{\circ} = \frac{\text{Perpendicular}}{\text{hypotenuse}} = \frac{BC}{AC} \\[1em] \Rightarrow \frac{1}{2} = \frac{BC}{6} \\[1em] \Rightarrow \frac{1}{2} \times 6 = BC \\[1em] \Rightarrow BC = 3 \text{ cm.}\end{aligned}$$

We know that,

Area of rectangle = length × width

= $3\sqrt3 \times 3$

= $9\sqrt3$ cm².

Hence, option 2 is the correct option.

Let the position of the aeroplane be A. Let h be the height of the aeroplane above the ground.

Let C and D be the positions of the two consecutive stones on the ground.

Let the distance from the closer stone C to the foot of the perpendicular B be x.

Then, CD = x + 1.

In △ABC,

$$\begin{aligned}\Rightarrow \tan 60^{\circ} = \frac{\text{Perpendicular}}{\text{Base}} = \frac{AB}{BC} \\[1em] \Rightarrow \sqrt3 = \frac{h}{x} \\[1em] \Rightarrow x = \frac{h}{\sqrt3} \text{ .....(1)}\end{aligned}$$

In △ABD,

$$\begin{aligned}\Rightarrow \tan 45^{\circ} = \frac{\text{Perpendicular}}{\text{Base}} = \frac{AB}{BD} \\[1em] \Rightarrow 1 = \frac{h}{x + 1} \\[1em] \Rightarrow x + 1 = h \text{ .....(2)}\end{aligned}$$

Substituting value of x from equation (2) in (1), we get :

$$\begin{aligned}\Rightarrow \frac{h}{\sqrt3} + 1 = h \\[1em] \Rightarrow h - \frac{h}{\sqrt{3}} = 1 \\[1em] \Rightarrow h \Big(1 - \frac{1}{\sqrt{3}} \Big) = 1 \\[1em] \Rightarrow h \Big( \frac{\sqrt{3} - 1}{\sqrt{3}} \Big) = 1 \\[1em] \Rightarrow h = \frac{\sqrt{3}}{\sqrt{3} - 1} \\[1em] \Rightarrow h = \frac{\sqrt{3}(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} \\[1em] \Rightarrow h = \frac{3 + \sqrt{3}}{(\sqrt{3})^2 - (1)^2} \\[1em] \Rightarrow h = \frac{3 + \sqrt{3}}{3 - 1} \\[1em] \Rightarrow h = \frac{3 + \sqrt{3}}{2} \text{ km}\end{aligned}$$

Hence, option 2 is the correct option.

Let AB be the tower of height h.

In △ABC,

$$\begin{aligned}\Rightarrow \tan 60^\circ = \frac{AB}{BC} \\[1em] \Rightarrow \sqrt{3} = \frac{h}{BC} \\[1em] \Rightarrow BC = \frac{h}{\sqrt{3}} ....(1)\end{aligned}$$

In △ABD,

$$\begin{aligned}\Rightarrow \tan 30^\circ = \frac{AB}{BD} \\[1em] \Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{BC + 20} \\[1em] \Rightarrow BC + 20 = h\sqrt{3} ....(2)\end{aligned}$$

Substituting value of BC from equation (1) in (2), we get :

$$\begin{aligned}\Rightarrow \frac{h}{\sqrt{3}} + 20 = h\sqrt{3} \\[1em] \Rightarrow \frac{h + 20\sqrt{3}}{\sqrt{3}} = h\sqrt{3} \\[1em] \Rightarrow h + 20\sqrt{3} = h(3) \\[1em] \\[1em] \Rightarrow 20\sqrt{3} = 2h \\[1em] \Rightarrow h = 10\sqrt{3} \text{ m}\end{aligned}$$

Hence, option 2 is the correct option.

Let A be the top of the lighthouse, C the initial position of ship and D be the position after 10 minutes.

From figure,

tan α = $\frac{AB}{BC}$

tan β = $\frac{AB}{BD}$

Since, BC is greater than BD.

∴ tan α < tan β

⇒ α < β.

Hence, option 2 is the correct option.

Let height of the tower (AB) = h.

Let BC = b and BD = a.

In △ABC,

$$\begin{aligned}\Rightarrow \tan 60^\circ = \frac{AB}{BC} \\[1em] \Rightarrow \tan 60^\circ = \frac{h}{b} ....(1)\end{aligned}$$

In △ABD,

$$\begin{aligned}\Rightarrow \tan 30^\circ = \frac{AB}{BD} \\[1em] \Rightarrow \tan 30^\circ = \frac{h}{a} \\[1em] \Rightarrow \tan (90^{\circ} - 60^\circ) = \frac{h}{a} \\[1em] \Rightarrow \cot 60^\circ = \frac{h}{a} \\[1em] \Rightarrow \frac{1}{\tan 60^\circ} = \frac{h}{a} \\[1em] \Rightarrow \frac{1}{\frac{h}{b}} = \frac{h}{a} \text{ [From (i)]} \\[1em] \Rightarrow \frac{b}{h} = \frac{h}{a} \\[1em] \Rightarrow h^2 = ab \\[1em] \Rightarrow h = \sqrt{ab}\end{aligned}$$

Hence, option 2 is the correct option.

Let AB be the cliff and two ships be at point C and D.

In triangle ABC,

$$\begin{aligned}\Rightarrow \tan 60^\circ = \frac{AB}{BC} \\[1em] \Rightarrow \sqrt3 = \frac{100}{BC} \\[1em] \Rightarrow BC = \frac{100}{\sqrt3} \text{ m}\end{aligned}$$

In triangle ABD,

$$\begin{aligned}\Rightarrow \tan 30^\circ = \frac{AB}{BD} \\[1em] \Rightarrow \frac{1}{\sqrt3} = \frac{100}{BD} \\[1em] \Rightarrow BD = 100\sqrt3 \text{ m}.\end{aligned}$$

Distance between the ships CD is,

$$\begin{aligned}\Rightarrow CD = BD - BC \\[1em] \Rightarrow CD = 100\sqrt{3} - \frac{100}{\sqrt{3}} \\[1em] \Rightarrow CD = 100 \Big( \sqrt{3} - \frac{1}{\sqrt{3}} \Big) \\[1em] \Rightarrow CD = 100 \Big(\frac{3 - 1}{\sqrt{3}}\Big) \\[1em] \Rightarrow CD = \frac{200}{\sqrt{3}} \\[1em] \Rightarrow CD = \frac{200}{1.732} = 115.47 \text{ m}.\end{aligned}$$

Hence, option 2 is the correct option.

Let height of the cliff be (AB) = 150 m.

Let C and D be the positions of the ships.

In triangle ABC,

$$\begin{aligned}\Rightarrow \tan(60^\circ) = \frac{AB}{BC} \\[1em] \Rightarrow \sqrt3 = \frac{150}{BC} \\[1em] \Rightarrow BC = \frac{150}{\sqrt3} = 50\sqrt3 \text{ m}\end{aligned}$$

In triangle ABD,

$$\begin{aligned}\Rightarrow \tan(45^\circ) = \frac{AB}{BD} \\[1em] \Rightarrow 1 = \frac{150}{BD} \\[1em] \Rightarrow BD = 150 \text{ m}.\end{aligned}$$

Distance the ship moved from position C to D,

CD = BD - BC

CD = 150 - $50\sqrt3$

CD = 50(3 - 1.732)

CD = 50(1.268)

CD = 63.4 m

Distance = 63.4 m = 0.0634 km

Time = 2 min = $\frac{1}{30}$ hr

$$\begin{aligned}\Rightarrow \text{Speed} = \frac{\text{Distance}}{\text{Time}} \\[1em] = \frac{0.0634}{\frac{1}{30}} \\[1em] = 0.0634 \times 30 \\[1em] = 1.9 \text{ km/hr}.\end{aligned}$$

Hence, option 1 is the correct option.

Height of the tower AB = 100 m

In triangle ABC,

$$\begin{aligned}\Rightarrow \tan 45^\circ = \frac{AB}{BC} \\[1em] \Rightarrow 1 = \frac{100}{BC} \\[1em] \Rightarrow BC = 100 \text{ m}.\end{aligned}$$

In triangle ABD,

$$\begin{aligned}\Rightarrow \tan 30^\circ = \frac{AB}{BD} \\[1em] \Rightarrow \frac{1}{\sqrt3} = \frac{100}{BD} \\[1em] \Rightarrow BD = 100\sqrt3 \text{ m}.\end{aligned}$$

The decrease in the shadow's length x = BD - BC

x = $100\sqrt3$ - 100

= $100(\sqrt3 - 1)$ m.

Hence, option 3 is the correct option.

Let AB be the height of tower = 40 m

Let the two points be P and Q on opposite sides of the tower.

In triangle ABP,

$$\begin{aligned}\Rightarrow \tan 45^\circ = \frac{AB}{BP} \\[1em] \Rightarrow 1 = \frac{40}{BP} \\[1em] \Rightarrow BP = 40 \text{ m}.\end{aligned}$$

In triangle ABQ,

$$\begin{aligned}\Rightarrow \tan 60^\circ = \frac{AB}{BQ} \\[1em] \Rightarrow \sqrt3 = \frac{40}{BQ} \\[1em] \Rightarrow BQ = \frac{40}{\sqrt3} \\[1em] \Rightarrow BQ = 23.09 \text{ m}.\end{aligned}$$

Distance between the two points = BP + BQ

= 40 + 23.09

= 63.09 = 63 m.

Hence, option 4 is the correct option.

Let height of lighthouse (CD) be h meters.

Let A and B be the boats approaching lighthouse.

In triangle ACD,

$$\begin{aligned}\Rightarrow \tan 30^\circ = \frac{h}{x} \\[1em] \Rightarrow \frac{1}{\sqrt3} = \frac{h}{x} \\[1em] \Rightarrow x = h\sqrt3.\end{aligned}$$

In triangle BCD,

$$\begin{aligned}\Rightarrow \tan 45^\circ = \frac{h}{y} \\[1em] \Rightarrow 1 = \frac{h}{y} \\[1em] \Rightarrow h = y.\end{aligned}$$

Given,

Distance between the two boats is 100 m.

x + y = 100

$$\begin{aligned}\Rightarrow h\sqrt3 + h = 100 \\[1em] \Rightarrow h(\sqrt{3} + 1) = 100 \\[1em] \Rightarrow h =\frac{100}{\sqrt3 + 1} \\[1em] \Rightarrow h = \frac{100(\sqrt{3} - 1)}{(\sqrt{3} + 1)(\sqrt{3} - 1)} \\[1em] \Rightarrow h = \frac{100(\sqrt{3} - 1)}{3 - 1} \\[1em] \Rightarrow h = \frac{100(\sqrt{3} - 1)}{2} \\[1em] \Rightarrow h = 50(\sqrt{3} - 1) \\[1em] \Rightarrow h = 50(1.732 - 1) \\[1em] \Rightarrow h = 50(0.732) \\[1em] \Rightarrow h = 36.6 \text{ m}.\end{aligned}$$

Hence, option 1 is the correct option.

Let height of the tower (AB) be h meters.

Let P and Q be two points on either sides of the tower.

In triangle ABP,

$\Rightarrow \tan \theta = \frac{h}{30}....(1)$

In triangle ABQ,

$$\begin{aligned}\Rightarrow \tan (90^{\circ} - \theta) = \frac{h}{40} \\[1em] \Rightarrow \cot \theta = \frac{h}{40} \\[1em] \Rightarrow \frac{1}{\tan \theta} = \frac{h}{40} \\[1em] \Rightarrow \frac{1}{\frac{h}{30}} = \frac{h}{40} \\[1em] \Rightarrow \frac{30}{h}= \frac{h}{40} \\[1em] \Rightarrow h^2 = 1200 \\[1em] \Rightarrow h = \sqrt{1200} \\[1em] \Rightarrow h = 34.64 \text{ m.}\end{aligned}$$

Hence, option 3 is the correct option.

Let AB(h) be the height of the lighthouse.

Let P and Q be the two ships on opposite sides of the lighthouse.

Let BP = x meters and BQ = y meters.

In triangle ABP,

$$\begin{aligned}\Rightarrow \tan \alpha = \frac{h}{x} \\[1em] \Rightarrow x = \frac{h}{\tan \alpha}.\end{aligned}$$

In triangle ABQ,

$$\begin{aligned}\Rightarrow \tan \beta = \frac{h}{y} \\[1em] \Rightarrow y = \frac{h}{\tan \beta}.\end{aligned}$$

The total distance between the ships is = x + y

$$\begin{aligned}= \frac{h}{\tan \alpha} + \frac{h}{\tan \beta} \\[1em] = h \Big( \frac{1}{\tan \alpha} + \frac{1}{\tan \beta} \Big) \\[1em] = h \Big( \frac{\tan \beta + \tan \alpha}{\tan \alpha \tan \beta} \Big) \\[1em] = \frac{h(\tan \alpha + \tan \beta)}{\tan \alpha \tan \beta}.\end{aligned}$$

Hence, option 3 is the correct option.

Let the tree before breaking be OP.

The tree breaks at point C. The top part falls and touches the ground at point B.

Let BC = h₁ and CP = h₂

In triangle CPB,

$$\begin{aligned}\Rightarrow \tan(30^\circ) = \frac{CP}{BP} \\[1em] \Rightarrow \frac{1}{\sqrt3} = \frac{h_2}{10} \\[1em] \Rightarrow h_2 = \frac{10}{\sqrt3} \text{ m.}\end{aligned}$$

In triangle ABC,

$$\begin{aligned}\Rightarrow \cos(30^\circ) = \frac{BP}{BC} \\[1em] \Rightarrow \frac{\sqrt3}{2} = \frac{10}{h_1} \\[1em] \Rightarrow h_1 = \frac{20}{\sqrt3} \text{ m.}\end{aligned}$$

The original height of the tree

$$\begin{aligned}= \frac{10}{\sqrt3} + \frac{20}{\sqrt3} \\[1em] = \frac{30}{\sqrt3} \\[1em] = \frac{30}{\sqrt3} \times \frac{\sqrt3}{\sqrt3} \\[1em] = 10\sqrt3.\end{aligned}$$

Hence, option 2 is the correct option.

Height of the pole (AE) = $\frac{10}{3}$ m

Height of the person (CD) = 2 m

AE = AB - BE

= $\frac{10}{3} - 2$

= $\frac{4}{3}$ m

Distance of man from pole (BD) = $\Big(\frac{4}{\sqrt{3}}\Big)$

From figure,

CE = BD = $\Big(\frac{4}{\sqrt{3}}\Big)$ m

In triangle AEC,

$$\begin{aligned}\Rightarrow \tan \theta = \frac{AE}{CE} \\[1em] \Rightarrow \tan \theta = \frac{\frac{4}{3}}{\frac{4}{\sqrt3}} \\[1em] \Rightarrow \tan \theta = \frac{1}{\sqrt3} \\[1em] \Rightarrow \tan \theta = \tan 30^{\circ} \\[1em] \Rightarrow \theta = 30^{\circ}.\end{aligned}$$

Hence, option 2 is the correct option.

Let AB be the height of first building and CD be the height of second building.

The distance between the buildings (BD) = 60 m.

CE is the difference in height between the two buildings.

In triangle AEC,

$$\begin{aligned}\Rightarrow \tan 30^{\circ} = \frac{CE}{AE} \\[1em] \Rightarrow \frac{1}{\sqrt3} = \frac{CE}{60} \\[1em] \Rightarrow CE = \frac{60}{\sqrt3} \\[1em] \Rightarrow CE = 34.64 \text{ m}.\end{aligned}$$

Height of first building,

AB = CD - CE

= 150 - 34.64

= 115.36 m.

Hence, option 1 is the correct option.

Let AB be the tower and CD be the column.

Let distance between tower (AB) and column (BD) = x

In triangle CBD,

$$\begin{aligned}\Rightarrow \tan 60^{\circ} = \frac{CD}{BD} \\[1em] \Rightarrow \sqrt3 = \frac{CD}{x} \\[1em] \Rightarrow x = \frac{CD}{\sqrt3} \text{ .....(1)}\end{aligned}$$

In triangle CAE,

$$\begin{aligned}\Rightarrow \tan 30^{\circ} = \frac{CE}{AE} \\[1em] \Rightarrow \frac{1}{\sqrt3} = \frac{CD - 25}{x} \\[1em] \Rightarrow x = (CD - 25)\sqrt3 \text{ .....(2)}\end{aligned}$$

From (1) and (2), we get :

$\frac{CD}{\sqrt{3}} = (CD - 25)\sqrt{3}$

CD = (CD - 25)3

CD = 3CD - 75

2CD = 75

CD = $\frac{75}{2}$

CD = 37.5 m

Hence, option 2 is the correct option.

Let BA be the flagstaff and BC be the building.

Let O be the position of observer.

In triangle OCB,

$$\begin{aligned}\Rightarrow \tan 45^{\circ} = \frac{BC}{OC} \\[1em] \Rightarrow 1 = \frac{BC}{72} \\[1em] \Rightarrow BC = 72 \text{ m}.\end{aligned}$$

In triangle OCA,

$$\begin{aligned}\Rightarrow \tan 60^{\circ} = \frac{AC}{OC} \\[1em] \Rightarrow \sqrt3 = \frac{AC}{72} \\[1em] \Rightarrow AC = 72\sqrt3 \text{ m.}\end{aligned}$$

The height of the flagstaff is

AB = AC - BC

= 72 $\sqrt{3}$ - 72

= 72($\sqrt{3}$ - 1)

= 72(1.732 - 1)

= 72(0.732)

= 52.7 m.

Hence, option 1 is the correct option.

Let the height of the tower (BC) be x and the height of the flagstaff (BA) be h.

Let P be the point on ground from foot of tower at distance d.

In triangle PCB,

$$\begin{aligned}\Rightarrow \tan \alpha = \frac{x}{d} \\[1em] \Rightarrow d = \frac{x}{\tan \alpha} \text{ .....(1)}\end{aligned}$$

In triangle PCA,

$$\begin{aligned}\Rightarrow \tan \beta = \frac{x + h}{d} \\[1em] \Rightarrow d = \frac{x + h}{\tan \beta} \text{ ........(2)}\end{aligned}$$

From (1) and (2), we get :

$\frac{x}{\tan \alpha} = \frac{x + h}{\tan \beta}$

x tan β = (x + h)tan α

x tan β = x tan α + h tan α

h tan α = x tan β - x tan α

h tan α = x(tan β - tan α)

x = $\frac{h \tan \alpha}{\tan \beta - \tan \alpha}$

Hence, option 4 is the correct option.

Let the height of the tower (BC) be H and AB be the height of the flag staff = $\frac{H}{5}$

Total height (AC) = H + $\frac{H}{5} = \frac{6H}{5}$

Let P be the point of observation at distance d from tower.

In triangle PCA,

$$\begin{aligned}\Rightarrow \tan 45^{\circ} = \frac{AC}{CD} \\[1em] \Rightarrow 1 = \frac{\frac{6H}{5}}{d} \\[1em] \Rightarrow d = \frac{6H}{5}.\end{aligned}$$

In triangle PCB,

$$\begin{aligned}\Rightarrow \tan \theta = \frac{H}{d} \\[1em] \Rightarrow \tan \theta = \frac{H}{\frac{6H}{5}} \\[1em] \Rightarrow \tan \theta = \frac{1}{\frac{6}{5}} \\[1em] \Rightarrow \tan \theta = \frac{5}{6}.\end{aligned}$$

Hence, option 2 is the correct option.

Let AB and CD be two poles of equal height (h).

BD = 30 m

Let BP = x m

In triangle ABP,

$$\begin{aligned}\Rightarrow \tan 60^{\circ} = \frac{AB}{BP} \\[1em] \Rightarrow \sqrt3 = \frac{h}{x} \\[1em] \Rightarrow x = \frac{h}{\sqrt3}\end{aligned}$$

In triangle PCD,

$$\begin{aligned}\Rightarrow \tan 30^{\circ} = \frac{h}{30 - x} \\[1em] \Rightarrow \frac{1}{\sqrt3} = \frac{h}{30 - x} \\[1em] \Rightarrow 30 - x = h\sqrt3 \\[1em] \Rightarrow 30 - \frac{h}{\sqrt3} = h\sqrt3 \\[1em] \Rightarrow \frac{30\sqrt3 - h}{\sqrt3} = h\sqrt3 \\[1em] \Rightarrow 30\sqrt3 - h = h\sqrt3(\sqrt3) \\[1em] \Rightarrow 30\sqrt3 - h = 3h \\[1em] \Rightarrow 30\sqrt3 = 3h + h \\[1em] \Rightarrow 30\sqrt3 = 4h \\[1em] \Rightarrow h = \frac{30\sqrt3}{4} \\[1em] \Rightarrow h = 7.5\sqrt3 \\[1em] \Rightarrow h = 7.5(1.732) \\[1em] \Rightarrow h = 12.99 \approx 13 \text{ m.}\end{aligned}$$

Hence, option 3 is the correct option.

Let the height of the shorter post (AB) be h meters and the height of the the taller post (CD) be 2h meters..

Let the distance between the post be k meters.

Let M be the point of observation,

BM = MD = $\frac{k}{2}$

In triangle ABM,

$$\begin{aligned}\Rightarrow \tan \theta = \frac{h}{\frac{k}{2}} \\[1em] \Rightarrow \tan \theta = \frac{2h}{k} \text{ ....(1)}\end{aligned}$$

In triangle CDM,

$$\begin{aligned}\Rightarrow \tan (90^{\circ} - \theta) = \frac{2h}{\frac{k}{2}} \\[1em] \Rightarrow \tan (90^{\circ} - \theta) = \frac{4h}{k} \\[1em] \Rightarrow \cot \theta = \frac{4h}{k} \\[1em] \Rightarrow \tan \theta = \frac{k}{4h} \text{ ....(2)}\end{aligned}$$

From (1) and (2), we get :

$\frac{2h}{k} = \frac{k}{4h}$

$$\begin{aligned}\Rightarrow 8h^2 = k^2 \\[1em] \Rightarrow h^2 = \frac{k^2}{8} \\[1em] \Rightarrow h = \sqrt{\frac{k^2}{8}} \\[1em] \Rightarrow h = \frac{k}{\sqrt{4 \times 2}} \\[1em] \Rightarrow h = \frac{k}{2\sqrt{2}} \text{ m}\end{aligned}$$

Hence, option 1 is the correct option.

Let AB and CD be two pillars.

Draw a line from A to meet CD at point E, AE = x

In triangle ADE,

$$\begin{aligned}\Rightarrow \tan 45^{\circ} = \frac{ED}{AE} \\[1em] \Rightarrow 1 = \frac{20}{x} \\[1em] \Rightarrow x = 20 \text{ m.}\end{aligned}$$

In triangle ACE,

$$\begin{aligned}\Rightarrow \tan 30^{\circ} = \frac{CE}{AE} \\[1em] \Rightarrow \frac{1}{\sqrt3} = \frac{CE}{20} \\[1em] \Rightarrow CE = \frac{20}{\sqrt3} \text{ m.}\end{aligned}$$

The total height of the second pillar is,

CD = ED + CE

$$\begin{aligned}\Rightarrow CD = 20 + \frac{20}{\sqrt{3}} \\[1em] \Rightarrow CD = 20 \Big(1 + \frac{1}{\sqrt{3}} \Big) \\[1em] \Rightarrow CD = 20 \Big( \frac{\sqrt{3} + 1}{\sqrt{3}} \Big) \\[1em] \Rightarrow CD = \frac{20}{\sqrt{3}}(\sqrt{3} + 1) \text{ m}.\end{aligned}$$

Hence, option 4 is the correct option.

Let A be the initial position of aeroplane and height AB = 3000 m, C be the final position and height CD = 3000 m.

O be the point of observation,

In triangle OBA,

$$\begin{aligned}\Rightarrow \tan 45^{\circ} = \frac{AB}{OB} \\[1em] \Rightarrow 1 = \frac{3000}{OB} \\[1em] \Rightarrow OB = 3000 \text{ m.}\end{aligned}$$

In triangle COD,

$$\begin{aligned}\Rightarrow \tan (30^{\circ}) = \frac{CD}{OD} \\[1em] \Rightarrow \frac{1}{\sqrt3} = \frac{3000}{OD} \\[1em] \Rightarrow OD = 3000 \sqrt3 \text{ m.}\end{aligned}$$

The distance the plane flew is,

BD = OD - OB

BD = 3000 $\sqrt{3}$ - 3000

BD = 3000($\sqrt{3}$ - 1)

BD = 3000(1.732 - 1) = 3000(0.732) = 2196 m.

$$\begin{aligned}\Rightarrow \text{Speed} = \frac{\text{Distance}}{\text{Time}} \\[1em] \Rightarrow \text{Speed} = \frac{2196}{15} \\[1em] \Rightarrow \text{Speed} = 146.4 \text{ m/s} \\[1em] \Rightarrow \text{Speed} = 146.4 \times 3.6 \\[1em] \Rightarrow \text{Speed} = 527.04 \approx 527 \text{ km/h}.\end{aligned}$$

Hence, option 4 is the correct option.

Directions:
Two pillars P₁ and P₂ of equal heights stand on either side of a road which is 100 m wide. At a point on the road between the pillars, the angles of elevation of the tops of the pillars P₁ and P₂ are 60° and 30° respectively.

Based on this information, answer the following questions:

31. The height of each pillar is:
(a) 25 m
(b) 36 m
(c) $25\sqrt{3}$ m
(d) $36\sqrt{3}$ m

32. The location of the point of observation is:
(a) 25 m from P₁
(b) 25 m from P₂
(c) $25\sqrt{3}$ m from P₁
(d) $25\sqrt{3}$ m from P₂

33. If a hook is fixed at the point of observation and strings are tied from the hook to the tops of both the towers, then the total length of string required is:
(a) 43.3 m
(b) 86.6 m
(c) 68.3 m
(d) 136.6 m

34. If a flagstaff is to be erected atop pillar P₂ such that the angle of elevation of its top from the point of observation is 45°, then the height of the flagstaff must be:
(a) $25\sqrt{3}$ m
(b) 50 m
(c) $25\sqrt{3}(\sqrt{3}-1)$ m
(d) $25(2-\sqrt{3})$ m

31. Let AP₁ and BP₂ be two pillars of height h.

Let the point of observation be O at distance x from P₁.

In triangle OAP₁,

$$\begin{aligned}\Rightarrow \tan 60^{\circ} = \frac{h}{x} \\[1em] \Rightarrow \sqrt3 = \frac{h}{x} \\[1em] \Rightarrow x = \frac{h}{\sqrt3}.\end{aligned}$$

In triangle OBP₂,

$$\begin{aligned}\Rightarrow \tan (30^{\circ}) = \frac{h}{100 - x} \\[1em] \Rightarrow \frac{1}{\sqrt3} = \frac{h}{100 - x} \\[1em] \Rightarrow 100 - x = h\sqrt3 \\[1em] \Rightarrow 100 - \frac{h}{\sqrt{3}} = h\sqrt{3} \\[1em] \Rightarrow 100 = h\sqrt{3} + \frac{h}{\sqrt{3}} \\[1em] \Rightarrow 100 = h \Big( \frac{3 + 1}{\sqrt{3}} \Big) = \frac{4h}{\sqrt{3}} \\[1em] \Rightarrow h = \frac{100\sqrt{3}}{4} = 25\sqrt{3} \text{ m}.\end{aligned}$$

Hence, option (c) is the correct option.

32. In triangle OAP₁,

$$\begin{aligned}\Rightarrow \tan 60^{\circ} = \frac{h}{x} \\[1em] \Rightarrow \sqrt3 = \frac{h}{x} \\[1em] \Rightarrow x = \frac{h}{\sqrt3} \\[1em] \Rightarrow x = \frac{25\sqrt{3}}{\sqrt{3}} = 25 \text{ m}\end{aligned}$$

Hence, option (a) is the correct option.

33. Let the length of strings be L₁ and L₂.

In triangle OAP₁,

$$\begin{aligned}\Rightarrow \sin 60^{\circ} = \frac{h}{L_1} \\[1em] \Rightarrow \frac{\sqrt3}{2} = \frac{25\sqrt3}{L_1} \\[1em] \Rightarrow L_1 = 50 \text{ m}.\end{aligned}$$

In triangle OBP₂,

$$\begin{aligned}\Rightarrow \sin (30^{\circ}) = \frac{h}{L_2} \\[1em] \Rightarrow \frac{1}{2} = \frac{25\sqrt3}{L_2} \\[1em] \Rightarrow L_2 = 50\sqrt3 = 86.6 \text{ m.}\end{aligned}$$

Total length = 50 + 86.6 = 136.6 m

Hence, option (d) is the correct option.

34. Let H be the height of flagstaff added to P₂

Total distance - Distance to P₁ = Distance to P₂

Distance to P₂ = 100 - 25 = 75 m.

Therefore,

$$\begin{aligned}\Rightarrow \tan (45^{\circ}) = \frac{H + 25\sqrt3}{75} \\[1em] \Rightarrow 1 = \frac{H + 25\sqrt3}{75} \\[1em] \Rightarrow 75 = 25\sqrt3 + H \\[1em] \Rightarrow H = 75 - 25\sqrt3 \\[1em] \Rightarrow H = 25\sqrt{3}(\sqrt{3} - 1) \text{ m.}\end{aligned}$$

Hence, option (c) is the correct option.

Directions:
The angle of elevation of the top of a building from the foot of a tower is 30°. The angle of elevation of the top of the tower from the foot of the building is 60°. The tower is 30 metres high.

Based on this information, answer the following questions:

35.The horizontal distance between the tower and the building is:
(a) 10 m
(b) 17.3 m
(c) 20 m
(d) 34.6 m

36.The height of the building is:
(a) 10 m
(b) $10\sqrt{3}$ m
(c) 15 m
(d) $15\sqrt{3}$ m

37. The straight line distance between the tops of the tower and the building is:
(a) $10\sqrt{3}$ m
(b) $10\sqrt{5}$ m
(c) $10\sqrt{7}$ m
(d) 30 m

38. A bird flew straight from the top of the tower to the foot of the building. What is the distance that the bird flew?
(a) 20 m
(b) 25 m
(c) $20\sqrt{3}$ m
(d) $20(\sqrt{3}-1)$ m

35. Let CD be the height of the tower and h be the height of the building (AB).

Let x (BD) be the horizontal distance between them.

In triangle CDB,

$$\begin{aligned}\Rightarrow \tan 60^{\circ} = \frac{30}{x} \\[1em] \Rightarrow \sqrt3 = \frac{30}{x} \\[1em] \Rightarrow x = \frac{30}{\sqrt3} \\[1em] \Rightarrow x = \frac{30\sqrt3}{3} \\[1em]\Rightarrow x = 10 \sqrt3 = 17.3 \text{ m}.\end{aligned}$$

Hence, option (b) is the correct option.

36.In triangle ADB,

$$\begin{aligned}\Rightarrow \tan 30^{\circ} = \frac{h}{x} \\[1em] \Rightarrow \frac{1}{\sqrt3} = \frac{h}{10\sqrt3} \\[1em] \Rightarrow h = 10 \text{ m}.\end{aligned}$$

Hence, option (a) is the correct option.

37. In right-angled triangle △AEC at E.

AE = BD = $10\sqrt3$ m

CE = CD - AB = 30 - 10 = 20 m

∴ AC² = AE² + CE²

AC² = ($10\sqrt3$)² + (20)²

AC² = 100(3) + 400

AC² = 300 + 400

AC² = 700

AC = $\sqrt{700} = 10 \sqrt7$ m

Hence, option (c) is the correct option.

38. In triangle CDB,

$$\begin{aligned}\Rightarrow \sin 60^{\circ} = \frac{CD}{BC} \\[1em] \Rightarrow \frac{\sqrt3}{2} = \frac{30}{BC} \\[1em] \Rightarrow BC = \frac{60}{\sqrt3} \\[1em] \Rightarrow BC = \frac{60\sqrt3}{3} \\[1em] \Rightarrow BC = 20\sqrt3 \text{ m.}\end{aligned}$$

Hence, option (c) is the correct option.

Directions : A TV tower stands vertically on the bank of a canal. From a point on the other bank directly opposite the tower, the angle of elevation of the top of the tower is 60°. From another point 20 m away from this point on the line joining this point to the foot of the tower, the angle of elevation of the top of the tower is 30°.

Based on this information, answer the following questions:

39.The height of the tower is:
(a) 10 m
(b) $10\sqrt{3}$ m
(c) 20 m
(d) $20\sqrt{3}$ m

40.The width of the canal is:
(a) 10 m
(b) $10\sqrt{3}$ m
(c) 20 m
(d) $20\sqrt{3}$ m

41.The straight line distances of the top of the tower from the two points of observation differ by:
(a) 7.32 m
(b) 14.64 m
(c) 17.32 m
(d) 20.64 m

42.How far away from the other bank must the point of observation be, so that the angle of elevation of the top of the tower is 45°? (a) 7.32 m
(b) 10 m
(c) 14.64 m
(d) 27.32 m

39. Let hbe the height of the tower (AB) and w be width of canal (BC).

Let D be another point 20 m away from C.

In triangle ABC,

$$\begin{aligned}\Rightarrow \tan 60^{\circ} = \frac{h}{w} \\[1em] \Rightarrow \sqrt3 = \frac{h}{w} \\[1em] \Rightarrow w = \frac{h}{\sqrt3} .....(1)\end{aligned}$$

In triangle ADB,

$$\begin{aligned}\Rightarrow \tan 30^{\circ} = \frac{h}{w + 20} \\[1em] \Rightarrow \frac{1}{\sqrt3} = \frac{h}{w + 20} \\[1em] \Rightarrow w + 20 = h\sqrt3 \\[1em] \Rightarrow \frac{h}{\sqrt3} + 20 = h\sqrt3 \\[1em] \Rightarrow \frac{h + 20\sqrt{3}}{\sqrt{3}} = h\sqrt{3} \\[1em] \Rightarrow h + 20\sqrt{3} = 3h \\[1em] \Rightarrow 20\sqrt{3} = 2h \\[1em] \Rightarrow h = 10\sqrt{3} \text{ m}.\end{aligned}$$

Hence, option (b) is the correct option.

40. In triangle ABC,

$$\begin{aligned}\Rightarrow \tan 60^{\circ} = \frac{h}{w} \\[1em] \Rightarrow \sqrt3 = \frac{h}{w} \\[1em] \Rightarrow w = \frac{h}{\sqrt3} \\[1em] \Rightarrow w = \frac{10\sqrt3}{\sqrt3} \\[1em] \Rightarrow w = 10 \text{ m.}\end{aligned}$$

Hence, option (a) is the correct option.

41. In triangle ABC,

$$\begin{aligned}\Rightarrow \sin 60^{\circ} = \frac{h}{AC} \\[1em] \Rightarrow \frac{\sqrt3}{2} = \frac{10\sqrt3}{AC} \\[1em] \Rightarrow AC = 20 \text{ m.}\end{aligned}$$

In triangle ABD,

$$\begin{aligned}\Rightarrow \sin 30^{\circ} = \frac{h}{AD} \\[1em] \Rightarrow \frac{1}{2} = \frac{10\sqrt3}{AD} \\[1em] \Rightarrow AD = 20\sqrt3 \text{ m.}\end{aligned}$$

The straight line distances of the top of the tower from the two points of observation differ by = AD - AC = $20\sqrt{3} - 20$

= $20(\sqrt{3} - 1)$

= 20(0.732) = 14.64 m.

Hence, option (b) is the correct option.

42. Let E be the new point of observation

In triangle ABE,

$$\begin{aligned}\Rightarrow \tan(45^\circ) = \frac{h}{BE} \\[1em] \Rightarrow 1 = \frac{10\sqrt{3}}{BE} \\[1em] \Rightarrow BE = 10\sqrt{3} \text{ m}\end{aligned}$$

Distance from the bank (CE) = BE - BC

CE = $10\sqrt{3}$ - 10

CE = 10($\sqrt{3}$ - 1)

CE = 10(0.732) = 7.32 m

Hence, option (a) is the correct option.

Let the height of the tower be h and the length of its shadow be x, and let the sun's altitude be θ.

$\tan \theta = \frac{h}{x}$. If h = x, then tan θ = 1, so θ = 45°.

∴ Assertion (A) is true.

The angle which the line of sight makes with the horizontal line, when the object is above the observer, is indeed the angle of elevation — this is the correct definition.

∴ Reason (R) is true.

However, the definition of the angle of elevation does not by itself explain why an equal shadow and height give 45°; that follows from tan θ = 1. So R is not the correct explanation of A.

Hence, option 2 is the correct option.

As the observer moves horizontally towards the base of the object, the horizontal distance decreases while the height of the object stays the same, so tan (angle of elevation) = $\frac{\text{height}}{\text{distance}}$ increases, and hence the angle of elevation increases.

∴ Assertion (A) is true.

Angle of elevation = angle of depression is actually true in the standard heights-and-distances context, but it does not explain why the angle of elevation increases when the observer moves closer.

∴ Reason (R) is true, but it is not the correct explanation of A.

Hence, option 2 is the correct option.

Let AB = Height of person and BC = Length of shadow.

∠ACB = θ (Angle of elevation of sun)

Given,

Length of shadow of a person (BC) = height of person (AB)

⇒ tan θ = $\frac{\text{Perpendicular}}{\text{Base}}$

⇒ tan θ = $\frac{\text{AB}}{\text{BC}}$

⇒ tan θ = $\frac{\text{AB}}{\text{AB}}$

⇒ tan θ = 1

⇒ tan θ = tan 45°

⇒ θ = 45°.

∴ Both (A) and (R) are true, and (R) is the correct explanation for (A).

Hence, option 1 is the correct option.

(a) According to question,

⇒ AB = $\sqrt{3} \times BC$

⇒ AB = $\sqrt{3}BC$

⇒ BC = $\frac{AB}{\sqrt{3}}$

From figure,

⇒ tan θ = $\frac{BC}{AB}$

⇒ tan θ = $\frac{BC}{\sqrt{3}BC}$

⇒ tan θ = $\frac{1}{\sqrt{3}}$

⇒ tan θ = tan 30°

⇒ θ = 30°.

Hence, θ = 30°.

(b) In right angle triangle ABC,

By pythagoras theorem,

⇒ AC² = BC² + AB²

⇒ 15² = $\Big(\frac{AB}{\sqrt{3}}\Big)^2$ + AB²

⇒ 225 = $\frac{AB^2}{3}$ + AB²

⇒ 225 = $\frac{AB^2 + 3AB^2}{3}$

⇒ 225 = $\frac{4AB^2}{3}$

⇒ AB² = $\frac{225 \times 3}{4}$

⇒ AB² = $\frac{675}{4}$

⇒ AB² = 168.75

⇒ AB = $\sqrt{168.75}$

⇒ AB = 12.99 m

Rounding off,

AB = 13 m.

Hence, AB = 13 m.

Let AC be the plank and BC be the height from where the drum is rolled down.

⇒ sin θ = $\frac{\text{Perpendicular}}{\text{Hypotenuse}}$

⇒ sin 10° = $\frac{BC}{AC}$

⇒ 0.1736 = $\frac{BC}{10}$

⇒ BC = 1.736 ~ 1.74 m

Hence, height from which the cylindrical drum was rolled down = 1.74 m.

(a) From figure,

tan 30° = $\frac{AB}{AR}$

Substituting values we get :

$$\begin{aligned}\Rightarrow \frac{1}{\sqrt{3}} = \frac{AB}{150} \\[1em] \Rightarrow AB = \frac{150}{\sqrt{3}} = \frac{150}{1.732} \\[1em] = 86.6 \text{ m}.\end{aligned}$$

Hence, height of the building = 86.6 m.

(b)

1. From figure,

tan 60° = $\frac{AB}{AN}$

Substituting values we get :

$$\begin{aligned}\Rightarrow \sqrt{3} = \frac{\frac{150}{\sqrt{3}}}{AN} \\[1em] \Rightarrow AN = \frac{150}{\sqrt{3}} \times \frac{1}{\sqrt{3}} \\[1em] \Rightarrow AN = \frac{150}{3} = 50\text{ m}.\end{aligned}$$

Hence, distance between Neha and foot of the building = 50 m.

2. From figure,

RN = AR - AN = 150 - 50 = 100 m.

Hence, distance between Rohit and Neha = 100 m.

3. From figure,

tan 60° = $\frac{ST}{NT}$

Substituting values we get :

$$\begin{aligned}\Rightarrow \sqrt{3} = \frac{30}{NT} \\[1em] \Rightarrow NT = \frac{30}{\sqrt{3}} \\[1em] \Rightarrow NT = \frac{30}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} \\[1em] \Rightarrow NT = \frac{30\sqrt{3}}{3} \\[1em] \Rightarrow NT = 10\sqrt{3} \\[1em] \Rightarrow NT = 10 \times 1.732 = 17.32 \text{ m}.\end{aligned}$$

Hence, distance between Neha and the tree = 17.32 m.

(iv) From figure,

AT = AN - NT = 50 - 17.32 = 32.68 m

Hence, distance between building and tree = 32.68 m.

As the satellite is vertically above the mid-point of the 1937 km distance between the peaks, the horizontal distance from the satellite to each peak is $\frac{1937}{2}$ = 968.5 km.

Thus,

DI = AG = 968.5 km and SI = PH = 968.5 km

(Taking $\sqrt{3}$ ≈ 1.7, so cos 30° ≈ 0.85.)

1. For Nanda Devi, the angle of depression is 30°.

So, angle of elevation also equals to 30°.

In triangle AFG,

$$\begin{aligned}\Rightarrow \cos 30° = \frac{AG}{AF} \\[1em] \Rightarrow 0.85 = \frac{968.5}{AF} \\[1em] \Rightarrow AF = \frac{968.5}{0.85} \\[1em] \Rightarrow AF \approx 1139.4 \text{ km}\end{aligned}$$

Hence, option (c) is the correct option.

2. For Mullayanagiri, the angle of depression is 60°.

So, angle of elevation also equals to 60°.

In triangle PFH,

$$\begin{aligned}\Rightarrow \cos 60° = \frac{PH}{PF} \\[1em] \Rightarrow 0.5 = \frac{968.5}{PF} \\[1em] \Rightarrow PF = \frac{968.5}{0.5} \\[1em] \Rightarrow PF = 1937 \text{ km}\end{aligned}$$

Hence, option (d) is the correct option.

3. The vertical drop from the satellite to the top of Nanda Devi is FG.

In triangle AFG,

$$\begin{aligned}\Rightarrow \tan 30° = \frac{FG}{AG} \\[1em] \Rightarrow FG = AG \times \tan 30° \\[1em] \Rightarrow FG = 968.5 \times \tan 30° \\[1em] \Rightarrow FG = \frac{968.5}{1.7} \\[1em] \Rightarrow FG \approx 569.7 \text{ km}\end{aligned}$$

Adding the height of Nanda Devi (7816 m = 7.816 km), the distance of the satellite from the ground is

$$\begin{aligned}\Rightarrow FI = FG + 7.816 \\[1em] \Rightarrow FI = 569.7 + 7.816 \\[1em] \Rightarrow FI \approx 577.52 \text{ km}\end{aligned}$$

Hence, option (a) is the correct option.

4. A man is standing at a distance of 7816 m from Nanda Devi, whose height is 7816 m.

Let θ be the angle of elevation. Then,

$$\begin{aligned}\Rightarrow \tan \theta = \frac{7816}{7816} \\[1em] \Rightarrow \tan \theta = 1 \\[1em] \Rightarrow \tan \theta = \tan 45° \\[1em] \Rightarrow \theta = 45°\end{aligned}$$

Hence, option (c) is the correct option.