Heights and Distances — Practice

Question 1

The angle of elevation of the top of a tower from a point 30 m from its base is 45°. The height of the tower is:
(a) 15 m (b) 30 m (c) 30√3 m (d) 60 m

Accepted Answer

(b) 30 m.

h = 30 × tan45° = 30 × 1 = 30 m (height equals base at 45°).

Question 2

The top of a tower of height 30√3 m subtends an angle of 60° at a point on the ground. The distance of the point from the base is:
(a) 30 m (b) 30√3 m (c) 45 m (d) 90 m

Accepted Answer

(a) 30 m.

base = h/tan60° = 30√3/√3 = 30 m.

Question 3

The length of the shadow of a tower is √3 times its height. The sun's angle of elevation is:
(a) 30° (b) 45° (c) 60° (d) 90°

Accepted Answer

(a) 30°.

tan θ = height/shadow = h/(√3 h) = 1/√3 → θ = 30°.

Question 4

As an observer walks 40 m toward a tower, the angle of elevation of its top changes from 30° to 60°. The height of the tower is:
(a) 20 m (b) 20√3 m (c) 40 m (d) 40√3 m

Accepted Answer

(b) 20√3 m.

Difference of bases = h√3 − h/√3 = 2h/√3 = 40 → h = 20√3 m (≈ 34.6).

Question 5

From the top of a 60 m high tower, the angle of depression of a car on the ground is 30°. The distance of the car from the base of the tower is:
(a) 60 m (b) 60√3 m (c) 30√3 m (d) 120 m

Accepted Answer

(b) 60√3 m.

distance = h/tan30° = 60 × √3 = 60√3 m (≈ 103.9).

Question 6

The angle of elevation of the top of a tower at a point 100 m from its base is 30°. The height of the tower is:
(a) 100 m (b) 100√3 m (c) 100/√3 m (d) 50 m

Accepted Answer

(c) 100/√3 m.

h = 100 × tan30° = 100/√3 m (≈ 57.7).

Question 7

From the top of a 100 m cliff, the angle of depression of a boat changes from 45° to 30° as the boat moves away. The distance travelled by the boat is:
(a) 100 m (b) 100√3 m (c) 100(√3 − 1) m (d) 100(√3 + 1) m

Accepted Answer

(c) 100(√3 − 1) m.

At 45° distance = 100; at 30° distance = 100√3. Boat moved 100√3 − 100 = 100(√3 − 1) m (≈ 73.2).

Question 8

Two poles of equal height stand on opposite sides of an 80 m wide road. From a point between them the angles of elevation of their tops are 60° and 30°. The height of each pole is:
(a) 20 m (b) 20√3 m (c) 40 m (d) 40√3 m

Accepted Answer

(b) 20√3 m.

Let x be the distance to the 60° pole: x√3 = (80 − x)/√3 → 3x = 80 − x → x = 20. Height = 20 tan60° = 20√3 m.

Practice — Heights and Distances

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