**Q) **From the top of a 7m high building, the angle of elevation of the top of a cable tower is 60 and the angle of depression of its foot is 30. Determine the height of the tower.

**Ans: **

Let’s draw a diagram with cable tower AB and Building PQ of 7m height. Angle of elevation to top of tower is given as 60° and angle of depression to tower’s foot is given as 30°. Let’s consider H be the height of cable tower and D be the distance between the tower and the building.

By interior angle property, we can say that:

∠PBQ = ∠CPB = 30°

Let’s start from Δ PQB, tan 30° =

D = 7√3 m

Now in Δ APC, tan 60° =

√3 =

H – 7 = (√3)(7√3)

H – 7 = 21

H = 28 m

**Therefore, the height of the cable tower is 28 m****.**