# From a point on the ground, the angles of elevation of the bottom and the top of a transmission tower fixed at the top of a 20 m high building are 45° and 60° respectively. Find the height of the tower

**Solution:**

Let the height of the building is BC, the height of the transmission tower which is fixed at the top of the building be AB.

D is the point on the ground from where the angles of elevation of the bottom B and the top A of the transmission tower AB are 45° and 60° respectively.

The distance of the point of observation D from the base of the building C is CD.

Combined height of the building and tower = AC = AB + BC

Trigonometric ratio involving sides AC, BC, CD, and ∠D (45° and 60°) is tan θ.

In ΔBCD,

tan 45° = BC/CD

1 = 20/CD

CD = 20

In ΔACD,

tan 60° = AC/CD

√3 = AC/20

AC = 20√3

Height of the tower, AB = AC - BC

AB = 20√3 - 20 m

= 20 (√3 - 1) m

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

**Video Solution:**

## From a point on the ground, the angles of elevation of the bottom and the top of a transmission tower fixed at the top of a 20 m high building are 45° and 60° respectively. Find the height of the tower.

Maths NCERT Solutions Class 10 Chapter 9 Exercise 9.1 Question 7

**Summary:**

If the angles of elevation of the bottom and the top of a transmission tower fixed at the top of a 20 m high building are 45° to 60° respectively, then the height of the tower is 20(√3 - 1) m.

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