ICSE Class 10 Maths Chapter 26 Heights and Distances

Chapter 26

Heights And Distances

Points To Remember

1. Line Of Sight. When the eye of a person at a point O looks at an object P, then the line OP is called the line of sight.

Angle Of Elevation. Suppose a man from a point O, looks up at an object P, placed above the level of his eye. Then, the angle which the line of sight makes with the horizontal through O, is called the angle of elevation of P as seen from O.

Let OA be a horizontal line. Let a man at O on the level ground be looking up towards an object P, say an aeroplane or the top of a tower or the flag at the top of a building.

Then, angle AOP is the angle of elevation of P as seen from O. (see fig.)

3. Angle Of Depression. Suppose a man from a point O, looks down at an object P, placed below the level of his eye. Then, the angle which the line of sight makes with the horizontal through O, is called the angle of depression of P as seen from O.

Let AO be a horizontal line. Let a man at O, on the top of a tower be looking down towards an object P, say a ship in the sea.

Then, angle AOP is the angle of depression of P as seen from O.

An Important Remark :

Angle of depression of P as seen from O = Angle of elevation of O as seen from P.

Therefore angle AOP = angle BPO (Alternate angles, as AO || PB)

Exercise 26

Q.1. The angle of elevation of the top of a pole from a point on the level ground, 15 m away from the foot of the pole is 30°. Find the height of the pole. [Take square root of 3 = 1.732]

Sol. Let P is the point and AB is the pole. Such that angle APB = 30°

PB = 15 m

Let AB = height of the pole = h m

tan theta = Perpendicular / Base = AB / PB

Therefore tan 30° = h / 15 implies h = 15 times tan 30°

implies h = 15 times 1 / square root of 3

= (15 times square root of 3) / (square root of 3 times square root of 3) = (15 square root of 3) / 3 = 5 square root of 3 m

= 5 times 1.732 = 8.660 = 8.66 m Ans.

Q.2. (i) From a point on the level ground, at a distance of 50 m from the foot of a tower, the angle of elevation of the top of the tower is 20°. Find the height of the tower.

(ii) 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. (2005)

Sol. Let P be the point on the ground and QR is the tower such that angle QPR = 20°, PR = 50 m and QR = h

Perpendicular / low, tan theta = Perpendicular / Base

tan 20° = QR / PR

implies tan 20° = h / 50 implies 0.3640 = h / 50

implies h = 50 times 0.3640 = 18.2000 = 18.2 m

Hence, height of the tower = 18.2 m Ans.

(ii) Let AB be cliff whose height is 92 m and C is buoy making depression angle of 20°.

angle ACB = 20°

Let, CB = x m.

In right triangle ABC,

cot theta = BC / AB implies cot 20° = x / 92

implies x = 92 cot 20°

= 92 times 2.7475 m

= 252.77 m

Distance of buoy from foot of hill = 252.77m

Q.3. From a point A on level ground, the angle of elevation of the top of a tower is 29°. If the tower is 25 m high, find the distance of A from the foot of the tower, to the nearest metre..

Sol. Let A be the point on the level ground and angle of elevation is 29°. and let PQ is the tower such that PQ = 25 m.

Now, tan theta = Perpendicular / Base

implies tan 29° = PQ / AP = 25 / x implies 0.5543 = 25 / x

implies x = 25 / 0.5543 implies x = 45.10

Hence, distance from the foot of the tower = 45 m. Ans.

Q.4. A vertical pole is 12 m high and the length of its shadow is 12 square root of 3 m. What is the angle of elevation of the sun ?

Sol. Let AB be the pole and BC be its shadow on the ground. Let angle of elevation be theta, then AB = 12 m, BC = 12 square root of 3 m and angle ACB = theta.

Therefore tan theta = Perpendicular / Base

implies tan theta = AB / BC = 12 / (12 square root of 3)

= 1 / square root of 3 = tan 30°

Therefore theta = 30° Ans.

Q.5. A kite is flying with a thread 150 m long. If the thread is assumed stretched straight and makes an angle of 55° with the horizontal, find the height of the kite above the ground.

Sol. Let A be the kite and AB is its height and AC be the thread which makes an angle of 55° with the ground (horizontal), then

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