ICSE Class 10 Maths Chapter 22 Heights and Distances

Heights and Distances

Angles of Elevation and Depression

Let AB be a tower (or pillar or minar, etc.) standing on a level ground and a man, standing at any point C on the level ground, is viewing an object at A.

The line CA, joining his eye to the object, is called the line of sight.

The angle, which the line of sight makes with the horizontal is called the angle of elevation.

In the given figure; angle ACB is the angle of elevation.

Similarly, if the man is at A and is viewing an object C on the level ground, then the angle, which the line of sight (AC) makes with horizontal, is called the angle of depression.

In the given figure; angle DAC is angle of depression.

Angle of elevation of point A as seen from point C is equal to the angle of depression of point C as seen from point A. i.e., ∠ACB = ∠DAC

Problem 1

The length of the shadow of a vertical tower is \(\sqrt{3}\) times its height. Find the angle of elevation of the sun.

Solution

Let the height of the tower be x m

Length of its shadow = \(\sqrt{3} x\) m

If θ is the angle of elevation of the sun, then

\(\tan \theta = \frac{x}{\sqrt{3}x} = \frac{1}{\sqrt{3}} = \tan 30°\)

Therefore θ = 30°

Problem 2

The angle of elevation of the top of a tower at a distance of 120 m from its foot on a horizontal plane is found to be 30°. Find the height of the tower.

Solution

Let AB be the tower and C be the point of observation

BC = 120 m

Since, angle of elevation is 30°

∠ACB = 30°

In △ ABC, \(\frac{AB}{120} = \tan 30°\)

\(AB = 120 \times \frac{1}{\sqrt{3}} = 69.28\) m

Problem 3

A guard observes an enemy boat, from an observation tower at a height of 180 m above sea level, to be at an angle of depression of 29°.

(i) Calculate, to the nearest metre, the distance of the boat from the foot of the observation tower.

(ii) After some time, it is observed that the boat is 200 m from the foot of the observation tower. Calculate the new angle of depression.

Solution

(i) Let P be the boat and AB be the observation tower. In right triangle ABP,

\(\tan 29° = \frac{AB}{PB}\)

\(0.5543 = \frac{180}{PB}\)

\(PB = \frac{180}{0.5543} = 325\) m (App.)

Alternative Method:

In △ ABP, ∠APB + ∠PAB = 90° => ∠PAB = 90° - 29° = 61°

\(\frac{BP}{AB} = \tan 61°\)

\(BP = 180 \times \tan 61° = 180 \times 1.804 = 325\) m (App.)

(ii) Let Q be the new position of the boat and θ be the new angle of depression

\(\tan \theta = \frac{AB}{QB} = \frac{180}{200} = 0.9000 = \tan 41° 59'\)

θ = 41°59'

Problem 4

Two people 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 people.

Solution

According to the given statement, the figure will be as shown alongside in which PQ is the tower so PQ = 70 m; and A and B be the positions of two people, such that ∠PAQ = 25° and ∠PBQ = 50°

In △ PAQ, \(\tan 25° = \frac{PQ}{AQ}\)

\(0.4663 = \frac{70}{AQ}\) i.e. \(AQ = \frac{70}{0.4663}\) m = 150.118 m

In △ PBQ, \(\tan 50° = \frac{PQ}{BQ}\)

\(1.1918 = \frac{70}{BQ}\) i.e. \(BQ = \frac{70}{1.1918}\) m = 58.735 m

The distance between the two people = AB = AQ - BQ = (150.118 - 58.735) m = 91.38 m

Alternative method:

In △ PAQ, ∠A = 25° => ∠APQ = 90° - 25° = 65°

\(\tan 65° = \frac{AQ}{PQ}\)

i.e. \(2.1445 = \frac{AQ}{70}\) => AQ = 2.1445 × 70 m = 150.115 m

In △ PBQ, ∠PBQ = 50° => ∠BPQ = 90° - 50° = 40°

\(\tan 40° = \frac{BQ}{PQ}\)

i.e. \(0.8391 = \frac{BQ}{70}\) => BQ = 0.8391 × 70 m = 58.737 m

The distance between the two people = AB = AQ - BQ = (150.115 - 58.737) m = 91.38 m

Teacher's Note

These angle measurement concepts are fundamental in architecture and construction, where builders must determine heights of structures and distances using ground-level observations.

Exercise 22(A)

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

2. 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.

3. 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, upto which the ladder reaches.

4. 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.

5. A kite is attached to a string. Find the length of the string, when the height of the kite is 60 m and the string makes an angle 30° with the ground.

6. 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.

7. 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?

8. The angle of elevation of the top of an unfinished tower at a point distance 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°?

9. 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°.

10. 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.

11. 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?

12. 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 the pole.

13. 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.

Problem 5

The length of the shadow of a vertical tower on level ground increases by 10 m, when the altitude of the sun changes from 45° to 30°. Calculate the height of the tower, correct to two decimal places.

Solution

Let AB be the tower. BC be its shadow when the sun's altitude is 45°, i.e. ∠ACB = 45°.

When the sun's altitude changes to 30°, the length of the shadow is BD and ∠ADB = 30°

Clearly, DC = 10 m.

In triangle ABC,

\(\tan 45° = \frac{AB}{BC}\) => \(1 = \frac{AB}{BC}\) => BC = AB

In triangle ABD,

\(\tan 30° = \frac{AB}{BD}\) => \(\frac{1}{\sqrt{3}} = \frac{AB}{BD}\) => BD = AB \(\sqrt{3}\)

Since BD - BC = DC

AB \(\sqrt{3}\) - AB = 10

AB (\(\sqrt{3}\) - 1) = 10

\(AB = \frac{10}{\sqrt{3} - 1} = \frac{10}{1.732 - 1} = \frac{10}{0.732} = 13.66\) m

OR,

\(AB = \frac{10}{\sqrt{3} - 1} = \frac{10}{\sqrt{3} - 1} \times \frac{\sqrt{3} + 1}{\sqrt{3} + 1} = \frac{10(1.732 + 1)}{3 - 1} = 5 \times 2.732 = 13.66\) m

This is a preview of the first 3 pages. To get the complete book, click below.