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# The angle of elevation of the top of a building from the foot of the tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 50 m high, find the height of the building

**Solution:**

Let the height of the tower be AB and the height of the building be CD.

The angle of elevation of the top of building D from the foot of tower B is 30° and the angle of elevation of the top of tower A from the foot of building C is 60°.

Distance between the foot of the tower and the building is BC.

Trigonometric ratio involving sides AB, CD, BC and angles ∠B and ∠C is tan θ.

In ΔABC,

tan 60° = AB/BC

√3 = 50/BC

BC = 50/√3 ....(i)

In ΔBCD,

tan 30° = CD / BC

1/√3 = CD / BC

1/√3 = CD / 50/√3 [from (i)]

CD = 1/√3 × 50/√3

CD = 50/3

Height of the building CD = 50/3 m.

**☛ Check: **NCERT Solutions for Class 10 Maths Chapter 9

**Video Solution:**

## The angle of elevation of the top of a building from the foot of the tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 50 m high, find the height of the building

Maths NCERT Solutions Class 10 Chapter 9 Exercise 9.1 Question 9

**Summary:**

If the angle of elevation of the top of a building from the foot of the tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60° and if the tower is 50 m high, then the height of the building is 50/3 m.

**☛ Related Questions:**

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