Heights & Distances — Exercise 22(A)

The measure of x is :

1. $15\sqrt{3}$ cm

2. 15 cm

3. $5\sqrt{3}$ cm

4. 5 cm

Answer

We know that,

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

From figure,

In △ ABC,

⇒ tan 60° = $\dfrac{AB}{BC}$

⇒ $\sqrt{3} = \dfrac{15}{x}$

⇒ x = $\dfrac{15}{\sqrt{3}}$

Rationalizing the denominator,

$\Rightarrow x = \dfrac{15}{\sqrt{3}} \times \dfrac{\sqrt{3}}{\sqrt{3}} \\[1em] \Rightarrow x = \dfrac{15\sqrt{3}}{3} \\[1em] \Rightarrow x = 5\sqrt{3}.$

Hence, Option 3 is the correct option.

The value of x is :

1. 40 m

2. 30 m

3. 20 m

4. $40\sqrt{3}$ m

Answer

We know that,

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

From figure,

In △ ABC,

⇒ cos 60° = $\dfrac{BC}{AB}$

⇒ $\dfrac{1}{2} = \dfrac{x}{40}$

⇒ x = $\dfrac{40}{2}$ = 20 m.

Hence, Option 3 is the correct option.

In the given figure, AB = BD, BC = 20 cm and ∠D = 45°, the length of AC is :

1. 54.64 cm

2. 48.28 cm

3. 40 cm

4. 14.64 cm

Answer

We know that,

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

From figure,

In △ BCD,

⇒ sin 45° = $\dfrac{BC}{BD}$

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

⇒ 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.

The measure of x correct to the nearest metre is :

1. 70 m

2. 35 m

3. 61 m

4. 140 m

Answer

We know that,

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

From figure,

In △ ABC,

⇒ cos 30° = $\dfrac{BC}{AB}$

⇒ $\dfrac{\sqrt{3}}{2} = \dfrac{x}{70}$

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

Hence, Option 3 is the correct option.

The length of DC is :

1. 10 m

2. $10(\sqrt{3} + 1)$ m

3. 20 m

4. $10(\sqrt{3} - 1)$ m

Answer

We know that,

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

From figure,

In △ ABC,

⇒ tan 45° = $\dfrac{AB}{BC}$

⇒ $1 = \dfrac{10}{BC}$

⇒ BC = 10 m.

In △ ABD,

⇒ tan 30° = $\dfrac{AB}{BD}$

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

⇒ 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.

The height of a tree is $\sqrt{3}$ times the length of its shadow. Find the angle of elevation of the sun.

Answer

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,

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

Hence, angle of elevation of sun is 60°.

The angle of elevation of the top of a tower from a point on the ground and at a distance of 160 m from its foot, is found to be 60°. Find the height of the tower.

Answer

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,

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

Hence, the height of the tower is 277.12 m.

A ladder is placed along a wall such that its upper end is resting against a vertical wall. The foot of the ladder is 2.4 m from the wall and the ladder is making an angle of 68° with the ground. Find the height, up to which the ladder reaches.

Answer

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,

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

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

Two persons are standing on the opposite sides of a tower. They observe the angles of elevation of the top of the tower to be 30° and 38° respectively. Find the distance between them, if the height of the tower is 50 m.

Answer

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,

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

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

In △PBQ,

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

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

Hence, the distance between two persons = 150.6 meters.

A boy, 1.6 m tall, is 20 m away from a tower and observes the angle of elevation of the top of the tower to be (i) 45° (ii) 60°. Find the height of the tower in each case.

Answer

(i) From figure,

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

In △ADE,

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

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,

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

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.

The upper part of a tree, broken over by the wind, makes an angle of 45° with the ground; and the distance from the root to the point where the top of the tree touches the ground, is 15 m. What was the height of the tree before it was broken?

Answer

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,

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

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.

The angle of elevation of the top of an unfinished tower from a point at a distance of 80 m from its base is 30°. How much higher must the tower be raised so that its angle of elevation at the same point may be 60° ?

Answer

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

From figure,

In △ABC,

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

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

From figure,

In △DBC,

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

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

Hence, the tower must be raised by 92.37 meters.

At a particular time, when the sun's altitude is 30°, the length of the shadow of a vertical tower is 45 m. Calculate :

(i) the height of the tower,

(ii) the length of the shadow of the same tower, when the sun's altitude is :

(a) 45° (b) 60°.

Answer

(i) Let AB be the tower.

From figure,

In △ABC,

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

Hence, the height of tower = 25.98 meters.

(ii) When sun's altitude is

(a) 45°

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

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

(b) 60°

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

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

Two vertical poles are on either side of a road. A 30 m long ladder is placed between the two poles. When the ladder rests against one pole, it makes angle 32° 24' with the pole and when it is turned to rest against another pole, it makes angle 32° 24' with the road. Calculate the width of the road.

Answer

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,

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

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,

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

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

Hence, width of road = 41.4 meters.

Two climbers are at points A and B on a vertical cliff face. To an observer C, 40 m from the foot of the cliff, on the level ground, A is at an elevation of 48° and B of 57°. What is the distance between the climbers?

Answer

Let P be the foot of cliff.

From figure,

In △BCP,

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

In △ACP,

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

Distance between climbers (BA) = BP - AP

= 61.57 - 44.40

= 17.17 meters.

Hence, distance between two climbers = 17.17 meters.

A man stands 9 m away from a flag-pole. He observes that angle of elevation of the top of the pole is 28° and the angle of depression of the bottom of the pole is 13°. Calculate the height of pole.

Answer

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

From figure,

In △ADB,

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

In △BDC,

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

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

Hence, the height of pole = 6.867 meters.

From the top of a cliff 92 m high, the angle of depression of a buoy is 20°. Calculate, to the nearest metre, the distance of the buoy from the foot of the cliff.

Answer

Let AB be the cliff and C be the buoy.

Given,

AB = 92 m

From figure,

In △ACB,

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

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