Solution:

Consider TR as the tower

PL as the pole on the same level

Ground PL = 20 m

From the point P construct PQ parallel to LR

∠TPQ = 60⁰ and ∠QPR = 30⁰

Here

∠PRL = ∠QPR = 30⁰ which are the alternate angles

Take LR = x and TR = h

TQ = TR – QR = (h – 20) m

In right triangle PRL

tan θ = PL/LR

Substituting the values

tan 30⁰ = 20/x

So we get

1/√3 = 20/x

x = 20√3 m

In right triangle PQT

tan 60⁰ = TQ/PQ

Substituting the values

√3 = (h – 20)/ x

√3 = (h – 20)/ 20√3

By cross multiplication

20√3 × √3 = h – 20

20 × 3 = h – 20

h = 60 + 20 = 80 m

Hence, the height of the tower is 80 m.