Two ships are sailing in the sea on the two sides of a lighthouse. The angles of elevation of the top of the lighthouse as observed from the two ships are 30° and 45° respectively. If the lighthouse is 100 m high, the distance between the two ships is :_________?

A. 173 m
B. 200 m
C. 273 m
D. 300 m

Let AB be the lighthouse and C and D be the
positions of the ships. Then,
AB = 100 m, ∠ACB = 300 and ∠ADB = 45°.
AB/AC = tan 30° = 1/√3
AC = AB X √3 = 100√3 m.
AB/AD = tan 45° = 1 ⇒ AD = AB = 100 m.
CD = (AC + AD) = (100√3 + 100) m
= 100 (√3 +1) m = (100 X 2.73) m = 273 m.

Correct Answer: 273 m ✔

Last Updated: February 07, 2020