Question

27. There are two temples one on each bank of river just opposite to each other. Once
temple is 50m high. From the top of this temple. The angles of depression of the top
and foot of other temple are 30 and 60' respectively. Find width of the river and the
height of other temple.​

Answer 1

Solution: Let's imagine you are standing at top of tower AB. Standing on Point A, you find angle of depression of foot 60° & top 30° of other tower named as DC.

Given: Length of AB = 50m.

→ tan60° = AB/BC

→ √3 = 50/BC

→ BC = 50/√3

Similarly, tan 30° = AE/DE

→ 1/√3 = AE/BC [DE = BC]

→ 1/√3 = AE/(50/√3)

→ 1/√3 × 50/√3 = AE

→ 50/3 = AE

→ 16.67 m = AE

Height of other temple = CD = BE

→ CD = AB - AE → CD = 50 - 16.67 m

→ CD = 33.33 m

Width of river = BC = 50/√3 m

→ BC = 50/√3 × √3/√3

→ BC = 50√3/3

→ BC = 28.87 m