# From the top of a 7 m high building, the angle of elevation of the top of a cable tower is 60° and the angle of depression of its foot is 45°. Determine the height of the tower

**Solution:**

Let the height of the tower be CE and the height of the building be AB. The angle of elevation from the top E of the tower to the top A of the building is 60° and the angle of depression from the bottom C of the tower to the top A of the building is 45°.

Draw AD || BC.

Then, ∠DAC = ∠ACB = 45° (alternate interior angles)

In ΔABC,

tan 45° = AB / BC

1 = 7/BC

BC = 7

ABCD is a rectangle,

Therefore, BC = AD = 7 and AB = CD = 7

In ΔADE,

tan 60° = ED/AD

√3 = ED/7

ED = 7√3

Height of tower = CE = ED + CD

= 7√3 + 7

= 7 (√3 + 1)

Height of the tower = 7 (√3 + 1) m.

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

**Video Solution:**

## From the top of a 7 m high building, the angle of elevation of the top of a cable tower is 60° and the angle of depression of its foot is 45°. Determine the height of the tower

Maths NCERT Solutions Class 10 Chapter 9 Exercise 9.1 Question 12

**Summary:**

If from the top of a 7 m high building, the angle of elevation of the top of a cable tower is 60° and the angle of depression of its foot is 45°, then the height of the tower is 7(1 + √3) m.

**☛ Related Questions:**

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