Question

From the top of a light house, the angle of depression of a boat is 15°. If the light house is 60 m high and its base is at sea level, the distance of the boat from the light house is

(a) \(\bigg(\frac{\sqrt3-1}{\sqrt3+1}\bigg)\)60 m

(b) \(\bigg(\frac{\sqrt3+1}{\sqrt3-1}\bigg)\)60 m

(c) \(\bigg(\frac{\sqrt3-1}{\sqrt3+1}\bigg)^2\)60 m

(d) \(\bigg(\frac{\sqrt3+1}{\sqrt3-1}\bigg)\)60 m

Answer 1

(b) \(\bigg(\frac{\sqrt3+1}{\sqrt3-1}\bigg)\)60 m

Let PQ be the lighthouse and R the position of the boat. 

Given, PQ = 60 m 

Angle of depression of boat from point P = 15° 

∴ ∠APR = 15° ⇒ ∠PRQ = 15°

Distance of boat from the lighthouse = RQ 

In rt. ΔPRQ, \(\frac{PQ}{RQ}\) = tan 15°

⇒ \(\frac{60}{RQ}\) = tan 15° ⇒ \(\frac{60}{RQ}\) = \(\frac{\sqrt3-1}{\sqrt3+1}\)

\(\bigg[\)tan 15° = tan (45° - 30°) = \(\frac{\text{tan}\,45°-\text{tan}\,30°}{1+\text{tan}\,45°.\text{tan}\,30°}\) = \(\frac{1-\frac{1}{\sqrt3}}{1+1.\frac{1}{\sqrt3}}\) = \(\frac{\sqrt3-1}{\sqrt3+1}\)\(\bigg]\)

⇒ RQ = \(\bigg(\frac{\sqrt3+1}{\sqrt3-1}\bigg)\). 60 meters.