Heights & Distances

Let the angle of elevation be θ as shown in the figure below:

Consider △ABC, AB be the height of electric pole and BC be the shadow. Since pole and it's shadow are perpendicular, ∠ABC = 90°.

From △ABC, we get

$$\begin{aligned}\Rightarrow \text{tan θ} = \frac{\text{AB}}{\text{BC}} \\[1em] \Rightarrow \text{tan θ} = \frac{10}{10\sqrt{3}} \\[1em] \Rightarrow \text{tan θ} = \frac{1}{\sqrt{3}} \\[1em] \Rightarrow \text{tan θ} = \text{tan 30°} \\[1em] \therefore \text {θ} = 30°.\end{aligned}$$

Hence, the elevation of the sun is 30°.

Let MP be the tower of height h metres and O be the point on the ground 150 m away from the foot of the tower.

Then, the angle of elevation = ∠MOP = ∠30° (given).

In △OMP, ∠OMP = 90°.

From △OMP, we get

$$\begin{aligned}\Rightarrow \text{tan 30°} = \frac{MP}{OM} \\[1em] \Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{150} \\[1em] \Rightarrow h = \frac{150}{\sqrt{3}} \\[1em] \Rightarrow h = 86.6\end{aligned}$$

Hence, the height of the tower = 86.6 m.

Let MP be the wall of height h metres and O be the point on the ground 1.5 m away from the foot of the wall.

Then, the angle of elevation = ∠MOP = ∠60° (given).

In △OMP, ∠OMP = 90°.

From △OMP, we get

$$\begin{aligned}\Rightarrow \text{tan 60°} = \frac{MP}{OM} \\[1em] \Rightarrow \sqrt{3} = \frac{h}{1.5} \\[1em] \Rightarrow h = 1.5 \times \sqrt{3} \\[1em] \Rightarrow h = 2.598 \approx 2.6\end{aligned}$$

Hence, the height of the wall = 2.6 m.

Let MP be the height of pole and OM be the length of shadow of pole.

Given, height of pole (MP) = shadow of pole (OM) = h metres.

Let the angle of elevation be θ

In △OMP, ∠OMP = 90°.

From △OMP, we get

$$\begin{aligned}\Rightarrow \text{tan θ} = \frac{MP}{OM} \\[1em] \Rightarrow \text{tan θ} = \frac{h}{h} \\[1em] \Rightarrow \text{tan θ} = 1 \\[1em] \Rightarrow \text{tan θ} = \text{tan } 45° \\[1em] \therefore θ = 45°.\end{aligned}$$

Hence, the angle of elevation of sun is 45°.

Let MO be the tower with O being the foot of tower.

Given, angle of elevation of the top of a tower is 30° and height of tower is 100m.

∴ ∠MPO = 30° and OM = 100 m.

In △POM, ∠POM = 90°.

From △POM, we get

$$\begin{aligned}\Rightarrow \text{tan 30°} = \frac{OM}{OP} \\[1em] \Rightarrow \frac{1}{\sqrt{3}} = \frac{100}{OP} \\[1em] \Rightarrow OP = 100 \times \sqrt{3} \\[1em] \Rightarrow OP = 173.2\end{aligned}$$

Hence, the point P is at a distance of 173.2 metres from the foot of the tower.

Let MP be the cliff and O be the buoy.

From figure,

Given angle of depression of a buoy is 20°.

∴ ∠POM = ∠OPN = 20° (Alternate angles are equal).

In △POM, ∠PMO = 90°.

From △POM, we get

$$\begin{aligned}\Rightarrow \text{tan 20°} = \frac{PM}{OM} \\[1em] \Rightarrow 0.3640 = \frac{92}{OM} \\[1em] \Rightarrow OM = \frac{92}{0.3640} \\[1em] \Rightarrow OM = 252.74 \approx 253\end{aligned}$$

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

Suppose boy is at point O and the kite is at point P.

Considering △POM,

∠PMO = 90°, ∠POM = 26° 32', OP = 100 m and MP = height of kite.

From △POM, we get

$$\begin{aligned}\Rightarrow \text{sin 26° 32}' = \frac{MP}{OP} \\[1em] \Rightarrow 0.4467 = \frac{MP}{100} \\[1em] \Rightarrow MP = 0.4467 \times 100 \\[1em] \Rightarrow MP = 44.67 \approx 44.7.\end{aligned}$$

Hence, the height of the kite is 44.7 metres.

Let MP be the pole and string is tied from point P on the pole to point O on the ground.

Considering △POM,

∠PMO = 90°, ∠POM = 45° (given) and MP = 10 m.

From △POM, we get

$$\begin{aligned}\Rightarrow \text{sin 45°} = \frac{MP}{OP} \\[1em] \Rightarrow \frac{1}{\sqrt{2}} = \frac{10}{OP} \\[1em] \Rightarrow OP = 10 \times \sqrt{2} \\[1em] \Rightarrow OP = 10 \times 1.414 \\[1em] \Rightarrow OP = 14.14\end{aligned}$$

Hence, the length of wire is 14.14 metres.

Let θ be the angle of elevation,

Given, cos θ = 0.53

∴ cos θ = cos 58°

⇒ θ = 58°

Let MP be the vertical tower and the man be standing at point O.

Considering △POM,

∠PMO = 90°, ∠POM = 58° and MP = 20 m.

From △POM, we get

$$\begin{aligned}\Rightarrow \text{tan 58°} = \frac{MP}{OM} \\[1em] \Rightarrow 1.6003 = \frac{20}{OM} \\[1em] \Rightarrow OM = \frac{20}{1.6003} \\[1em] \Rightarrow OM = 12.49 \approx 12.5\end{aligned}$$

Hence, the man is at a distance of 12.5 metres from the foot of tower.

(i) Let ACB be the tree. When broken at point C by the storm, let its top A touch the ground so that ∠CAB = 38° 30' and AB = 6 m.

From right angled △ABC, we get

$$\begin{aligned}\Rightarrow \text{tan 38° 30'} = \frac{BC}{AB} \\[1em] \Rightarrow 0.7954 = \frac{BC}{6} \\[1em] \Rightarrow BC = 0.7954 \times 6 \\[1em] \Rightarrow BC = 4.77\end{aligned}$$

Hence, the tree is broken at a height of 4.77 m

(ii) From right angled △ABC, we get

$$\begin{aligned}\Rightarrow \text{cos 38° 30}' = \frac{AB}{AC} \\[1em] \Rightarrow 0.7826 = \frac{6}{AC} \\[1em] \Rightarrow AC = \frac{6}{0.7826} \\[1em] \Rightarrow AC = 7.67\end{aligned}$$

∴ The height of the tree = BC + AC = 4.77 + 7.67 = 12.44

Hence, the original height of the tree is 12.44 metres.

Let CD be an observer of height 1.5 m which is 20.5 m away from a tower AB of height 22 m.

CD = 1.5 m and DB = 20.5 m

From C, draw CE ⊥ AB, then CDBE is a rectangle.

AE = AB - BE = AB - CD = 22 - 1.5 = 20.5 m
CE = DB = 20.5 m.

Let angle of elevation be θ,

From right angled △ACE, we get

$$\begin{aligned}\Rightarrow \text{tan θ} = \frac{AE}{CE} \\[1em] \Rightarrow \text{tan θ} = \frac{20.5}{20.5} \\[1em] \Rightarrow \text{tan θ} = 1 \\[1em] \Rightarrow \text{tan θ} = \text{tan 45°} \\[1em] \therefore \text{θ} = 45°.\end{aligned}$$

Hence, the angle of elevation of the top of the tower from the eye of the observer is 45°.

Let height of tower PT be h meters.

Considering right angled △PQR, we get

$$\begin{aligned}\Rightarrow \text{tan 60°} = \frac{QR}{PQ} \\[1em] \Rightarrow \text{tan 60°} = \frac{50}{PQ} \\[1em] \Rightarrow \sqrt{3} = \frac{50}{PQ} \\[1em] \Rightarrow PQ = \frac{50}{\sqrt{3}} \\[1em]\end{aligned}$$

Now considering right angled △PQT, we get

$$\begin{aligned}\Rightarrow \text{tan 30°} = \frac{h}{PQ} \\[1em] \Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{\frac{50}{\sqrt{3}}} \\[1em] \Rightarrow \frac{1}{\sqrt{3}} = \frac{\sqrt{3} \times h}{50} \\[1em] \Rightarrow h = \frac{50}{\sqrt{3} \times \sqrt{3}} \\[1em] \Rightarrow h = \frac{50}{3} \\[1em] \Rightarrow h = 16.7.\end{aligned}$$

On correcting to nearest meter, h = 17 m.

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

Let QR be the tall building and S be the point at which helicopter is present.

Considering right angled △PQR, we get

$$\begin{aligned}\Rightarrow \text{tan 30°} = \frac{QR}{PQ} \\[1em] \Rightarrow \text{tan 30°} = \frac{10}{PQ} \\[1em] \Rightarrow \frac{1}{\sqrt{3}} = \frac{10}{PQ} \\[1em] \Rightarrow PQ = 10\sqrt{3}\end{aligned}$$

Now considering right angled △PQS, we get

$$\begin{aligned}\Rightarrow \text{tan 60°} = \frac{QS}{PQ} \\[1em] \Rightarrow \sqrt{3} = \frac{QS}{10\sqrt{3}} \\[1em] \Rightarrow QS = 10\sqrt{3} \times \sqrt{3} \\[1em] \Rightarrow QS = 30.\end{aligned}$$

Hence, the height of helicopter above the ground is 30 m.

Let the aeroplane at a height of 3125 m be at a point R and the plane above be at point S.

Considering right angled △PQR, we get

$$\begin{aligned}\Rightarrow \text{tan 30°} = \frac{QR}{PQ} \\[1em] \Rightarrow \text{tan 30°} = \frac{3125}{PQ} \\[1em] \Rightarrow \frac{1}{\sqrt{3}} = \frac{3125}{PQ} \\[1em] \Rightarrow PQ = 3125\sqrt{3}\end{aligned}$$

Now considering right angled △PQS, we get

$$\begin{aligned}\Rightarrow \text{tan 60°} = \frac{QS}{PQ} \\[1em] \Rightarrow \sqrt{3} = \frac{QS}{3125\sqrt{3}} \\[1em] \Rightarrow QS = 3125\sqrt{3} \times \sqrt{3} \\[1em] \Rightarrow QS = 9375\end{aligned}$$

Distance between two aeroplanes = QS - QR = 9375 - 3125 = 6250 m.

Hence, the distance between two planes = 6250 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,

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

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,

$$\begin{aligned}\Rightarrow \text{tan 60°} = \frac{QR}{SQ} \\[1em] \Rightarrow \sqrt{3} = \frac{QR}{QR - 20} \\[1em] \Rightarrow \sqrt{3}(QR - 20) = QR \\[1em] \Rightarrow \sqrt{3}QR - 20\sqrt{3} = QR \\[1em] \Rightarrow \sqrt{3}QR - QR = 20\sqrt{3}\\[1em] \Rightarrow QR(1.732 - 1) = 20 \times 1.732 \\[1em] \Rightarrow 0.732\text{ QR} = 34.64 \\[1em] \Rightarrow QR = \frac{34.64}{0.732} \\[1em] \Rightarrow QR = 47.32\end{aligned}$$

On correcting to 2 significant figures QR = 47.

Hence, the height of the tower is 47 m.

Let the height of the tower BD be h metres and the length of its shadow be d 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., BD = h meters, AB = d meters and CA = 10 metres.

From right angled △ABD, we get

$$\begin{aligned}\Rightarrow \text{tan 45°} = \frac{BD}{AB} \\[1em] \Rightarrow 1 = \frac{h}{d} \\[1em] \Rightarrow h = d\end{aligned}$$

From right angled △BCD, we get

$$\begin{aligned}\Rightarrow \text{tan 30°} = \frac{BD}{BC} \\[1em] \Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{AC + AB} \\[1em] \Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{10 + d} \\[1em] \Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{10 + h} [\because h = d] \\[1em] \Rightarrow 10 + h = \sqrt{3}h \\[1em] \Rightarrow \sqrt{3}h - h = 10 \\[1em] \Rightarrow h(\sqrt{3} - 1) = 10 \\[1em] \Rightarrow 0.732\text{ h} = 10 \\[1em] \Rightarrow h = \frac{10}{0.732} \\[1em] \Rightarrow h = 13.66\end{aligned}$$

Hence, the height of the tower is 13.66 meters.

Let R be the top of the tower and Q the foot. P and T 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,

∠RPQ = ∠SRP = 30°      (Alternate angles are equal)
∠RTQ = ∠SRT = 45°      (Alternate angles are equal)
PT = 1 km
TQ = PQ - PT = PQ - 1      (Eq 1)

From right angled △PQR, we get

$$\begin{aligned}\Rightarrow \text{tan 30°} = \frac{QR}{PQ} \\[1em] \Rightarrow \frac{1}{\sqrt{3}} = \frac{QR}{PQ} \\[1em] \Rightarrow PQ = \sqrt{3} \text{ QR} \\[1em] \Rightarrow QR = \frac{PQ}{\sqrt{3}}\text{ ......(Eq 2)}\end{aligned}$$

From right angled △TQR, we get

$$\begin{aligned}\Rightarrow \text{tan 45°} = \frac{QR}{TQ} \\[1em] \Rightarrow 1 = \frac{QR}{TQ} \\[1em] \Rightarrow TQ = QR \\[1em] \Rightarrow PQ - 1 = QR \text{ (Using Eq 1)}\\[1em] \Rightarrow PQ - 1 = \frac{PQ}{\sqrt{3}} \text{ (Using Eq 2)}\\[1em] \Rightarrow \sqrt{3}PQ - \sqrt{3} = PQ \\[1em] \Rightarrow \sqrt{3}PQ - PQ = \sqrt{3} \\[1em] \Rightarrow 0.732PQ = 1.732 \\[1em] \Rightarrow PQ = \frac{1.732}{0.732} \\[1em] \Rightarrow PQ = 2.366\end{aligned}$$

Using Eq 1,

TQ = PQ - 1 = 2.366 - 1 = 1.366.

Hence, the distance of two kilometer stones from hill are 1.366 km and 2.366 km.

Let QR be the tower and man be initially at point P after moving 60 m let it reach point S.

PS = 60 m

From figure,

PQ = PS + SQ = (SQ + 60) m.     (Eq 1)

Considering right angled △SQR, we get

$$\begin{aligned}\Rightarrow \text{tan 60°} = \frac{QR}{SQ} \\[1em] \Rightarrow \sqrt{3} = \frac{QR}{SQ} \\[1em] \Rightarrow QR = \sqrt{3} \text{ SQ} \text{ .....(Eq 2)}\end{aligned}$$

Considering right angled △PQR, we get

$$\begin{aligned}\Rightarrow \text{tan 30°} = \frac{QR}{PQ} \\[1em] \Rightarrow \frac{1}{\sqrt{3}} = \frac{QR}{PQ} \\[1em] \Rightarrow PQ = \sqrt{3} \text{ QR} \\[1em] \Rightarrow SQ + 60 = \sqrt{3} \text{ QR} \text{ ........(Eq 3)} \\[1em]\end{aligned}$$

Putting value of QR from Eq 2, in Eq 3 we get,

$$\begin{aligned}\Rightarrow SQ + 60 = \sqrt{3} \times \sqrt{3} \text{ SQ} \\[1em] \Rightarrow SQ + 60 = 3 SQ \\[1em] \Rightarrow 2SQ = 60 \\[1em] \Rightarrow SQ = 30.\end{aligned}$$

From Eq 2,

$$\begin{aligned}\Rightarrow QR = \sqrt{3}SQ \\[1em] \Rightarrow QR = \sqrt{3} \times 30 \\[1em] \Rightarrow QR = 51.96\end{aligned}$$

Correcting upto nearest meter QR = 52.

Hence, the height of building is 52 meters.

Let the height of tower QR be h meters and the angle of elevation be θ₁ and θ₂ at points P and S respectively.

So,

$\text{tan }θ_1 = \frac{5}{12} \text{ and } \text{tan }θ_2 = \frac{3}{4}$

From figure,

QR = h meters
PQ = PS + QS = (192 + QS) meters.

Considering right angled △PQR, we get

$$\begin{aligned}\Rightarrow \text{tan }θ_1 = \frac{QR}{PQ} \\[1em] \Rightarrow \frac{5}{12} = \frac{h}{192 + QS} \\[1em] \Rightarrow 5(192 + QS) = 12h \\[1em] \Rightarrow 5QS + 960 = 12h \text{ .......(Eq 1)}\end{aligned}$$

Considering right angled △SQR, we get

$$\begin{aligned}\Rightarrow \text{tan }θ_2 = \frac{QR}{QS} \\[1em] \Rightarrow \frac{3}{4} = \frac{h}{QS} \\[1em] \Rightarrow 3QS = 4h \\[1em] \Rightarrow QS = \frac{4h}{3}\text{ .......(Eq 2)}\end{aligned}$$

Putting value of QS from Eq 2 in Eq 1 we get,

$$\begin{aligned}\Rightarrow 5 \times \frac{4h}{3} + 960 = 12h \\[1em] \Rightarrow \frac{20h}{3} + 960 = 12h \\[1em] \Rightarrow \frac{20h + 2880}{3} = 12h \\[1em] \Rightarrow 20h + 2880 = 36h \\[1em] \Rightarrow 36h - 20h = 2880 \\[1em] \Rightarrow 16h = 2880 \\[1em] \Rightarrow h = \frac{2880}{16} \\[1em] \Rightarrow h = 180.\end{aligned}$$

Hence, the height of the tower is 180 meters.

(i) From figure,

Considering right angled △AFT, we get

$$\begin{aligned}\Rightarrow \text{tan x°} = \frac{TF}{AF} \\[1em] \Rightarrow \frac{2}{5} = \frac{TF}{200} \\[1em] \Rightarrow TF = \frac{2 \times 200}{5} \\[1em] \Rightarrow TF = 80\end{aligned}$$

Hence, the height of the tower is 80 meters.

(ii) From figure,

BF = AF - AB = 200 - 80 = 120 meters.

Considering right angled △BFT, we get

$$\begin{aligned}\Rightarrow \text{tan y°} = \frac{TF}{BF} \\[1em] \Rightarrow \text{tan y°} = \frac{80}{120} \\[1em] \Rightarrow \text{tan y°} = 0.667 \\[1em] \Rightarrow \text{tan y°} = \text{tan 33° 41}' \\[1em] \Rightarrow \text{y°} = 33° 41'\end{aligned}$$

Rounding off to nearest degree, y = 34°.

Hence, angle y = 34°.

From figure,

∠ACB = ∠EAC = 45° (Alternate angles are equal)
∠ADB = ∠EAD = 60° (Alternate angles are equal)

Considering right angled △ABC, we get

$$\begin{aligned}\Rightarrow \text{tan 45°} = \frac{AB}{CB} \\[1em] \Rightarrow 1 = \frac{300}{CB} \\[1em] \Rightarrow CB = 300\end{aligned}$$

Considering right angled △ADB, we get

$$\begin{aligned}\Rightarrow \text{tan 60°} = \frac{AB}{DB} \\[1em] \Rightarrow \sqrt{3} = \frac{300}{DB} \\[1em] \Rightarrow DB = \frac{300}{\sqrt{3}} \\[1em] \Rightarrow DB = 173.2\end{aligned}$$

Distance between two objects (CD) = CB - DB = 300 - 173.2 = 126.8.

Rounding off to nearest meter CD = 127 m.

Hence, the distance between two objects = 127 meters.

Let AB be the first tower and CD be the second tower. From C draw a line parallel to AD and perpendicular to AB meeting AB at point E.

Now ADCE forms a rectangle,

EC = AD = 140
AE = DC = 60.

Considering right angled △BCE, we get

$$\begin{aligned}\Rightarrow \text{tan 30°} = \frac{BE}{EC} \\[1em] \Rightarrow \frac{1}{\sqrt{3}} = \frac{BE}{140} \\[1em] \Rightarrow BE = \frac{140}{\sqrt{3}} \\[1em] \Rightarrow BE = 80.83\end{aligned}$$

From figure,

AB = AE + BE = 60 + 80.83 = 140.83

Hence, the height of the first tower is 140.83 meters.

Let AB be the tower of length 80 m and the ships be at point C and D.

From figure,

∠ADB = ∠EAD = 30° (Alternate angles are equal)
∠ACB = ∠EAC = 40° (Alternate angles are equal)

Considering right angled △ADB, we get

$$\begin{aligned}\Rightarrow \text{tan 30°} = \frac{AB}{DB} \\[1em] \Rightarrow \frac{1}{\sqrt{3}} = \frac{80}{DB} \\[1em] \Rightarrow DB = 80 \times \sqrt{3} = 138.56\end{aligned}$$

Considering right angled △ACB, we get

$$\begin{aligned}\Rightarrow \text{tan 40°} = \frac{AB}{BC} \\[1em] \Rightarrow 0.8391 = \frac{80}{BC} \\[1em] \Rightarrow BC = \frac{80}{0.8391} \\[1em] \Rightarrow BC = 95.34\end{aligned}$$

Distance between two ships (DC) = DB - BC = 138.56 - 95.34 = 43.22 meters.

Rounding off to nearest meter DC = 43 meters.

Hence, the distance between two ships is 43 meters.

Let the height of pillar (CD) be h meters.

Considering right angled △ACD, we get

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

From figure,

BC = AB - AC = (15 - h) meters

Considering right angled △BCD, we get

$$\begin{aligned}\Rightarrow \text{tan 60°} = \frac{CD}{BC} \\[1em] \Rightarrow \sqrt{3} = \frac{h}{15 - h} \\[1em] \Rightarrow \sqrt{3}(15 - h) = h \\[1em] \Rightarrow 15\sqrt{3} - \sqrt{3}h = h \\[1em] \Rightarrow 15\sqrt{3} = \sqrt{3}h + h \\[1em] \Rightarrow 25.98 = 2.732h \\[1em] \Rightarrow h = \frac{25.98}{2.732} \\[1em] \Rightarrow h = 9.51\end{aligned}$$

Hence, the height of the pillar is 9.51 meters.

From figure,

Considering right angled △ACD, we get

$$\begin{aligned}\Rightarrow \text{tan 30°} = \frac{CD}{AC} \\[1em] \Rightarrow \frac{1}{\sqrt{3}} = \frac{10}{AC} \\[1em] \Rightarrow AC = 10 \times \sqrt{3} = 17.32\end{aligned}$$

Considering right angled △BCD, we get

$$\begin{aligned}\Rightarrow \text{tan 60°} = \frac{CD}{BC} \\[1em] \Rightarrow \sqrt{3} = \frac{10}{BC} \\[1em] \Rightarrow BC = \frac{10}{\sqrt{3}} \\[1em] \Rightarrow BC = 5.77\end{aligned}$$

AB = AC - BC = 17.32 - 5.77 = 11.55 meters.

Hence, the distance between A and B = 11.55 meters.

Let CD be the light house, 60 m tall and ships at point A and B.

From figure,

∠DAC = ∠XDA = 60° (Alternate angles are equal)
∠DBC = ∠YDB = 45° (Alternate angles are equal)

Considering right angled △ACD, we get

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

Considering right angled △BCD, we get

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

Distance between two ships (AB) = AC + BC = 34.64 + 60 = 94.64 meters.

Rounding off to nearest meter, AB = 95 meters.

Hence, the distance between two ships is 95 meters.

Let aeroplane be at point D and boats be at point A and B. Since, aeroplane is at an altitude of 250 m therefore,

∴ CD = 250 m.

From figure,

∠DAC = ∠XDA = 45° (Alternate angles are equal)
∠DBC = ∠YDB = 60° (Alternate angles are equal)

Considering right angled △BCD, we get

$$\begin{aligned}\Rightarrow \text{tan 60°} = \frac{CD}{BC} \\[1em] \Rightarrow \sqrt{3} = \frac{250}{BC} \\[1em] \Rightarrow BC = \frac{250}{\sqrt{3}} = 144.34\end{aligned}$$

Considering right angled △ACD, we get

$$\begin{aligned}\Rightarrow \text{tan 45°} = \frac{CD}{AC} \\[1em] \Rightarrow 1 = \frac{250}{AC} \\[1em] \Rightarrow AC = 250\end{aligned}$$

Width of the river (AB) = AC + BC = 144.34 + 250 = 394.34 meters.

Rounding off to nearest meter AB = 394 meters.

Hence, the width of the river is 394 meters.

(i) Let the rocks be at point A and D.

From figure a,

∠CAB = ∠ECA = 12° 20' (Alternate angles are equal)
∠CDB = ∠ECD = 16° (Alternate angles are equal)

Considering right angled △ABC, we get

$$\begin{aligned}\Rightarrow \text{tan 12° 20}' = \frac{BC}{AB} \\[1em] \Rightarrow 0.2186 = \frac{126}{AB} \\[1em] \Rightarrow AB = \frac{126}{0.2186} = 576.29\end{aligned}$$

Considering right angled △BCD, we get

$$\begin{aligned}\Rightarrow \text{tan 16°} = \frac{BC}{BD} \\[1em] \Rightarrow 0.2867 = \frac{126}{BD} \\[1em] \Rightarrow BD = \frac{126}{0.2867} = 439.48\end{aligned}$$

Distance between two rocks (AD) = AB - BD = 576.29 - 439.48 = 136.81

Hence, the distance between two rocks when they are on same side of the tower is 136.81 meters.

(ii) Let the rocks be at point A and B.

From figure b,

∠CAD = ∠XCA = 12° 20' (Alternate angles are equal)
∠CBD = ∠YCB = 16° (Alternate angles are equal)

Considering right angled △ADC, we get

$$\begin{aligned}\Rightarrow \text{tan 12° 20}' = \frac{CD}{AD} \\[1em] \Rightarrow 0.2186 = \frac{126}{AD} \\[1em] \Rightarrow AD = \frac{126}{0.2186} = 576.29\end{aligned}$$

Considering right angled △BCD, we get

$$\begin{aligned}\Rightarrow \text{tan 16°} = \frac{CD}{DB} \\[1em] \Rightarrow 0.2867 = \frac{126}{DB} \\[1em] \Rightarrow DB = \frac{126}{0.2867} = 439.48\end{aligned}$$

Distance between two rocks (AB) = AD + DB = 576.29 + 439.48 = 1015.7

Hence, the distance between two rocks when they are on opposite sides of the tower is 1015.7 meters.

Let AB be the lamp post and CD the height of man.

BD is the distance of man from the foot of the lamp and FD is the shadow of man.

CE || DB.

Take AB = x and CD = 1.8 m
EB = CD = 1.8 m
CE = DB = 3.6 m
AE = (x - 1.8) m
Shadow (FD) = 5.4 m

Considering right angled △ACE, we get

$$\begin{aligned}\Rightarrow \text{tan θ} = \frac{AE}{CE} \\[1em] \Rightarrow \text{tan θ} = \frac{x - 1.8}{3.6} \text{ ......(Eq 1)}\end{aligned}$$

Considering right angled △CFD, we get

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

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.

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

From figure,

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

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

Let BD = x meters.

From figure,

EC = BD = x meters.

EB = CD = 20 meters.

In △ AEC,

⇒ tan 45° = $\frac{AE}{EC}$

⇒ 1 = $\frac{AE}{x}$

⇒ AE = x meters.

In △ ABD,

⇒ tan 60° = $\frac{AB}{BD}$

$$\begin{aligned}\Rightarrow \sqrt{3} = \frac{AE + EB}{BD} \\[1em] \Rightarrow \sqrt{3} = \frac{x + 20}{x} \\[1em] \Rightarrow x\sqrt{3} = x + 20 \\[1em] \Rightarrow x\sqrt{3} - x = 20 \\[1em] \Rightarrow x(\sqrt{3} - 1) = 20 \\[1em] \Rightarrow x = \frac{20}{(\sqrt{3} - 1)} \\[1em] \Rightarrow x = \frac{20}{1.732 - 1} \\[1em] \Rightarrow x = \frac{20}{0.732} \\[1em] \Rightarrow x = 27.32 \text{ meters}\end{aligned}$$

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.

Let the height of tower (BC) be h meters and BD be the pole of height 5 meters above it.

From figure,

∠BAC = ∠EBA = 45° (Alternate angles are equal)

DC = DB + BC = 5 + h.

Considering right angled △BCA, we get

$$\begin{aligned}\Rightarrow \text{tan 45°} = \frac{BC}{AC} \\[1em] \Rightarrow 1 = \frac{h}{AC} \\[1em] \Rightarrow AC = h \text{ ......(Eq 1)}\end{aligned}$$

Considering right angled △DCA, we get

$$\begin{aligned}\Rightarrow \text{tan 60°} = \frac{DC}{AC} \\[1em] \Rightarrow \sqrt{3} = \frac{h + 5}{AC}\end{aligned}$$

Putting value of AC from Eq 1 in above equation we get,

$$\begin{aligned}\Rightarrow \sqrt{3} = \frac{h + 5}{h} \\[1em] \Rightarrow \sqrt{3}h = h + 5 \\[1em] \Rightarrow \sqrt{3}h - h = 5 \\[1em] \Rightarrow 0.732\text{ h} = 5 \\[1em] \Rightarrow h = \frac{5}{0.732} \\[1em] \Rightarrow h = 6.83\end{aligned}$$

Hence, the height of the tower is 6.83 meters.

Let AB be the pole and CD be the tower. Let length of tower (CD) be h metres.

Let distance between pole and tower (BD) be x meters.

From figure,

ABDE is a rectangle so,

DE = AB = 20 meters
AE = BD = x meters
CE = CD - DE = (h - 20) meters
∠EAD = ∠ADB = 30° (Alternate angles are equal)

Considering right angled △ABD we get,

$$\begin{aligned}\Rightarrow \text{tan 30°} = \frac{AB}{BD} \\[1em] \Rightarrow \frac{1}{\sqrt{3}} = \frac{20}{x} \\[1em] \Rightarrow x = 20\sqrt{3}\end{aligned}$$

Considering right angled △ACE we get,

$$\begin{aligned}\Rightarrow \text{tan 60°} = \frac{CE}{AE} \\[1em] \Rightarrow \sqrt{3} = \frac{h - 20}{BD} \\[1em] \Rightarrow \sqrt{3} = \frac{h - 20}{x} \\[1em] \Rightarrow \sqrt{3} = \frac{h - 20}{20\sqrt{3}} \\[1em] \Rightarrow 20\sqrt{3} \times \sqrt{3} = h - 20 \\[1em] \Rightarrow 60 = h - 20 \\[1em] \Rightarrow h = 80.\end{aligned}$$

Hence, the height of the tower is 80 meters.

Let AB be the building and CD be the monument. Let length of monument (CD) be h metres.

Let distance between building and monument (BD) be x meters.

From figure,

ABDE is a rectangle so,

DE = AB = 20 meters
AE = BD = x meters
CE = CD - DE = (h - 20) meters
∠EAD = ∠ADB = 15°

Considering right angled △ABD we get,

$$\begin{aligned}\Rightarrow \text{tan 15°} = \frac{AB}{BD} \\[1em] \Rightarrow 0.2679 = \frac{20}{x} \\[1em] \Rightarrow x = \frac{20}{0.2679} \\[1em] \Rightarrow x = 74.65\end{aligned}$$

Considering right angled △ACE we get,

$$\begin{aligned}\Rightarrow \text{tan 45°} = \frac{CE}{AE} \\[1em] \Rightarrow 1 = \frac{h - 20}{x} \\[1em] \Rightarrow x = h - 20 \\[1em] \Rightarrow 74.65 = h - 20 \\[1em] \Rightarrow h = 74.65 + 20 \\[1em] \Rightarrow h = 94.65\end{aligned}$$

Hence, the height of the monument is 94.65 meters.

Let the initial length of shadow be x meters and height of tower be h meters.

Considering right angled △DBC, we get

$$\begin{aligned}\Rightarrow \text{tan 45°} = \frac{BC}{DB} \\[1em] \Rightarrow 1 = \frac{h}{x} \\[1em] \Rightarrow x = h \text{ ......(Eq 1)}\end{aligned}$$

Considering right angled △ABC we get,

$$\begin{aligned}\Rightarrow \text{tan 30°} = \frac{BC}{AB} \\[1em] \Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{x + 10} \\[1em] \Rightarrow x + 10 = \sqrt{3}h \\[1em] \Rightarrow h + 10 = \sqrt{3}h \text{ .....(Using Eq 1)} \\[1em] \Rightarrow \sqrt{3}h - h = 10 \\[1em] \Rightarrow 0.732h = 10 \\[1em] \Rightarrow h = \frac{10}{0.732} \\[1em] \Rightarrow h = 13.7\end{aligned}$$

Hence, the height of tower is 13.7 meters.

Speed of aircraft = 360 km/h

Distance covered in 20 seconds = $\frac{360 \times 20}{60 \times 60} = 2$ km

Let aeroplane be flying at a height of h km.

E is the fixed point on ground and A is the initial position of aircraft and C is the position after 20 seconds.

Considering right angled △EDC we get,

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

From figure,

$$\begin{aligned}\Rightarrow ED = EB + BD \\[1em] \Rightarrow h\sqrt{3} = EB + 2 \\[1em] \Rightarrow EB = h\sqrt{3} - 2 \text{ .......(Eq 1)}\end{aligned}$$

Considering right angled △AEB we get,

$$\begin{aligned}\Rightarrow \text{tan 45°} = \frac{AB}{EB} \\[1em] \Rightarrow 1 = \frac{h}{EB} \\[1em] \Rightarrow EB = h \text{ .......(Eq 2)}\end{aligned}$$

Comparing Eq 1 and Eq 2 we get,

$$\begin{aligned}h = h\sqrt{3} - 2 \\[1em] \sqrt{3}h - h = 2 \\[1em] 0.732h = 2 \\[1em] h = \frac{2}{0.732} \\[1em] h = 2.732 \text{ km} = 2732 \text{ m}\end{aligned}$$

Hence, the aircraft is flying at a height of 2732 meters.

Let ∠BCP = α and ∠ACP = β

From figure,

⇒ α + 54° = 90°

⇒ α = 90° - 54° = 36°.

⇒ β + 42° = 90°

⇒ β = 90° - 42° = 48°.

⇒ tan α = $\frac{BP}{CP}$

⇒ tan 36° = $\frac{BP}{100}$

⇒ 0.7265 = $\frac{BP}{100}$

⇒ BP = 0.7265 × 100 = 72.65 m

⇒ tan β = $\frac{AP}{CP}$

⇒ tan 48° = $\frac{AP}{100}$

⇒ 1.1106 = $\frac{AP}{100}$

⇒ AP = 1.1106 × 100 = 111.06 m

(a) From figure,

AB = AP + BP = 72.65 + 111.06 = 183.71 m

Hence, the distance between two ships = 183.71 m.

(b) On rounding off,

AB = 184 m.

Hence, the distance between two ships = 184 m.

From figure,

⇒ tan 52° = $\frac{CD}{AC}$

⇒ 1.28 = $\frac{100}{AC}$

⇒ AC = $\frac{100}{1.28}$ = 78.125 m

⇒ tan 45° = $\frac{CD}{BC}$

⇒ 1 = $\frac{100}{BC}$

⇒ BC = 100 m

⇒ AB = AC + BC = 78.125 + 100 = 178.125 m

Hence, AB = 178 m, to the nearest metre.

Let the kite be at point B.

Considering right angled △ABC we get,

$$\begin{aligned}\text{sin 60} = \frac{BC}{AB} \\[1em] \frac{\sqrt{3}}{2} = \frac{40\sqrt{3}}{AB} \\[1em] AB = \frac{40 \sqrt{3} \times 2}{\sqrt{3}} \\[1em] AB = 80\end{aligned}$$

Hence, Option 1 is the correct option.

From figure,

∠ACB = ∠DAC = 30° (Alternate angles are equal)

Considering right angled △ABC we get,

$$\begin{aligned}\Rightarrow \text{tan 30°} = \frac{AB}{BC} \\[1em] \Rightarrow \frac{1}{\sqrt{3}} = \frac{75}{BC} \\[1em] \Rightarrow BC = 75\sqrt{3}\end{aligned}$$

Hence, Option 3 is the correct option.

Let AB be the ladder and θ be the angle of elevation.

Considering right angled △ABC we get,

$$\begin{aligned}\Rightarrow \text{cos θ} = \frac{7}{14} \\[1em] \Rightarrow \text{cos θ} = \frac{1}{2} \\[1em] \Rightarrow \text{cos θ} = \text{cos 60°} \\[1em] \Rightarrow θ = 60°.\end{aligned}$$

Hence, Option 4 is the correct option.

Let AB be the lighthouse and C be the point 80 m away from its foot.

Given, AB = 80 m

BC = 80 m

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

⇒ tan θ = $\frac{80}{80}$

⇒ tan θ = 1

⇒ tan θ = tan 45°

⇒ θ = 45°.

Hence, option 2 is the correct option.

Let the angle of elevation be θ and AB be the pole.

Considering right angled △ABC we get,

$$\begin{aligned}\Rightarrow \text{tan θ} = \frac{6}{2\sqrt{3}} \\[1em] \Rightarrow \text{tan θ} = \frac{3}{\sqrt{3}} \\[1em] \Rightarrow \text{tan θ} = \sqrt{3} \\[1em] \Rightarrow \text{tan θ} = \text{tan 60°} \\[1em] \Rightarrow θ = 60°.\end{aligned}$$

Hence, Option 1 is the correct option.

Let the angle of elevation be θ and height of tower be h meters.

So, shadow of tower = h$\sqrt{3}$ meters

Considering right angled △ABC we get,

$$\begin{aligned}\Rightarrow \text{tan θ} = \frac{AB}{BC} \\[1em] \Rightarrow \text{tan θ} = \frac{h}{h\sqrt{3}} \\[1em] \Rightarrow \text{tan θ} = \frac{1}{\sqrt{3}} \\[1em] \Rightarrow \text{tan θ} = \text{tan 30°} \\[1em] \Rightarrow θ = 30°.\end{aligned}$$

Hence, Option 2 is the correct option.

Considering right angled △ABC we get,

$$\begin{aligned}\Rightarrow \text{sin 30°} = \frac{BC}{AC} \\[1em] \Rightarrow \frac{1}{2} = \frac{BC}{8} \\[1em] \Rightarrow BC = 4 \text{ cm}\end{aligned}$$

Similarly,

$$\begin{aligned}\Rightarrow \text{cos 30°} = \frac{AB}{AC} \\[1em] \Rightarrow \frac{\sqrt{3}}{2} = \frac{AB}{8} \\[1em] \Rightarrow AB = 4\sqrt{3} \text{ cm}\end{aligned}$$

Area of right angle triangle = A

$$\begin{aligned}\therefore A = \frac{1}{2} \times \text{ base} \times \text{ height} \\[1em] \Rightarrow A = \frac{1}{2} \times 4\sqrt{3} \times 4 \\[1em] \Rightarrow A = 8\sqrt{3}\text{ cm}^2\end{aligned}$$

Hence, Option 3 is the correct option.

From figure,

⇒ tan α = $\frac{AB}{BC}$

⇒ tan β = $\frac{AB}{BD}$

⇒ tan γ = $\frac{AB}{BE}$

Since, BC < BD < BE.

∴ tan α > tan β > tan γ.

∴ tan α > tan β.

Hence, Option 1 is the correct option.

From figure,

⇒ sin 60° = $\frac{QR}{PR}$

⇒ $\frac{\sqrt{3}}{2} = \frac{9}{PR}$

⇒ PR = $\frac{9 \times 2}{\sqrt{3}} = 3\sqrt{3} \times 2 = 6\sqrt{3}$ cm.

Hence, Option 2 is the correct option.

Consider the below figure:

(i) Let after moving 50 m towards tower from point A, the person reaches point D and height of tower be h meters.

From figure,

AD = 50 m, AB = AD + DB = (50 + DB) m

Considering right angled triangle △ABC,

$$\begin{aligned}\Rightarrow \text{tan 30°} = \frac{BC}{AB} \\[1em] \Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{50 + DB} \\[1em] \Rightarrow 50 + DB = h\sqrt{3} \text{ .......( Eq 1)}\end{aligned}$$

Considering right angled triangle △BCD,

$$\begin{aligned}\Rightarrow \text{tan 60°} = \frac{BC}{DB} \\[1em] \Rightarrow \sqrt{3} = \frac{h}{DB} \\[1em] \Rightarrow h = DB\sqrt{3} \text{ .......( Eq 2)}\end{aligned}$$

Putting value of h from Eq 2 in Eq 1 we get,

$$\begin{aligned}\Rightarrow 50 + DB = DB\sqrt{3} \times \sqrt{3} \\[1em] \Rightarrow 50 + DB = 3DB \\[1em] \Rightarrow 2DB = 50 \\[1em] \Rightarrow DB = 25 \\[1em] \therefore h = DB\sqrt{3} = 25 \times 1.732 = 43.3 \text{ m}\end{aligned}$$

Hence, the height of tower is 43.3 m.

(ii) From figure,

Distance of tower from A (AB) = AD + DB = 50 + 25 = 75 m.

Hence, the distance of tower from A is 75 m.

Let the plane 3000 m high be at point B and plane below it be at point D.

From figure,

Considering right angled triangle △ABC,

$$\begin{aligned}\Rightarrow \text{tan 60°} = \frac{BC}{AC} \\[1em] \Rightarrow \sqrt{3} = \frac{3000}{AC} \\[1em] \Rightarrow AC = \frac{3000}{\sqrt{3}} \\[1em] \Rightarrow AC = 1732 \text{ m}.\end{aligned}$$

Considering right angled triangle △ADC,

$$\begin{aligned}\Rightarrow \text{tan 45°} = \frac{DC}{AC} \\[1em] \Rightarrow 1 = \frac{DC}{1732} \\[1em] \Rightarrow DC = 1732 \text{ m}.\end{aligned}$$

Distance between two planes (BD) = BC - DC = 3000 - 1732 = 1268 m.

Hence, the vertical distance between two planes is 1268 m.

Let CD be the tower of height h meters and BD the flagstaff.

A be point on the ground from where the angles of elevation of the top and bottom of the flagstaff are 45° and 36° respectively.

From figure,

BC = BD + DC = (7 + h) meters.

Considering right angled triangle △ABC,

$$\begin{aligned}\Rightarrow \text{tan 45°} = \frac{BC}{AC} \\[1em] \Rightarrow 1 = \frac{7 + h}{AC} \\[1em] \Rightarrow AC = h + 7 \text{ ......(Eq 1)}.\end{aligned}$$

Considering right angled triangle △ADC,

$$\begin{aligned}\Rightarrow \text{tan 36°} = \frac{DC}{AC} \\[1em] \Rightarrow 0.7265 = \frac{h}{AC} \\[1em] \Rightarrow AC = \frac{h}{0.7265}\end{aligned}$$

Putting value of AC in Eq 1 we get,

$$\begin{aligned}\Rightarrow \frac{h}{0.7265} = h + 7 \\[1em] \Rightarrow h = 0.7265(h + 7) \\[1em] \Rightarrow h = 0.7265h + 5.0855 \\[1em] \Rightarrow h - 0.7265\text{ h} = 5.0855 \\[1em] \Rightarrow 0.2735\text{ h} = 5.0855 \\[1em] \Rightarrow h = \frac{5.0855}{0.2735} \\[1em] \Rightarrow h = 18.6 \text{ m}.\end{aligned}$$

Hence, the height of tower is 18.6 m.

Let AD be man and BC be tower of height h meters.

From figure,

ABED is a rectangle.

BE = AD = 1.6 m
DE = AB = 20 m.

CE = BC - BE = (h - 1.6) m.

Considering right angled triangle △DCE,

$$\begin{aligned}\Rightarrow \text{tan 60°} = \frac{CE}{DE} \\[1em] \Rightarrow \sqrt{3} = \frac{h - 1.6}{20} \\[1em] \Rightarrow h - 1.6 = 20\sqrt{3} \\[1em] \Rightarrow h - 1.6 = 34.64 \\[1em] \Rightarrow h = 36.24 \text{ m}.\end{aligned}$$

Hence, the height of tower is 36.24 m.

Let AD be man, BC be tower and θ be the angle of elevation.

From figure,

ABED is a rectangle.

BE = AD = 1.54 m
DE = AB = 2.1 m.

CE = BC - BE = (3.64 - 1.54) = 2.1 m.

Considering right angled triangle △DCE,

$$\begin{aligned}\Rightarrow \text{tan θ} = \frac{CE}{DE} \\[1em] \Rightarrow \text{tan θ} = \frac{2.1}{2.1} \\[1em] \Rightarrow \text{tan θ} = 1 \\[1em] \Rightarrow \text{tan θ} = \text{tan 45°} \\[1em] \Rightarrow θ = 45°\end{aligned}$$

Hence, the angle of elevation is 45°.

Let initially aeroplane be at point B and after 10 seconds it is at point C.

Since, aeroplane is flying horizontally 1 km above the ground so, BE = CD = 1 km.

Considering right angled triangle △ABE,

$$\begin{aligned}\Rightarrow \text{tan 60°} = \frac{BE}{AE} \\[1em] \Rightarrow \sqrt{3} = \frac{1}{AE} \\[1em] \Rightarrow AE = \frac{1}{\sqrt{3}} \\[1em] \Rightarrow AE = 0.577 \text{ km}.\end{aligned}$$

From figure,

AD = AE + ED = (0.577 + ED) km.

Considering right angled triangle △ACD,

$$\begin{aligned}\Rightarrow \text{tan 30°} = \frac{CD}{AD} \\[1em] \Rightarrow \frac{1}{\sqrt{3}} = \frac{1}{0.577 + ED} \\[1em] \Rightarrow 0.577 + ED = \sqrt{3} \\[1em] \Rightarrow ED = 1.732 - 0.577 \\[1em] \Rightarrow ED = 1.155 \text{ km}.\end{aligned}$$

Aeroplane covers 1.155 km in 10 seconds.

Time = 10 seconds = $\frac{10}{60 \times 60} = \frac{10}{3600} = \frac{1}{360}$ hours.

Speed = $\frac{\text{Distance}}{\text{Time}} = \frac{1.155}{\frac{1}{360}} = 1.155 \times 360 =$ 415.66 km/h.

Hence, the speed of aeroplane is 415.66 km/h.

Let A be the man on the deck of the ship B and CE is the cliff.

AB = 16 m and angle of elevation of the top of cliff is 45° and angle of depression of base of cliff is 30°.

Let CE = h, AD = x, then
CD = h - 16, AD = BE = x.

Now in right angled triangle △CAD,

$$\begin{aligned}\text{tan 45°} = \frac{CD}{AD} \\[1em] 1 = \frac{h - 16}{x} \\[1em] x = h - 16 \text{ ......(i)}\end{aligned}$$

Again in right angled triangle △ADE,

$$\begin{aligned}\text{tan 30°} = \frac{DE}{AD} = \frac{16}{x} \\[1em] \frac{1}{\sqrt{3}} = \frac{16}{x} \\[1em] x = 16\sqrt{3} \\[1em] x = 27.71 \text{ m} \text{ ......(ii)}\end{aligned}$$

From (i) and (ii),

$$\begin{aligned}\Rightarrow h - 16 = 27.71 \\[1em] \Rightarrow h = 27.71 + 16 \\[1em] \Rightarrow h = 43.71 \text{ m}\end{aligned}$$

Hence, the distance of cliff from the ship is 27.71 m and height of cliff is 43.71 m.

Let XY be tree of h meters.

From figure,

Considering right angled triangle △XQY,

$$\begin{aligned}\Rightarrow \text{tan 45°} = \frac{XY}{YQ} \\[1em] \Rightarrow 1 = \frac{h}{YQ} \\[1em] \Rightarrow YQ = h \text{ ....(Eq 1)}\end{aligned}$$

Considering right angled triangle △XPY,

$$\begin{aligned}\Rightarrow \text{tan 30°} = \frac{XY}{PY} \\[1em] \Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{100 - YQ} \\[1em] \Rightarrow 100 - YQ = h\sqrt{3}\end{aligned}$$

Putting value of YQ from Eq 1 in above equation,

$$\begin{aligned}\Rightarrow 100 - h = 1.732h \\[1em] \Rightarrow 1.732h + h = 100 \\[1em] \Rightarrow 2.732h = 100 \\[1em] \Rightarrow h = \frac{100}{2.732} = 36.6 \text{ m}.\end{aligned}$$

Hence, the height of tree is 36.6 m.

Let P be the man standing on the deck of the ship which is 20 m above sea level and B is the bird.

Let bird be flying above h meters from deck of the ship.

∴ BC = h

Let the shadow be created at point R, the shadow will be created vertically opposite at same distance as bird is from ship.

∴ AR = AB = AC + BC = h + 20.

From figure,

In right angled triangle △PCB,

$$\begin{aligned}\Rightarrow \text{tan 30°} = \frac{BC}{PC} \\[1em] \Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{x} \\[1em] \Rightarrow x = h\sqrt{3} ......(i)\end{aligned}$$

In right angled triangle △PCR,

$$\begin{aligned}\Rightarrow \text{tan 60°} = \frac{CR}{CP} \\[1em] \Rightarrow \sqrt{3} = \frac{h + 40}{x} \\[1em] \Rightarrow \frac{h + 40}{h\sqrt{3}} = \sqrt{3} \quad \text{ [From (i)]}\\[1em] \Rightarrow h + 40 = \sqrt{3} \times \sqrt{3}h \\[1em] \Rightarrow h + 40 = 3h \\[1em] \Rightarrow 3h - h = 40 \\[1em] \Rightarrow 2h = 40 \\[1em] \Rightarrow h = 20.\end{aligned}$$

From sea level, height of bird (AB) = h + 20 = 20 + 20 = 40 m.

Hence, the height of bird from sea level is 40 m.

In △APB,

⇒ tan 42° = $\frac{AB}{AP}$

⇒ 0.9004 = $\frac{AB}{100}$

⇒ AB = 0.9004 × 100 = 90.04 m

In △APF,

⇒ tan 54° = $\frac{AF}{AP}$

⇒ 1.3764 = $\frac{AF}{100}$

⇒ AF = 1.3764 × 100 = 137.64 m

From figure,

BF = AF - AB = 137.64 - 90.04 = 47.60 ≈ 48 m.

Hence, height of flagstaff = 48 m.