Heights & Distances — Exercise 23

The angle of elevation of the top of a pole from a point on the level ground and 15 m away from the pole is 30°. Find the height of the pole.

Answer

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

BC = 15 m , ∠ACB = 30°

In triangle ABC,

We know that,

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

Hence, height of the pole is 8.66 m.

From the top of a cliff, 50 m high, the angle of depression of a buoy is 30°. Calculate to the nearest metre, the distance of the buoy from the foot of the cliff.

Answer

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,

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

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

A vertical pole is 12 m high and the length of its shadow is $12\sqrt{3}$ m. What is the angle of elevation of the sun?

Answer

Let AB be the height of pole = 12 m.

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

In triangle ABC,

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

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

A kite is flying with a thread 80 m long. If the thread is assumed stretched straight and makes an angle of 60° with the horizontal, find the height of the kite above the ground.

Answer

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,

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

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

The length of a string between a kite and a point on the ground is 85 m. If the string makes an angle θ with the level ground such that tan θ = $\Big(\dfrac{15}{8}\Big)$, how high is the kite?

Answer

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

Given,

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

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

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

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

In triangle ABC,

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

Hence, height of kite from ground = 75 m.

A vertical tower is 20 m high. A man standing at some distance from the tower knows that the cosine of the angle of elevation of the top of the tower is 0.53. How far is he standing from the foot of the tower?

Answer

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,

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

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

At a point on a level ground, the angle of elevation of the top of a tower is θ such that tan θ = $\Big(\dfrac{7}{12}\Big)$. On walking 64 m towards the tower, the angle of elevation is φ, where tan φ = $\Big(\dfrac{3}{4}\Big)$. Find the height of the tower.

Answer

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,

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

In right angled triangle ABC,

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

Hence, height of the tower is 168 m.

From two points A and B on the same side of a building, the angles of elevation of the top of the building are 30° and 60° respectively. If the height of the building is 10 m, find the distance between A and B, correct to two decimal places.

Answer

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,

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

In right angled triangle CBD,

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

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

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

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

The shadow of a vertical tower AB on level ground is increased by 10 m, when the altitude of the sun changes from 45° to 30°, as shown in the figure. Find the height of the tower and give your answer correct to $\Big(\dfrac{1}{10}\Big)$ of a metre.

Answer

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

$\Rightarrow \tan 45° = \dfrac{BA}{AD} \\[1em] \Rightarrow 1 = \dfrac{h}{x} \\[1em] \Rightarrow h = x$

From right angled △BCA, we get

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

Hence, the height of the tower is 13.7 m.

The angles of elevation of the top of a tower from two points on the ground at distances a metres and b metres from the base of the tower and in the same straight line with it are complementary. Prove that the height of the tower is $\sqrt{ab}$ metres.

Answer

Given,

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

From figure,

In △ABD,

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

Multiplying (1) by (2) we get,

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

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

A man in a boat rowing away from a lighthouse, 150 m high, takes 1.5 minutes to change the angle of elevation of the top of the lighthouse from 60° to 45°. Find the speed of the boat.

Answer

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,

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

In △ABC,

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

⇒ 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.

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

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

Two pillars of equal heights stand on either side of a roadway, which is 120 m wide. At a point on the road lying between the pillars, the elevations of the pillars are 60° and 30° respectively. Find the height of each pillar and the position of the point.

Answer

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,

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

In △CDE,

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

We know that,

⇒ BD = 120 m

⇒ BE + ED = 120 m

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

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

From equation (1),

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

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

The angle of elevation of the top Q of a vertical tower PQ from a point X on the ground is 60°. At a point Y, 40 m vertically above X, the angle of elevation is 45°. Find the height of tower PQ and the distance XQ.

Answer

Given,

XY = 40 m

⇒ PQ = h (let)

⇒ ∠PXQ = 60° and ∠RYQ = 45°

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

In △RYQ,

$\Rightarrow \tan 45° = \dfrac{RQ}{RY} \\[1em] \Rightarrow 1 = \dfrac{h - 40}{RY} \\[1em] \Rightarrow RY = h - 40.$

From figure,

PX = RY = h - 40

In △PXQ,

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

In △QPX,

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

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

A man 1.8 m tall stands at a distance of 3.6 m from a lamp post and casts a shadow of 5.4 m on the ground. Find the height of the lamp post.

Answer

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 :

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

Considering right angled △DCE, we get :

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

Comparing Eq 1 and Eq 2 we get,

$\Rightarrow \dfrac{x - 1.8}{3.6} = \dfrac{1}{3} \\[1em] \Rightarrow 3x - 5.4 = 3.6 \\[1em] \Rightarrow 3x = 5.4 + 3.6 \\[1em] \Rightarrow 3x = 9 \\[1em] \Rightarrow x = 3.$

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

In the adjoining figure, a man stands on the ground at a point A, which is on the same horizontal plane as B, the foot of the vertical pole BC. The height of the pole is 10 m. The man’s eye is 2 m above the ground. He observes the angle of elevation of C, the top of the pole, as x°, where tan x° = $\Big(\dfrac{2}{5}\Big)$. Calculate the distance AB in metres.

Answer

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,

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

From figure,

AB = DE = 20 m.

Hence, AB = 20 m.

From a window A, 10 m above the ground, the angle of elevation of the top C of a tower is x°, where tan x = $\Big(\dfrac{5}{2}\Big)$ and the angle of depression of the foot D of the tower is y°, where tan y = $\Big(\dfrac{1}{4}\Big)$.

See the figure given alongside. Calculate the height CD of the tower in metres.

Answer

⇒ AB = DE = 10 m.

In ∆AED,

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

In ∆AEC,

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

From figure,

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

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

An aeroplane at an altitude of 900 m finds that two ships are sailing towards it in the same direction. The angles of depression of the ships, as observed from the plane, are 60° and 30° respectively. Find the distance between the ships.

Answer

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,

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

In ∆ABD,

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

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

In the given figure, AB is a tower and two objects C and D are located on the ground on the same side of AB. When observed from the top B of the tower, their angles of depression are 45° and 60° respectively. Find the distance between the objects, if the height of the tower is 180 m.

Answer

Considering right angled △ABC, we get

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

Considering right angled △ADB, we get

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

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

Hence,the distance between two objects = 76.08 meters.

From the top of a church spire 96 m high, the angles of depression of two vehicles on a road, at the same level as the base of the spire and on the same side of it are x° and y°, where tan x° = $\Big(\dfrac{1}{4}\Big)$ and tan y° = $\Big(\dfrac{1}{7}\Big)$. Calculate the distance between the vehicles.

Answer

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

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

Considering right angled △ABC, we get

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

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

Hence, distance between the vehicles is 288 m.

Two men standing on the same side of a tower in a straight line with it, measure the angles of elevation of the top of the tower as 25° and 50° respectively. If the height of the tower is 70 m, find the distance between the two men.

Answer

From figure,

CD is the distance between two persons.

In △ABC, we get

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

In △ABD, we get

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

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

Hence, distance between the two men is 91.38 m.

In the given figure, AB represents a pole and CD represents a 60 m high tower, both of which are standing on the same horizontal plane. From the top of the tower, the angles of depression of the top and the foot of the pole are 30° and 60° respectively. Calculate:

(i) the horizontal distance between the pole and the tower,

(ii) the height of the pole.

Answer

(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 :

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

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

(ii) In △DEB,

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

Height of the pole AB = CE = 60 - DE

= 60 - x

= 60 - 20

= 40 m.

Hence, height of the pole = 40 m.

From a boat, 200 m away from a vertical cliff, the angles of elevation of the top and the foot of a vertical pillar at the edge of the cliff are 36° and 34° respectively. Find:

(i) the height of the cliff, and

(ii) the height of the pillar.

Answer

(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,

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

Hence, height of the cliff 134.9 m.

(ii) In △AOC,

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

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

Hence, height of the pillar 10.3 m.

A man on the top of a vertical observation tower observes a car moving at a uniform speed coming directly towards it. If it takes 10 minutes for the angle of depression to change from 30° to 45°, how soon after this will the car reach the observation tower? Give your answer to the nearest second.

Answer

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,

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

In △ABC,

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

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

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

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

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

The angle of elevation of a stationary cloud from a point 25 m above a lake is 30° and the angle of depression of its reflection in the lake is 60°. What is the height of the cloud above the lake-level?

Answer

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,

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

In △ABD,

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

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.

In the given diagram, AB is a vertical tower 100 m away from the foot of a 30 storied building CD. The angles of depression from the point C and E, (E being the mid-point of CD), are 35° and 14° respectively. (Use mathematical table for the required values rounded off correct to two places of decimals only.)

Find the height of the:

(a) tower AB

(b) building CD

Answer

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 = $\dfrac{CD}{2} = \dfrac{H}{2}$

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,

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

In ΔAEP,

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

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

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

⇒ 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.

The angle of elevation from a point P of the top of a tower QR, 50 m high, is 60° and that of the tower PT from a point Q is 30°. Find the height of the tower PT, correct to the nearest metre.

Answer

Let height of tower PT be h meters.

Considering right angled ΔPQR, we get

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

Now considering right angled ΔPQT, we get

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

On correcting to nearest meter, h = 17 m.

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

A man observes the angle of elevation of the top of the tower to be 45°. He walks towards it in a horizontal line through its base. On covering 20 m, the angle of elevation changes to 60°. Find the height of the tower correct to 2 significant figures.

Answer

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,

$\Rightarrow \tan 45° = \dfrac{\text{perpendicular}}{\text{base}} \\[1em] \Rightarrow 1 = \dfrac{QR}{PQ} \\[1em] \Rightarrow PQ = QR.$

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,

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

On correcting to 2 significant figures QR = 47.

Hence, the height of the tower is 47 m.

From the top of a cliff 60 m high, the angles of depression of two boats are 30° and 60° respectively. Find the distance between the boats, when the boats are:

(i) on the same side of the cliff,

(ii) on the opposite sides of the cliff.

Answer

(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

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

From right angled ΔPQR, we get

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

Distance between the boats,

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

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,

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

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

From the top of a hill the angles of depression of two consecutive kilometer stones, due east are found to be 30° and 45° respectively. Find the distance of the two stones from the foot of the hill.

Answer

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

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

From right angled ΔADB, we get

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

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.

An aeroplane at an altitude of 1500 m finds that two ships are sailing towards it in the same direction. The angles of depression as observed from the aeroplane are 45° and 30° respectively. Find the distance between the two ships.

Answer

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,

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

In right angled ΔOAP,

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

AQ = y = 1500 m

PQ = x = 1095 m

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

An aeroplane at an altitude of 250 m observes the angle of depression of two boats on the opposite banks of a river to be 45° and 60° respectively. Find the width of the river. Write the answer to the nearest whole number.

Answer

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

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

Considering right angled ΔABD, we get

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

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.

From the top of a tower, 100 m high, a man observes the angles of depression of two ships A and B, on opposite sides of the tower as 45° and 38° respectively. If the foot of the tower and the ships are in the same horizontal line, find the distance between the two ships A and B.

Answer

Let CD be the tower.

From figure,

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

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

In ΔACD,

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

In ΔBCD,

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

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.

The horizontal distance between two towers is 120 m. The angle of elevation of the top and angle of depression of the bottom of the first tower as observed from the second tower is 30° and 24° respectively. Find the heights of the two towers. Give your answer correct to 3 significant figures.

Answer

From figure,

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

From figure,

AC = ED = 120 m.

In ΔBED,

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

In ΔEDA,

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

⇒ 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.

From the top of a cliff, the angle of depression of the top and bottom of a tower are observed to be 45° and 60° respectively. If the height of the tower is 20 m, find:

(i) the height of the cliff,

(ii) the distance between the cliff and the tower.

Answer

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° = $\dfrac{AE}{EC}$

⇒ 1 = $\dfrac{AE}{x}$

⇒ AE = x meters

In ΔABD,

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

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.

Two lamp posts AB and CD, each of height 100 m, are on either side of the road. P is a point on the road between the two lamp posts. The angles of elevation of the top of the lamp post from the point P are 60° and 40°. Find the distances PB and PD.

Answer

Given,

AB = 100 m

CD = 100 m

∠APB = 40°

∠CPD = 60°

Let PB = x and PD = y.

In triangle ABP,

In triangle ABP,

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

PB = x = 119.17 m

In triangle CDP,

We know that,

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

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

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