Question

From the top of a tower 100 m high the angle of  elevation of the top of a hill is found to be 30°. From the bottom of this tower, the angle of elevation of the top of the hill is found to be 45°, then find the height of the hill.

Answer 1

Let AB is a tower of height 100 m and height of bill CD is h meter. The angle of elevation of hill from top of tower and bottom of tower is 30° and 45° respectively

i.e., ∠CAE = 30° and ∠CBD = 45°

AB = ED = 100 m

CE = CD – ED

= (h – 100)m

BD = x m (Let)

From right angled ∆CEA,

Again From right angled ∆CDB,

tan 45° = CD/BD

l = h/x

h = x m ….(ii)

Put the value of x from eq. (i) in eq. (ii),

= 50(3 + 1.732)

= 50 × 4.73

= 236.5 m

Hence, height of hill = 236.5 m