Heights & Distances

We know that,

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

From figure,

In △ ABC,

⇒ tan 60° = $\frac{AB}{BC}$

⇒ $\sqrt{3} = \frac{15}{x}$

⇒ x = $\frac{15}{\sqrt{3}}$

Rationalizing the denominator,

$$\begin{aligned}\Rightarrow x = \frac{15}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} \\[1em] \Rightarrow x = \frac{15\sqrt{3}}{3} \\[1em] \Rightarrow x = 5\sqrt{3}.\end{aligned}$$

Hence, Option 3 is the correct option.

We know that,

cos θ = $\frac{\text{Base}}{\text{Hypotenuse}}$

From figure,

In △ ABC,

⇒ cos 60° = $\frac{BC}{AB}$

⇒ $\frac{1}{2} = \frac{x}{40}$

⇒ x = $\frac{40}{2}$ = 20 m.

Hence, Option 3 is the correct option.

We know that,

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

From figure,

In △ BCD,

⇒ sin 45° = $\frac{BC}{BD}$

⇒ $\frac{1}{\sqrt{2}} = \frac{20}{BD}$

⇒ BD = $20\sqrt{2}$ m.

From figure,

⇒ AB = BD = $20\sqrt{2}$ m

⇒ AC = AB + BC = $20\sqrt{2} + 20$ = 28.28 + 20 = 48.28 cm.

Hence, Option 2 is the correct option.

We know that,

cos θ = $\frac{\text{Base}}{\text{Hypotenuse}}$

From figure,

In △ ABC,

⇒ cos 30° = $\frac{BC}{AB}$

⇒ $\frac{\sqrt{3}}{2} = \frac{x}{70}$

⇒ x = $\frac{70\sqrt{3}}{2} = 35\sqrt{3} = 60.62 \approx 61$ m.

Hence, Option 3 is the correct option.

We know that,

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

From figure,

In △ ABC,

⇒ tan 45° = $\frac{AB}{BC}$

⇒ $1 = \frac{10}{BC}$

⇒ BC = 10 m.

In △ ABD,

⇒ tan 30° = $\frac{AB}{BD}$

⇒ $\frac{1}{\sqrt{3}} = \frac{10}{BD}$

⇒ BD = $10\sqrt{3}$ m.

From figure,

DC = BD - BC = $10\sqrt{3} - 10 = 10(\sqrt{3} - 1)$ m.

Hence, Option 4 is the correct option.

Let AB be the tree and BC be the shadow of tree.

Let the length of the shadow (BC) of the tree be x meters.

So, the height of the tree (AB) = $\sqrt{3}x$ meters.

If θ is the angle of elevation of the sun, then we have :

In △ABC,

$$\begin{aligned}\Rightarrow \text{tan θ} = \frac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \text{tan θ} = \frac{AB}{BC} \\[1em] \Rightarrow \text{tan θ} = \frac{\sqrt{3}x}{x} \\[1em] \Rightarrow \text{tan θ} = \sqrt{3} \\[1em] \Rightarrow \text{tan θ} = \text{tan 60°} \\[1em] \Rightarrow θ = 60°.\end{aligned}$$

Hence, angle of elevation of sun is 60°.

Let AB be the tower and C be the point at a distance of 160 m from foot of tower.

Let,

AB = h meters.

From figure,

In △ABC,

$$\begin{aligned}\Rightarrow \text{tan θ} = \frac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \text{tan 60°} = \frac{AB}{BC} \\[1em] \Rightarrow \sqrt{3} = \frac{h}{160} \\[1em] \Rightarrow h = 160\sqrt{3} \\[1em] \Rightarrow h = 160 \times 1.732 = 277.12 \text{ m}.\end{aligned}$$

Hence, the height of the tower is 277.12 m.

Let AC be the ladder and height of wall upto which ladder reaches be h meters.

∴ AB = h meters and BC = 2.4 meters.

Given,

Ladder is making an angle of 68° with the ground.

From figure,

In △ABC,

$$\begin{aligned}\Rightarrow \text{tan θ} = \frac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \text{tan 68°} = \frac{AB}{BC} \\[1em] \Rightarrow 2.475 = \frac{h}{2.4} \\[1em] \Rightarrow h = 2.4 \times 2.475 \\[1em] \Rightarrow h = 5.94 \text{ m}.\end{aligned}$$

Hence, the ladder reaches upto a height of 5.94 m.

Let PQ be the tower.

Let one of the persons, A be at a distance of x meters and the second person B be at a distance of y metres from the foot of the tower (Q).

Given, that angle of elevation of A is 30°.

From figure,

In △PQA,

$$\begin{aligned}\Rightarrow \text{tan 30°} = \frac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \frac{1}{\sqrt{3}} = \frac{PQ}{AQ} \\[1em] \Rightarrow AQ = PQ\sqrt{3} = 50\sqrt{3} \\[1em] \Rightarrow x = 50 \times 1.732 \\[1em] \Rightarrow x = 86.6 \text{ meters}.\end{aligned}$$

Given, that angle of elevation of B is 38°.

In △PBQ,

$$\begin{aligned}\Rightarrow \text{tan 38°} = \frac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow 0.7813 = \frac{PQ}{BQ} \\[1em] \Rightarrow BQ = \frac{50}{0.7813} \\[1em] \Rightarrow y = 64 \text{ meters}.\end{aligned}$$

AB = x + y = 86.6 + 64 = 150.6 meters.

Hence, the distance between two persons = 150.6 meters.

(i) From figure,

When angle of elevation is 45°, then CE is the tower.

In △ADE,

$$\begin{aligned}\Rightarrow \text{tan 45°} = \frac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow 1 = \frac{DE}{AD} \\[1em] \Rightarrow AD = DE \\[1em] \Rightarrow DE = 20 \text{ meters}.\end{aligned}$$

From figure,

CD = AB.

CE = CD + DE = 1.6 + 20 = 21.6 meters.

Hence, height of tower when angle of elevation is 45° is 21.6 meters.

(ii) From figure,

When angle of elevation is 60°.

In △ADF,

$$\begin{aligned}\Rightarrow \text{tan 60°} = \frac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \sqrt{3} = \frac{DF}{AD} \\[1em] \Rightarrow DF = AD\sqrt{3} \\[1em] \Rightarrow DF = 20 \times 1.732 = 34.64 \text{ meters}.\end{aligned}$$

From figure,

CD = AB.

CF = CD + DF = 1.6 + 34.64 = 36.24 meters.

Hence, height of tower when angle of elevation is 60° is 36.24 meters.

Let A be the point from where tree breaks and C be the point where above part of tree touches the ground.

From figure,

In △ABC,

$$\begin{aligned}\text{tan 45°} = \frac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow 1 = \frac{AB}{BC} \\[1em] \Rightarrow AB = BC = 15 \text{ meters}.\end{aligned}$$

In right angle triangle ABC,

⇒ AC² = AB² + BC²

⇒ AC² = 15² + 15²

⇒ AC² = 225 + 225

⇒ AC² = 450

⇒ AC = $\sqrt{450} = 15\sqrt{2}$ meters.

Height of tree = AB + AC = 15 + $15\sqrt{2}$

= 15 + 21.21

= 36.21 meters.

Hence, height of tree before it was broken = 36.21 meters.

Let AB be the unfinished tower and C be the point 80 m from base of tower.

From figure,

In △ABC,

$$\begin{aligned}\Rightarrow \text{tan 30°} = \frac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \frac{1}{\sqrt{3}} = \frac{AB}{BC} \\[1em] \Rightarrow AB = \frac{BC}{\sqrt{3}} \\[1em] \Rightarrow AB = \frac{80}{\sqrt{3}} \\[1em] \Rightarrow AB = \frac{80}{1.732} \\[1em] \Rightarrow AB = 46.19\text{ meters}.\end{aligned}$$

Let tower be raised to point D in order to make angle of elevation 60°.

From figure,

In △DBC,

$$\begin{aligned}\Rightarrow \text{tan 60°} = \frac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \sqrt{3} = \frac{BD}{BC} \\[1em] \Rightarrow BD = BC\sqrt{3} \\[1em] \Rightarrow BD = 80\sqrt{3} \\[1em] \Rightarrow BD = 138.56\text{ metres}.\end{aligned}$$

AD = BD - AB = 138.56 - 46.19 = 92.37 meters.

Hence, the tower must be raised by 92.37 meters.

(i) Let AB be the tower.

From figure,

In △ABC,

$$\begin{aligned}\text{tan 30°} = \frac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \frac{1}{\sqrt{3}} = \frac{AB}{BC} \\[1em] \Rightarrow \frac{h}{45} = \frac{1}{\sqrt{3}} \\[1em] \Rightarrow h = \frac{45}{\sqrt{3}} \\[1em] \Rightarrow h = \frac{45}{1.732} \\[1em] \Rightarrow h = 25.98 \text{ meters}.\end{aligned}$$

Hence, the height of tower = 25.98 meters.

(ii) When sun's altitude is

(a) 45°

$$\begin{aligned}\Rightarrow \text{tan 45°} = \frac{AB}{BC} \\[1em] \Rightarrow 1 = \frac{h}{BC} \\[1em] \Rightarrow BC = h = 25.98 \text{ meters}.\end{aligned}$$

Hence, length of shadow of tower when the sun's altitude is 45° = 25.98 meters.

(b) 60°

$$\begin{aligned}\Rightarrow \text{tan 60°} = \frac{AB}{BC} \\[1em] \Rightarrow \sqrt{3} = \frac{h}{BC} \\[1em] \Rightarrow BC = \frac{h}{\sqrt{3}} \\[1em] \Rightarrow BC = \frac{25.98}{1.732} \\[1em] \Rightarrow BC = 15\text{ meters}.\end{aligned}$$

Hence, length of shadow of tower when the sun's altitude is 60° = 15 meters.

Let AP and CQ be two poles.

When ladder is at position AB resting on pole AP.

Then, ∠BAP = 32° 24'

From figure,

In △ABP,

$$\begin{aligned}\text{sin 32° 24}' = \frac{\text{Perpendicular}}{\text{Hypotenuse}} \\[1em] \Rightarrow 0.536 = \frac{BP}{AB} \\[1em] \Rightarrow BP = AB \times 0.536 \\[1em] \Rightarrow BP = 30 \times 0.536 \\[1em] \Rightarrow BP = 16.08 \text{ meters}.\end{aligned}$$

When ladder is at position BC resting on pole CQ.

Then it makes angle 32° 24' with road.

∴ ∠CBQ = 32° 24'

From figure,

In △BQC,

$$\begin{aligned}\text{cos 32° 24}' = \frac{\text{Base}}{\text{Hypotenuse}} \\[1em] \Rightarrow 0.844 = \frac{BQ}{BC} \\[1em] \Rightarrow BQ = BC \times 0.844 \\[1em] \Rightarrow BQ = 30 \times 0.844 \\[1em] \Rightarrow BQ = 25.32\text{ meters}.\end{aligned}$$

Width of road = BP + BQ = 16.08 + 25.32 = 41.4 meters.

Hence, width of road = 41.4 meters.

Let P be the foot of cliff.

From figure,

In △BCP,

$$\begin{aligned}\text{tan 57°} = \frac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow 1.539 = \frac{BP}{PC} \\[1em] \Rightarrow BP = PC \times 1.539 \\[1em] \Rightarrow BP = 40 \times 1.539 \\[1em] \Rightarrow BP = 61.57 \text{ meters}.\end{aligned}$$

In △ACP,

$$\begin{aligned}\text{tan 48°} = \frac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow 1.110 = \frac{AP}{PC} \\[1em] \Rightarrow AP = PC \times 1.110 \\[1em] \Rightarrow AP = 40 \times 1.110 \\[1em] \Rightarrow AP = 44.40 \text{ meters}.\end{aligned}$$

Distance between climbers (BA) = BP - AP

= 61.57 - 44.40

= 17.17 meters.

Hence, distance between two climbers = 17.17 meters.

Let AC be the pole and D be the point where man stands.

From figure,

In △ADB,

$$\begin{aligned}\Rightarrow \text{tan 28°} = \frac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow 0.532 = \frac{AB}{BD} \\[1em] \Rightarrow AB = BD \times 0.532 \\[1em] \Rightarrow AB = 9 \times 0.532 = 4.788 \text{ m}.\end{aligned}$$

In △BDC,

$$\begin{aligned}\Rightarrow \text{tan 13°} = \frac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow 0.231 = \frac{BC}{BD} \\[1em] \Rightarrow BC = BD \times 0.231 \\[1em] \Rightarrow BC = 9 \times 0.231 = 2.079 \text{ m}.\end{aligned}$$

AC = AB + BC = 4.788 + 2.079 = 6.867 meters.

Hence, the height of pole = 6.867 meters.

Let AB be the cliff and C be the buoy.

Given,

AB = 92 m

From figure,

In △ACB,

$$\begin{aligned}\text{tan 20°} = \frac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \text{tan 20°} = \frac{AB}{BC} \\[1em] \Rightarrow 0.364 = \frac{92}{BC}\\[1em] \Rightarrow BC = \frac{92}{0.364} \\[1em] \Rightarrow BC = 252.7 ≈ 253 \text{ meters}.\end{aligned}$$

Hence, the distance of the buoy from the foot of the cliff is 253 meters.

We know that,

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

From figure,

In △ ABD,

⇒ tan 45° = $\frac{AD}{BD}$

⇒ $1 = \frac{1}{BD}$

⇒ BD = 1 m.

In △ ACD,

⇒ tan 30° = $\frac{AD}{CD}$

⇒ $\frac{1}{\sqrt{3}} = \frac{1}{CD}$

⇒ CD = $\sqrt{3}$ m.

From figure,

BC = BD + CD = $(1 + \sqrt{3})$ m.

Hence, Option 3 is the correct option.

We know that,

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

From figure,

In △ ABC,

⇒ tan 50° = $\frac{AB}{BC}$

⇒ tan 50° = $\frac{h}{6}$

⇒ h = 6 tan 50°

⇒ h = 6 tan (90° - 40°)

⇒ h = 6 cot 40° ........(1)

In △ ABD,

⇒ tan 40° = $\frac{AB}{BD}$

⇒ tan 40° = $\frac{h}{8}$

⇒ h = 8 tan 40° ........(2)

Multiplying equation (1) and (2), we get :

⇒ h × h = 6 cot 40° × 8 tan 40°

⇒ h² = 8 × 6 × cot 40° $\times \frac{1}{\text{cot 40°}}$

⇒ h² = 8 × 6

⇒ h = $\sqrt{8 \times 6}$.

Hence, Option 1 is the correct option.

From figure,

BCDE is a rectangle.

∴ BC = DE = 8 m and BE = DC = 30 m.

We know that,

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

From figure,

In △ ABE,

⇒ tan 45° = $\frac{AB}{BE}$

⇒ $1 = \frac{AB}{30}$

⇒ AB = 30 m.

From figure,

AC = AB + BC = 30 + 8 = 38 m.

Hence, Option 2 is the correct option.

We know that,

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

From figure,

In △ ABC,

⇒ tan 60° = $\frac{AB}{BC}$

⇒ $\sqrt{3} = \frac{20}{BC}$

⇒ BC = $\frac{20}{\sqrt{3}}$ cm.

In rectangle BCDE,

Opposite sides of rectangle are equal.

∴ DE = BC = $\frac{20}{\sqrt{3}}$ cm

From figure,

In △ AED,

⇒ tan 30° = $\frac{AE}{DE}$

⇒ $\frac{1}{\sqrt{3}} = \frac{AE}{\frac{20}{\sqrt{3}}}$

⇒ AE = $\frac{1}{\sqrt{3}} \times \frac{20}{\sqrt{3}} = \frac{20}{3} = 6\frac{2}{3}$ cm.

Hence, Option 3 is the correct option.

We know that,

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

From figure,

In △ ABC,

⇒ tan 60° = $\frac{AB}{BC}$

⇒ $\sqrt{3} = \frac{15}{BC}$

⇒ BC = $\frac{15}{\sqrt{3}}$

Rationalizing the denominator,

$$\begin{aligned}\Rightarrow \text{BC } = \frac{15}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} \\[1em] \Rightarrow \text{BC } = \frac{15\sqrt{3}}{3} \\[1em] \Rightarrow \text{BC } = 5\sqrt{3}.\end{aligned}$$

In △ ABD,

⇒ tan 30° = $\frac{AB}{BD}$

⇒ $\frac{1}{\sqrt{3}} = \frac{15}{BD}$

⇒ BD = $15\sqrt{3}$ cm.

From figure,

DC = BD - BC = $15\sqrt{3} - 5\sqrt{3} = 10\sqrt{3}$ cm.

Hence, Option 4 is the correct option.

From figure,

In △ABD,

$$\begin{aligned}\Rightarrow \text{tan 30°} = \frac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \frac{1}{\sqrt{3}} = \frac{AB}{BD} \\[1em] \Rightarrow AB = \frac{BD}{\sqrt{3}} ........(1)\end{aligned}$$

In △ABC,

$$\begin{aligned}\Rightarrow \text{tan 45°} = \frac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow 1 = \frac{AB}{BC} \\[1em] \Rightarrow AB = BC ..........(2)\end{aligned}$$

From (1) and (2),

$$\begin{aligned}\Rightarrow BC = \frac{BD}{\sqrt{3}} \\[1em] \Rightarrow \sqrt{3}BC = BD \\[1em] \Rightarrow \sqrt{3}BC = BC + CD \\[1em] \Rightarrow \sqrt{3}BC - BC = 30 \\[1em] \Rightarrow BC(\sqrt{3} - 1) = 30 \\[1em] \Rightarrow BC = \frac{30}{\sqrt{3} - 1} \\[1em] \Rightarrow BC = \frac{30}{1.732 - 1} \\[1em] \Rightarrow BC = \frac{30}{0.732} \\[1em] \Rightarrow BC = 40.98 \text{ meters}.\end{aligned}$$

∴ AB = X = 40.98 meters .....[From (2)]

Hence, X = 40.98 meters.

Let AB be the tree.

Let the two points be C and D such that CD = 20 m, ∠ADB = 30° and ∠ACB = 60°.

In ∆ABC,

$$\begin{aligned}\Rightarrow \text{tan 60°} = \frac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \sqrt{3} = \frac{AB}{BC} \\[1em] \Rightarrow AB = BC\sqrt{3} ..........(1)\end{aligned}$$

In ∆ABD,

$$\begin{aligned}\Rightarrow \text{tan 30°} = \frac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \frac{1}{\sqrt{3}} = \frac{AB}{BD} \\[1em] \Rightarrow AB = \frac{BD}{\sqrt{3}} ..........(2)\end{aligned}$$

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

$$\begin{aligned}\Rightarrow BC\sqrt{3} = \frac{BD}{\sqrt{3}} \\[1em] \Rightarrow BD = 3BC \\[1em] \Rightarrow BC + CD = 3BC \\[1em] \Rightarrow CD = 2BC \\[1em] \Rightarrow BC = \frac{CD}{2} \\[1em] \Rightarrow BC = \frac{20}{2} \\[1em] \Rightarrow BC = 10 \text{ meters}.\end{aligned}$$

AB = $BC \times \sqrt{3}$ = BC × 1.732 = 17.32 meters.

Hence, the height of the tree is 17.32 metres.

(i) Let's consider AB to be the lighthouse.

Given, depression angles are 48° and 36°.

When ships are on the same side,

In ∆ABC,

$$\begin{aligned}\text{tan 48°} = \frac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow 1.1106 = \frac{AB}{BC} \\[1em] \Rightarrow 1.1106 = \frac{100}{BC} \\[1em] \Rightarrow BC = \frac{100}{1.1106} \\[1em] \Rightarrow BC = 90.04 \text{ meters}.\end{aligned}$$

In ∆ABD,

$$\begin{aligned}\text{tan 36°} = \frac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow 0.7265 = \frac{AB}{BD} \\[1em] \Rightarrow 0.7265 = \frac{100}{BD} \\[1em] \Rightarrow BD = \frac{100}{0.7265} \\[1em] \Rightarrow BD = 137.64 \text{ meters}.\end{aligned}$$

Distance between the two ships (CD) = BD – BC = 137.64 - 90.04

= 47.6 ≈ 48 m.

Hence, distance between ships when on the same side = 48 m.

(ii) Let's consider AB to be the lighthouse.

Given, depression angles are 48° and 36°.

As, alternate angles are equal.

∴ ∠ADB = ∠QAD = 36° and ∠ACB = ∠PAC = 48°.

When ships are on the opposite side,

In ∆ABC,

$$\begin{aligned}\text{tan 48°} = \frac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow 1.1106 = \frac{AB}{BC} \\[1em] \Rightarrow 1.1106 = \frac{100}{BC} \\[1em] \Rightarrow BC = \frac{100}{1.1106} \\[1em] \Rightarrow BC = 90.04 \text{ meters}.\end{aligned}$$

In ∆ABD,

$$\begin{aligned}\text{tan 36°} = \frac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow 0.7265 = \frac{AB}{BD} \\[1em] \Rightarrow 0.7265 = \frac{100}{BD} \\[1em] \Rightarrow BD = \frac{100}{0.7265} \\[1em] \Rightarrow BD = 137.64 \text{ meters}.\end{aligned}$$

Distance between the two ships (CD) = BD + BC = 137.64 + 90.04

= 227.68 ≈ 228 m.

Hence, the distance between two ships, when on opposite side = 228 m.

Let AB and CD be the two towers of height h meters. Let P be a point in the roadway BD such that BD = 150 m, ∠APB = 60° and ∠CPD = 30°.

In ∆ABP,

$$\begin{aligned}\text{tan 60°} = \frac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \sqrt{3} = \frac{AB}{BP} \\[1em] \Rightarrow BP = \frac{AB}{\sqrt{3}} \\[1em] \Rightarrow BP = \frac{h}{\sqrt{3}}\space ............(1)\end{aligned}$$

In ∆CDP,

$$\begin{aligned}\text{tan 30°} = \frac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \frac{1}{\sqrt{3}} = \frac{CD}{PD} \\[1em] \Rightarrow PD = \sqrt{3}CD \\[1em] \Rightarrow PD = \sqrt{3}h\space ............(2)\end{aligned}$$

We know that,

⇒ BD = 150 m

⇒ BP + PD = 150 m

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

$$\begin{aligned}\Rightarrow \frac{h}{\sqrt{3}} + \sqrt{3}h = 150 \\[1em] \Rightarrow \frac{h + 3h}{\sqrt{3}} = 150 \\[1em] \Rightarrow \frac{4h}{\sqrt{3}} = 150 \\[1em] \Rightarrow 4h = 150 \times \sqrt{3} \\[1em] \Rightarrow h = \frac{150 \times 1.732}{4} \\[1em] \Rightarrow h = 37.5 \times 1.732 \\[1em] \Rightarrow h = 64.95 \text{ meters}.\end{aligned}$$

From equation (1),

BP = $\frac{h}{\sqrt{3}} = \frac{64.95}{1.732}$ = 37.5 meters.

Hence, height of each pillar is 64.95 m and the point P is 37.5 m from the pillar AB.

From figure,

DE = CB = 15 m.

In ∆AED,

$$\begin{aligned}\text{tan 22°} = \frac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow 0.404 = \frac{AE}{DE} \\[1em] \Rightarrow 0.404 = \frac{AE}{15} \\[1em] \Rightarrow AE = 15 \times 0.404 \\[1em] \Rightarrow AE = 6.06 \text{ meters}.\end{aligned}$$

In ∆ABC,

$$\begin{aligned}\text{tan 47°} = \frac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow 1.0724 = \frac{AB}{BC} \\[1em] \Rightarrow 1.0724 = \frac{AB}{15} \\[1em] \Rightarrow AB = 15 \times 1.0724 \\[1em] \Rightarrow AB = 16.086 \text{ meters}.\end{aligned}$$

From figure,

CD = BE = AB - AE

= 16.086 - 6.06

= 10.03 meters.

Hence, CD = 10.03 meters.

(i) Let AB be the the tower, C be the first point of observation and D be the second point.

From figure,

BC = ED = a (let) and BE = CD = 30 m.

In △ABC,

$$\begin{aligned}\text{tan 60°} = \frac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \sqrt{3} = \frac{AB}{BC}\\[1em] \Rightarrow AB = \sqrt{3}BC \\[1em] \Rightarrow AB = \sqrt{3}a ..........(1)\end{aligned}$$

In △AED,

$$\begin{aligned}\text{tan 45°} = \frac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow 1 = \frac{AE}{ED}\\[1em] \Rightarrow AE = ED \\[1em] \Rightarrow AE = a ..........(2)\end{aligned}$$

We know that,

⇒ CD = AB - AE

⇒ 30 = $\sqrt{3}a - a$

⇒ 30 = $a(\sqrt{3} - 1)$

⇒ a = $\frac{30}{\sqrt{3} - 1} = \frac{30}{1.732 - 1} = \frac{30}{0.732}$ = 40.98 metres.

From equation (1),

AB = $\sqrt{3}a = 1.732 \times$ 40.98 = 70.98 meters.

Hence, height of tower = 70.98 meters.

(ii) From part (i),

ED = a = 40.98 meters.

Hence, horizontal distance from the points of observation is 40.98 meters.

Let CD be the cliff so CD = 60 meters and AB be the tower.

Since, alternate angles are equal,

∴ ∠ECA = ∠CAF = 30° and ∠ECB = ∠CBD = 60°.

Let AF = BD = a meters.

In △BCD,

$$\begin{aligned}\text{tan 60°} = \frac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \sqrt{3} = \frac{CD}{BD}\\[1em] \Rightarrow CD = \sqrt{3}BD \\[1em] \Rightarrow CD = \sqrt{3}a \\[1em] \Rightarrow 60 = \sqrt{3}a \\[1em] \Rightarrow a = \frac{60}{\sqrt{3}} \\[1em] \Rightarrow a = \frac{60}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} \\[1em] \Rightarrow a = \frac{60\sqrt{3}}{3} \\[1em] \Rightarrow a = 20\sqrt{3} \text{ meters}.\end{aligned}$$

In △AFC,

$$\begin{aligned}\text{tan 30°} = \frac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \frac{1}{\sqrt{3}} = \frac{CF}{AF}\\[1em] \Rightarrow CF = \frac{AF}{\sqrt{3}} \\[1em] \Rightarrow CF = \frac{a}{\sqrt{3}} \\[1em] \Rightarrow CF = \frac{20\sqrt{3}}{\sqrt{3}} \\[1em] \Rightarrow CF = 20 \text{ meters}.\end{aligned}$$

From figure,

⇒ AB = DF = CD - CF

⇒ AB = CD - CF

⇒ AB = 60 - 20 = 40 meters.

Hence, the height of tower = 40 meters.

Let CD be the cliff and A be the position of the ship when angle of elevation is 30° and B be the position when angle of elevation is 60°.

(i) From figure,

$$\begin{aligned}\text{tan 60°} = \frac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \sqrt{3} = \frac{CD}{BD} \space .............(1) \\[1em] \text{tan 30°} = \frac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \frac{1}{\sqrt{3}} = \frac{CD}{AD} \space .............(2)\end{aligned}$$

Dividing equation (1) by (2), we get :

$$\begin{aligned}\Rightarrow \frac{\sqrt{3}}{\frac{1}{\sqrt{3}}} = \frac{\frac{CD}{BD}}{\frac{CD}{AD}} \\[1em] \Rightarrow \sqrt{3} \times \sqrt{3} = \frac{AD}{BD} \\[1em] \Rightarrow \frac{AD}{BD} = 3 \\[1em] \Rightarrow AD = 3BD \\[1em] \Rightarrow AB + BD = 3BD \\[1em] \Rightarrow AB = 2BD.\end{aligned}$$

Boat reaches from point A to B in 3 minutes.

Thus, boat covers distance AB in 3 minutes or it covers distance

⇒ 2 BD in 3 minutes

⇒ BD in $\frac{3}{2}$ = 1.5 minutes.

Hence, it will takes 1.5 minutes more to reach the shore.

(ii) In △ADC,

$$\begin{aligned}\text{tan 30°} = \frac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \frac{1}{\sqrt{3}} = \frac{CD}{AD} \\[1em] \Rightarrow AD = \sqrt{3}CD\end{aligned}$$

In △BDC,

$$\begin{aligned}\text{tan 60°} = \frac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \sqrt{3} = \frac{CD}{BD} \\[1em] \Rightarrow BD = \frac{CD}{\sqrt{3}}\end{aligned}$$

From figure,

$$\begin{aligned}\Rightarrow AB = AD - BD \\[1em] \Rightarrow AB = \sqrt{3}CD - \frac{CD}{\sqrt{3}} \\[1em] \Rightarrow AB = \frac{3CD - CD}{\sqrt{3}} \\[1em] \Rightarrow AB = \frac{2CD}{\sqrt{3}} .........(1)\end{aligned}$$

Let speed of boat be a meter/second so in 3 minutes boat will travel :

Distance (AB) = Speed × Time

AB = a × 3 × 60

AB = 180a meters .........(2)

From (1) and (2) we get,

$$\begin{aligned}\Rightarrow \frac{2CD}{\sqrt{3}} = 180a \\[1em] \Rightarrow \frac{2 \times 500}{\sqrt{3}} = 180a \\[1em] \Rightarrow \frac{1000}{1.732} = 180a \\[1em] \Rightarrow a = \frac{1000}{180 \times 1.732} \\[1em] \Rightarrow a = \frac{1000}{311.76} = 3.21 \text{ m/s}.\end{aligned}$$

Hence, speed of boat = 3.21 m/s.

Let man in the boat be originally at point C and after 2 minutes it reaches the point D and AB be the lighthouse.

AB = 150 meters.

In △ABC,

$$\begin{aligned}\text{tan 60°} = \frac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \sqrt{3} = \frac{AB}{BC} \\[1em] \Rightarrow AB = \sqrt{3} BC \\[1em] \Rightarrow BC = \frac{AB}{\sqrt{3}} \\[1em] \Rightarrow BC = \frac{150}{\sqrt{3}} \\[1em] \Rightarrow BC = \frac{150}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} \\[1em] \Rightarrow BC = \frac{150\sqrt{3}}{3} \\[1em] \Rightarrow BC = 50\sqrt{3} \text{ meters}.\end{aligned}$$

In △ABD,

$$\begin{aligned}\text{tan 45°} = \frac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow 1 = \frac{AB}{BD} \\[1em] \Rightarrow AB = BD \\[1em] \Rightarrow BD = 150 \text{ metres}.\end{aligned}$$

CD = BD - BC = 150 - $50\sqrt{3}$

= 150 - 86.6

= 63.4 meters.

In 2 minutes boat covers 63.4 meters or boat covers 63.4 meters in 120 seconds.

Speed = $\frac{\text{Distance}}{\text{Time}} = \frac{63.4}{120}$ = 0.53 m/sec.

Hence, the speed of boat = 0.53 m/sec.

Let CD be the tree and B be the position of the person when angle of elevation is 60° and A be the position when angle of elevation is 30°.

(i) In △BCD,

$$\begin{aligned}\text{tan 60°} = \frac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \sqrt{3} = \frac{CD}{BC} \\[1em] \Rightarrow BC = \frac{CD}{\sqrt{3}}.\end{aligned}$$

In △ACD,

$$\begin{aligned}\text{tan 30°} = \frac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \frac{1}{\sqrt{3}} = \frac{CD}{AC} \\[1em] \Rightarrow AC = CD\sqrt{3}.\end{aligned}$$

From figure,

AB = AC - BC

$$\begin{aligned}\Rightarrow 40 = CD\sqrt{3} - \frac{CD}{\sqrt{3}} \\[1em] \Rightarrow \frac{3CD - CD}{\sqrt{3}} = 40 \\[1em] \Rightarrow \frac{2CD}{\sqrt{3}} = 40 \\[1em] \Rightarrow CD = \frac{40\sqrt{3}}{2} = 20\sqrt{3} \\[1em] \Rightarrow CD = 20 \times 1.732 = 34.64 \text{ m}.\end{aligned}$$

Hence, height of tree = 34.64 meters.

(ii) From part (i), we get :

BC = $\frac{CD}{\sqrt{3}} = \frac{20\sqrt{3}}{\sqrt{3}}$

= 20 meters.

Hence, width of river = 20 meters.

Let AB be the first tower and CD be the second tower and ∠EDA = 45° is the angle of depression.

Given, angle of depression of the top of the first tower as seen from the top of the second tower is 45°.

We know that,

Alternate angles are equal.

∴ ∠DAF = ∠EDA = 45°.

From figure,

AF = BC = 75 m.

In △ADF,

$$\begin{aligned}\text{tan 45°}= \frac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow 1 = \frac{DF}{AF} \\[1em] \Rightarrow DF = AF = 75 \text{ m}.\end{aligned}$$

From figure,

AB = FC = CD - DF = 160 - 75 = 85 m.

Hence, height of first tower = 85 m.

Let CD be the tower of height h meters and BC be shadow when angle of elevation is 45° and AC be the shadow when angle of elevation is 30°.

In △ACD,

$$\begin{aligned}\text{tan 30°} = \frac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \frac{1}{\sqrt{3}} = \frac{CD}{AC} \\[1em] \Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{AC} \\[1em] \Rightarrow h = \frac{AC}{\sqrt{3}} \text{ m} .........(1)\end{aligned}$$

In △BCD,

$$\begin{aligned}\text{tan 45°} = \frac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow 1 = \frac{CD}{BC} \\[1em] \Rightarrow BC = CD = h\end{aligned}$$

AC = AB + BC = (2y + h) meters.

Substituting value of AC in equation 1, we get :

$$\begin{aligned}\Rightarrow h = \frac{2y + h}{\sqrt{3}} \\[1em] \Rightarrow \sqrt{3}h = 2y + h \\[1em] \Rightarrow \sqrt{3}h - h = 2y \\[1em] \Rightarrow h(\sqrt{3} - 1) = 2y \\[1em] \Rightarrow h = \frac{2y}{\sqrt{3} - 1}.\end{aligned}$$

Multiplying numerator and denominator by $(\sqrt{3} + 1)$.

$$\begin{aligned}\Rightarrow h = \frac{2y}{\sqrt{3} - 1} \times \frac{\sqrt{3} + 1}{\sqrt{3} + 1} \\[1em] \Rightarrow h = \frac{2y(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} \\[1em] \Rightarrow h = \frac{2y(\sqrt{3} + 1)}{(\sqrt{3})^2 - 1^2} \\[1em] \Rightarrow h = \frac{2y(\sqrt{3} + 1)}{3 - 1} \\[1em] \Rightarrow h = \frac{2y(\sqrt{3} + 1)}{2} \\[1em] \Rightarrow h = y(\sqrt{3} + 1).\end{aligned}$$

Hence, proved that the height of tower = $y(\sqrt{3} + 1)$ meters.

Let aeroplane be originally at point E and after 10 seconds it reaches point D.

In △ABE,

$$\begin{aligned}\text{tan 60°} = \frac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \sqrt{3} = \frac{BE}{AB} \\[1em] \Rightarrow \sqrt{3} = \frac{1}{AB}\\[1em] \Rightarrow AB = \frac{1}{\sqrt{3}} \text{ km}.\end{aligned}$$

In △ACD,

$$\begin{aligned}\text{tan 30°} = \frac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \frac{1}{\sqrt{3}} = \frac{CD}{AC} \\[1em] \Rightarrow AC = \sqrt{3}CD \\[1em] \Rightarrow AC = \sqrt{3} \times 1 = \sqrt{3} \text{ km}.\end{aligned}$$

From figure,

DE = BC.

BC = AC - AB = $\sqrt{3} - \frac{1}{\sqrt{3}}$

= $\frac{3 - 1}{\sqrt{3}}$

= $\frac{2}{1.732}$

= 1.1547 km.

∴ DE = 1.1547 km.

∴ Aeroplane travels 1.1547 km in 10 seconds.

Time = 10 seconds = $\frac{10}{3600} = \frac{1}{360}$ hours.

Speed = $\frac{\text{Distance}}{\text{Time}} = \frac{1.1547}{\frac{1}{360}}$

= 1.1547 × 360

= 415.69 km/hr.

Hence, speed of aeroplane = 415.69 km/hr.

Let C and D be the position of two kilometer stones and AB be the hill.

In △ABD,

$$\begin{aligned}\text{tan 30°} = \frac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \frac{1}{\sqrt{3}} = \frac{AB}{BD} \\[1em] \Rightarrow BD = \sqrt{3} AB .........(1)\end{aligned}$$

In △ABC,

$$\begin{aligned}\text{tan 45°} = \frac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow 1 = \frac{AB}{BC} \\[1em] \Rightarrow BC = AB ............(2)\end{aligned}$$

From figure,

⇒ CD = BD - BC

⇒ 1 = $\sqrt{3}AB - AB$

⇒ 1 = $AB(\sqrt{3} - 1)$

⇒ AB = $\frac{1}{\sqrt{3} - 1} = \frac{1}{0.732}$ = 1.366 km.

From equation (2),

BC = AB = 1.366 km

BD = BC + CD = 1.366 + 1 = 2.366 km.

Hence, kilometer stones are at a distance of 1.366 and 2.366 km.

We know that,

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

From figure,

In △ BCD,

⇒ tan 30° = $\frac{BD}{CD}$

⇒ $\frac{1}{\sqrt{3}} = \frac{BD}{10}$

⇒ BD = $\frac{10}{\sqrt{3}}$ m.

In △ ACD,

⇒ tan 60° = $\frac{AD}{CD}$

⇒ $\sqrt{3} = \frac{AD}{10}$

⇒ AD = $10\sqrt{3}$ cm.

From figure,

AB = AD + BD = $10\sqrt{3} + \frac{10}{\sqrt{3}}$

$= \frac{10\sqrt{3} \times \sqrt{3} + 10}{\sqrt{3}} = \frac{30 + 10}{\sqrt{3}} = \frac{40}{\sqrt{3}}$ m.

Hence, Option 1 is the correct option.

We know that,

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

From figure,

In △ ABC,

⇒ tan x° = $\frac{AB}{CB}$

⇒ $\frac{5}{2} = \frac{100}{CB}$

⇒ CB = $\frac{2 \times 100}{5} = \frac{200}{5}$ = 40 m.

Hence, Option 2 is the correct option.

In rectangle BCDE,

Opposite sides of rectangle are equal.

∴ BE = DC = 60 m

We know that,

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

From figure,

In △ ABE,

⇒ tan 30° = $\frac{AB}{BE}$

⇒ $\frac{1}{\sqrt{3}} = \frac{AB}{60}$

⇒ AB = $\frac{60}{\sqrt{3}} = 20\sqrt{3}$ m.

In △ EBC,

⇒ tan 45° = $\frac{BC}{BE}$

⇒ $1 = \frac{BC}{60}$

⇒ BC = 60 m.

From figure,

AC = AB + BC = $(20\sqrt{3} + 60)$ m.

Hence, Option 3 is the correct option.

Since, BCDE is a square.

∴ BE = CD = 90 cm.

We know that,

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

From figure,

∠AEB = ∠EFD = 45° (Corresponding angles are equal)

In △ ABE,

⇒ tan 45° = $\frac{AB}{BE}$

⇒ $1 = \frac{AB}{90}$

⇒ AB = 90 cm.

From figure,

AC = AB + BC = 90 + 90 = 180 cm.

We know that,

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

In △ ACF,

⇒ sin 45° = $\frac{AC}{AF}$

⇒ $\frac{1}{\sqrt{2}} = \frac{180}{AF}$

⇒ AF = $180\sqrt{2}$ cm.

Hence, Option 4 is the correct option.

We know that,

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

From figure,

In △ ABC,

⇒ tan 45° = $\frac{AB}{BC}$

⇒ $1 = \frac{AB}{45}$

⇒ AB = 45 m.

In △ ABD,

⇒ tan 30° = $\frac{AB}{BD}$

⇒ $\frac{1}{\sqrt{3}} = \frac{45}{BD}$

⇒ BD = $45\sqrt{3}$ m.

From figure,

DC = BD - BC = $45\sqrt{3} - 45 = 45(\sqrt{3} - 1)$ m.

Hence, Option 3 is the correct option.

Given, the ratio of the height of a vertical pole to the length of its shadow is 3 : $\sqrt{3}$.

Let AB be the pole and BC be the shadow and angle of elevation of Sun be θ.

$$\begin{aligned}\Rightarrow \frac{\text{Height of a vertical pole}}{\text{Height of shadow}} = \frac{3}{\sqrt{3}}\\[1em] \Rightarrow \frac{\text{AB}}{\text{BC}} = \sqrt{3}\\[1em] \Rightarrow \text{tan θ} = \sqrt{3}\\[1em] \Rightarrow \text{tan θ} = \text{tan 60°} \\[1em] \Rightarrow θ = 60°\end{aligned}$$

∴ Both A and R are true and R is correct reason for A.

Hence, option 3 is the correct option.

Let AB be the wall and AC be the ladder. Let distance between foot of ladder and wall be x.

So, length of ladder (AC) = 2x.

According to the Pythagoras theorem :

⇒ Hypotenuse² = Base² + Height²

⇒ AC² = BC² + AB²

⇒ (2x)² = x² + AB²

⇒ 4x² = x² + AB²

⇒ AB² = 4x² - x²

⇒ AB² = 3x²

⇒ AB = $\sqrt{3x^2}$

⇒ AB = $\sqrt{3}x$.

Let angle between ladder and wall be θ.

We know that,

$$\begin{aligned}\Rightarrow \text{sin θ} = \frac{\text{Perpendicular side}}{\text{Hypotenuse}} \\[1em] \Rightarrow \text{sin θ} = \frac{BC}{AC} \\[1em] \Rightarrow \text{sin θ} = \frac{x}{2x} \\[1em] \Rightarrow \text{sin θ} = \frac{1}{2} \\[1em]\end{aligned}$$

⇒ sin θ = sin 30°

⇒ θ = 30°.

Assertion (A) is false.

If ladder makes angle θ with the wall then sin θ = $\frac{\text{base}}{\text{hypotenuse}} = \frac{x}{2x} = \frac{1}{2}$.

Here base refers side opposite to angle θ i.e. BC.

Reason (R) is true.

Hence, option 2 is the correct option.

Given,

tan A = $1\frac{2}{3} = \frac{5}{3}$ and tan B = $\frac{1}{3}$.

From figure,

$$\begin{aligned}\Rightarrow \text{tan A} = \frac{h}{9} \\[1em] \Rightarrow \frac{5}{3} = \frac{h}{9} \\[1em] \Rightarrow h = \frac{5 \times 9}{3} \\[1em] \Rightarrow h = 5 \times 3 \\[1em] \Rightarrow h = 15\text{ cm} \\[1em] \Rightarrow \text{tan B} = \frac{h}{x + 9}\\[1em] \Rightarrow \frac{1}{3} = \frac{15}{x + 9}\\[1em] \Rightarrow x + 9 = 15 \times 3 \Rightarrow x + 9 = 45 \Rightarrow x = 45 - 9 \Rightarrow x = 36 \text{ cm}.\end{aligned}$$

∴ Both the statements are true.

Hence, option 1 is the correct option.

From figure,

In triangle BDC,

$$\begin{aligned}\Rightarrow \text{tan 45°} = \frac{BC}{DC}\\[1em] \Rightarrow \text{tan 45°} = \frac{BC}{100}\\[1em] \Rightarrow BC = 100\text{ tan 45°}\\[1em] \text{In triangle ADC,}\\[1em] \Rightarrow \text{tan 60°} = \frac{AC}{DC}\\[1em] \Rightarrow \text{tan 60°} = \frac{AB + BC}{DC}\\[1em] \Rightarrow \text{tan 60°} = \frac{AB + 100\text{ tan 45°}}{100}\\[1em] \Rightarrow 100\text{ tan 60°} = AB + 100\text{ tan 45°}\\[1em] \Rightarrow AB = 100\text{ tan 60°} - 100\text{ tan 45°}\end{aligned}$$

∴ Statement 1 is false, and statement 2 is true.

Hence, option 4 is the correct option.

(i) From figure,

BE = CD = 20 m and DE = CB = 5 m.

In △ABE,

$$\begin{aligned}\text{tan 32°} = \frac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow 0.6249 = \frac{AE}{BE} \\[1em] \Rightarrow AE = 0.6249 \times BE \\[1em] \Rightarrow AE = 0.6249 \times 20 \\[1em] \Rightarrow AE = 12.498 \text{ m}.\end{aligned}$$

AD = AE + DE = 12.498 + 5 = 17.498 ≈ 17.5 m.

Hence, AD = 17.5 meters.

(ii) We know that,

An exterior angle is equal to the sum of two opposite interior angles.

∴ ∠ACD = ∠ABC + ∠BAC

Also, ∠ABC = ∠BAC (As, angles opposite to equal sides are equal)

∴ ∠ACD = 2∠ABC

⇒ 2∠ABC = 48°

⇒ ∠ABC = 24°.

In △ABD,

$$\begin{aligned}\text{sin 24°} = \frac{\text{Perpendicular}}{\text{Hypotenuse}} \\[1em] \Rightarrow 0.4067 = \frac{AD}{AB} \\[1em] \Rightarrow AD = 0.4067 \times AB \\[1em] \Rightarrow AD = 0.4067 \times 30 \\[1em] \Rightarrow AD = 12.20 \text{ m}.\end{aligned}$$

Hence, AD = 12.20 meters.

In ∆PSB,

$$\begin{aligned}\text{sin 60°} = \frac{\text{Perpendicular}}{\text{Hypotenuse}} \\[1em] \Rightarrow \frac{\sqrt{3}}{2} = \frac{PS}{PB} \\[1em] \Rightarrow PB = \frac{2PS}{\sqrt{3}} \\[1em] \Rightarrow PB = \frac{2 \times 1}{1.732} = 1.155 \text{ m}.\end{aligned}$$

In ∆APQ,

∠APQ = ∠ABR = 60° (Corresponding angles are equal.)

$$\begin{aligned}\text{cos 60°} = \frac{\text{Base}}{\text{Hypotenuse}} \\[1em] \Rightarrow \frac{1}{2} = \frac{PQ}{AP} \\[1em] \Rightarrow AP = 2PQ \\[1em] \Rightarrow AP = 2 \text{ m}.\end{aligned}$$

From figure,

AB = AP + PB = 2 + 1.155 = 3.155 m.

Hence, AB = 3.155 meters.

In ∆ABD,

$$\begin{aligned}\text{tan 35°} = \frac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow 0.7002 = \frac{AD}{BD} \\[1em] \Rightarrow BD = \frac{AD}{0.7002} \\[1em] \Rightarrow BD = \frac{20}{0.7002} \\[1em] \Rightarrow BD = 28.563 \text{ m}.\end{aligned}$$

In ∆ACD,

$$\begin{aligned}\text{tan 42°} = \frac{\text{Perpendicular}}{Base} \\[1em] \Rightarrow 0.9004 = \frac{CD}{AD} \\[1em] \Rightarrow CD = AD \times 0.9004 \\[1em] \Rightarrow CD = 20 \times 0.9004 \\[1em] \Rightarrow CD = 18.008 \text{ m}.\end{aligned}$$

From figure,

BC = BD - CD = 28.563 - 18.008 = 10.55 meters.

Hence, BC = 10.55 meters.

In △ACD,

$$\begin{aligned}\text{cos 30°} = \frac{\text{Base}}{\text{Hypotenuse}} \\[1em] \Rightarrow \frac{\sqrt{3}}{2} = \frac{AD}{CD} \\[1em] \Rightarrow AD = \frac{\sqrt{3}}{2} \times CD \\[1em] \Rightarrow AD = \frac{\sqrt{3}}{2} \times 6 \\[1em] \Rightarrow AD = 3\sqrt{3} \\[1em] \Rightarrow AD = 5.196 \text{ m}.\end{aligned}$$

In △BDE,

$$\begin{aligned}\text{sin 47°} = \frac{\text{Perpendicular}}{\text{Hypotenuse}} \\[1em] \Rightarrow 0.73 = \frac{BD}{DE} \\[1em] \Rightarrow BD = 0.73 \times DE \\[1em] \Rightarrow BD = 0.73 \times 5 = 3.65 \text{ m}.\end{aligned}$$

From figure,

AB = AD + BD = 5.196 + 3.65 = 8.846 = 8.85 meters.

Hence, AB = 8.85 meters.

Given,

CA = CB = 15 cm and ∠ACB = 131°.

Construct a perpendicular CP from center C to the chord AB.

We know that perpendicular form center to the chord bisects the chord.

Then, CP bisects AB.

In △ACP and △BCP,

∠APC = ∠BPC = 90°

CP = CP [∵ Common Side]

AP = PB [∵ CP bisects AB]

∴ △ACP ≅ △BCP by SAS axiom.

∴ ∠ACP = ∠BCP = $\frac{131°}{2}$ = 65.5° [By C.P.C.T.]

In △ACP,

$$\begin{aligned}\text{sin 65.5°} = \frac{\text{Perpendicular}}{\text{Hypotenuse}} \\[1em] \Rightarrow 0.91 = \frac{AP}{AC} \\[1em] \Rightarrow AP = 0.91 \times AC \\[1em] \Rightarrow AP = 0.91 \times 15 = 13.65 \text{ cm}.\end{aligned}$$

(i) From figure,

AB = AP + PB = 2AP

= 2 × 13.65

= 27.30 cm

Hence, AB = 27.30 cm.

(ii) In △ACP,

$$\begin{aligned}\text{cos 65.5°} = \frac{\text{Base}}{\text{Hypotenuse}} \\[1em] \Rightarrow 0.415 = \frac{CP}{AC} \\[1em] \Rightarrow CP = 0.415 \times 15 \\[1em] \Rightarrow CP = 6.225 \text{ cm}.\end{aligned}$$

Hence, CP = 6.225 cm.

Let's assume AB to be the vertical tower and C and D be the two points such that CD = 192 m.

Let ∠ACB = θ and ∠ADB = α

Given,

$$\begin{aligned}\Rightarrow \text{tan θ} = \frac{5}{12} \\[1em] \Rightarrow \frac{\text{Perpendicular}}{\text{Base}} = \frac{5}{12} \\[1em] \Rightarrow \frac{AB}{BC} = \frac{5}{12} \\[1em] \Rightarrow AB = \frac{5}{12}BC .........(1)\end{aligned}$$

Also given,

$$\begin{aligned}\Rightarrow \text{tan α} = \frac{3}{4} \\[1em] \Rightarrow \frac{\text{Perpendicular}}{\text{Base}} = \frac{3}{4} \\[1em] \Rightarrow \frac{AB}{BD} = \frac{3}{4} \\[1em] \Rightarrow AB = \frac{3}{4}BD .........(2)\end{aligned}$$

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

$$\begin{aligned}\Rightarrow \frac{5}{12}BC = \frac{3}{4}BD \\[1em] \Rightarrow \frac{BC}{BD} = \frac{3 \times 12}{4 \times 5} \\[1em] \Rightarrow \frac{BD + CD}{BD} = \frac{36}{20} \\[1em] \Rightarrow \frac{BD + CD}{BD} = \frac{9}{5} \\[1em] \Rightarrow 5(BD + CD) = 9BD \\[1em] \Rightarrow 5BD + 5CD = 9BD \\[1em] \Rightarrow 9BD - 5BD = 5CD \\[1em] \Rightarrow 4BD = 5CD \\[1em] \Rightarrow BD = \frac{5 \times 192}{4} \\[1em] \Rightarrow BD = \frac{960}{4} \\[1em] \Rightarrow BD = 240 \text{ m}.\end{aligned}$$

BC = BD + DC = 240 + 192 = 432 m.

From equation (1),

AB = $\frac{5}{12} \times BC = \frac{5}{12} \times 432$

= 5 × 36

= 180 m.

Hence, the height of the tower is 180 m.

Let AB be the tower of height x meters, surmounted by a vertical flagstaff AD of height h meters (given). Let C be a point on the plane such that ∠ACB = α, ∠DCB = β and AD = h.

In ∆ABC,

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

In ∆DBC,

$$\begin{aligned}\Rightarrow \text{tan β} = \frac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \text{tan β} = \frac{BD}{BC} \\[1em]\end{aligned}$$

Substituting value of BC from (1) in above equation :

$$\begin{aligned}\Rightarrow \text{tan β} = \frac{BD}{\frac{x}{\text{tan α}}} \\[1em] \Rightarrow \text{tan β} = \frac{BA + AD}{\frac{x}{\text{tan α}}} \\[1em] \Rightarrow \text{tan β} \times \frac{x}{\text{tan α}} = x + h \\[1em] \Rightarrow x \text{tan β} = \text{tan α}(x + h) \\[1em] \Rightarrow x \text{tan β} = x \text{tan α} + h \text{tan α} \\[1em] \Rightarrow x \text{tan β} - x \text{tan α} = h \text{tan α} \\[1em] \Rightarrow x(\text{tan β} - \text{tan α}) = h \text{tan α} \\[1em] \Rightarrow x = \frac{\text{h tan α}}{\text{tan β} - \text{tan α}}.\end{aligned}$$

Hence, proved that the height of the tower = $\frac{\text{h tan α}}{\text{tan β} - \text{tan α}}.$

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.

(i) In ∆CED,

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

From figure,

AB = DE = 20 m.

Hence, AB = 20 m.

(ii) Let A'D' be the new position of the man and θ be the angle of elevation of the top of the tower.

Such that, D'E = 15 m

In ∆CED',

$$\begin{aligned}\text{tan θ} = \frac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \text{tan θ} = \frac{CE}{D'E} \\[1em] \Rightarrow \text{tan θ} = \frac{8}{15} \\[1em] \Rightarrow \text{tan θ} = 0.533 \\[1em] \Rightarrow \text{tan θ} = \text{tan } 28° \\[1em] \Rightarrow \text{θ} = 28°.\end{aligned}$$

Hence, angle of elevation of the top of the pole when the man is standing 15 metres from the pole is 28°.

From figure,

⇒ AB = DE = 10 m.

In ∆AED

$$\begin{aligned}\Rightarrow \text{tan y°} = \frac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \text{tan y°} = \frac{DE}{AE} \\[1em] \Rightarrow \frac{1}{4} = \frac{DE}{AE} \\[1em] \Rightarrow AE = 4DE = 4 \times 10 = 40 \text{ m}.\end{aligned}$$

In ∆AEC,

$$\begin{aligned}\Rightarrow \text{tan x°} = \frac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \text{tan x°} = \frac{CE}{AE} \\[1em] \Rightarrow \frac{5}{2} = \frac{CE}{AE} \\[1em] \Rightarrow CE = AE \times \frac{5}{2} = 40 \times \frac{5}{2} = 100 \text{ m}.\end{aligned}$$

From figure,

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

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

Let angle of elevation be θ.

According to question,

⇒ cos θ = 0.53

⇒ cos θ = cos 58°

⇒ θ = 58°.

⇒ cos² θ = 0.2809

⇒ 1 - sin² θ = 0.2809

⇒ sin² θ = 1 - 0.2809

⇒ sin² θ = 0.7191

⇒ sin θ = $\sqrt{0.7191}$

⇒ sin θ = 0.848

⇒ tan θ = $\frac{\text{sin θ}}{\text{cos θ}} = \frac{0.848}{0.53}$ = 1.6

$$\begin{aligned}\therefore \frac{\text{Perpendicular}}{\text{Base}} = 1.6 \\[1em] \Rightarrow \frac{AB}{BC} = 1.6 \\[1em] \Rightarrow \frac{20}{BC} = 1.6 \\[1em] \Rightarrow BC = \frac{20}{1.6} = 12.5 \text{ m}.\end{aligned}$$

The man is standing at a distance of 12.5 meters.

(i) Let AB be the pole and CD be the tower.

Given,

Length of pole (AB) = 20 m.

From figure,

CE = AB = 20 m.

In △AEC,

$$\begin{aligned}\text{tan 30°} = \frac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \frac{1}{\sqrt{3}} = \frac{CE}{AE} \\[1em] \Rightarrow AE = \sqrt{3} CE \\[1em] \Rightarrow AE = 20\sqrt{3} \text{ m}.\end{aligned}$$

In △AED,

$$\begin{aligned}\text{tan 60°} = \frac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \sqrt{3} = \frac{DE}{AE} \\[1em] \Rightarrow DE = AE\sqrt{3} \\[1em] \Rightarrow DE = 20\sqrt{3} \times \sqrt{3} = 20 \times 3 = 60 \text{ m}.\end{aligned}$$

From figure,

CD = DE + CE = 60 + 20 = 80 m.

Hence, the height of the tower = 80 m.

(ii) Given,

Length of tower (CD) = 75 m

From figure,

AE = BC = y (let)

In △AEC,

$$\begin{aligned}\text{tan 30°} = \frac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \frac{1}{\sqrt{3}} = \frac{CE}{AE} \\[1em] \Rightarrow CE = \frac{AE}{\sqrt{3}} \\[1em] \Rightarrow CE = \frac{y}{\sqrt{3}} \text{ m}.\end{aligned}$$

In △AED,

$$\begin{aligned}\text{tan 60°} = \frac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \sqrt{3} = \frac{DE}{AE} \\[1em] \Rightarrow DE = AE\sqrt{3} \\[1em] \Rightarrow DE = y\sqrt{3} \text{ m}.\end{aligned}$$

From figure,

⇒ CD = CE + DE

$$\begin{aligned}\Rightarrow \frac{y}{\sqrt{3}} + \sqrt{3}y = 75 \\[1em] \Rightarrow \frac{y + 3y}{\sqrt{3}} = 75 \\[1em] \Rightarrow \frac{4y}{\sqrt{3}} = 75 \\[1em] \Rightarrow y = \frac{75\sqrt{3}}{4} \text{ m}.\end{aligned}$$

We know that,

CE = $\frac{y}{\sqrt{3}} = \frac{\frac{75\sqrt{3}}{4}}{\sqrt{3}} = \frac{75}{4}$ = 18.75 meters.

Hence, height of pole = 18.75 meters.

(i) We know that,

Alternate angles are equal.

∴ ∠ACB = ∠EAC = 60°

In △ABC,

$$\begin{aligned}\text{tan 60°} = \frac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \sqrt{3} = \frac{AB}{BC} \\[1em] \Rightarrow BC = \frac{AB}{\sqrt{3}} = \frac{60}{1.732} = 34.64 \text{ m}.\end{aligned}$$

Hence, horizontal distance between AB and CD = 34.64 meters.

(ii) We know that,

From figure,

FD = BC = 34.64 m

As, alternate angles are equal.

∴ ∠ADF = ∠EAD = 30°

In △AFD,

$$\begin{aligned}\text{tan 30°} = \frac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \frac{1}{\sqrt{3}} = \frac{AF}{FD} \\[1em] \Rightarrow AF = \frac{FD}{\sqrt{3}} \\[1em] \Rightarrow AF = \frac{34.64}{1.732} = 20 \text{ m}.\end{aligned}$$

From figure,

BF = AB - AF = 60 - 20 = 40 m.

∴ CD = 40 m.

Hence, height of lamp post = 40 m.

Let D be the position of aeroplane and A and B be the positions of boat.

As, angle of depression are 60° and 45°.

Since, alternate angles are equal.

∴ ∠DAC = EDA = 60° and ∠DBC = FDB = 45°.

In △ACD,

$$\begin{aligned}\text{tan 60°} = \frac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \sqrt{3} = \frac{CD}{AC} \\[1em] \Rightarrow AC = \frac{CD}{\sqrt{3}} \\[1em] \Rightarrow AC = \frac{250}{1.732} \\[1em] \Rightarrow AC = 144.34 \text{ m}.\end{aligned}$$

In △BCD,

$$\begin{aligned}\text{tan 45°} = \frac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow 1 = \frac{CD}{BC} \\[1em] \Rightarrow BC = CD = 250 \text{ m}.\end{aligned}$$

AB = AC + BC = 144.34 + 250 = 394.34 ≈ 394 m.

Hence, width of river = 394 m.

From figure,

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

From figure,

EC = BD = 120 m.

In △AEC,

$$\begin{aligned}\text{tan 30°} = \frac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \frac{1}{\sqrt{3}} = \frac{AE}{EC} \\[1em] \Rightarrow AE = \frac{EC}{\sqrt{3}} \\[1em] \Rightarrow AE = \frac{120}{\sqrt{3}} \\[1em] \Rightarrow AE = \frac{120}{1.732} = 69.28 \text{ m}.\end{aligned}$$

In △EBC,

$$\begin{aligned}\text{tan 24°} = \frac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow 0.445 = \frac{EB}{EC} \\[1em] \Rightarrow EB = EC \times 0.445 \\[1em] \Rightarrow EB = 120 \times 0.445 = 53.4 \text{ m}.\end{aligned}$$

From figure,

⇒ CD = EB = 53.4 meters.

⇒ AB = AE + EB = 69.28 + 53.4 = 122.68 ≈ 123 meters.

Hence, height of two towers = 123 meters and 53.4 meters.

Let CD be the lighthouse.

As, angle of depression are 60° and 45°.

Since, alternate angles are equal.

∴ ∠DAC = ∠EDA = 60° and ∠DBC = ∠FDB = 45°.

In △ACD,

$$\begin{aligned}\text{tan 60°} = \frac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \sqrt{3} = \frac{CD}{AC} \\[1em] \Rightarrow AC = \frac{CD}{\sqrt{3}} \\[1em] \Rightarrow AC = \frac{60}{1.732} \\[1em] \Rightarrow AC = 34.64 \text{ m}.\end{aligned}$$

In △BCD,

$$\begin{aligned}\text{tan 45°} = \frac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow 1 = \frac{CD}{BC} \\[1em] \Rightarrow BC = CD = 60 \text{ m}.\end{aligned}$$

AB = AC + BC = 34.64 + 60 = 94.64 ≈ 95 m.

Hence, the distance between two ships = 95 m.

Let height of lighthouse (CD) be h meters.

Let A and B be the boats approaching lighthouse.

From figure,

∠CAD = ∠ACX = 30° and ∠CBD = ∠BCY = 45°

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.