Q) From the top of a 45 m high light house, the angles of depression of two ships, on the opposite side of it, are observed to be 30° and 60°. If the line joining the ships passes through the foot of the light house, find the distance between the ships. (Use √3 = 1.73)

Ans: Let’s make a diagram for the given question:

From the  top of  a  45 m high  light house,  the  angles of depression of  two ships

Let’s start from Δ CPD, tan ∠CPD = tan 30° = \frac{CD}{PD}

\frac{1}{\sqrt 3} = \frac{45}{PD}

∴ PD = 45 √3 m

Next, we take in Δ CQD, tan ∠CQD = tan 60° = \frac{CD}{QD}

∴ √3 = \frac{45}{QD}

∴ QD = \frac{45}{\sqrt 3} = 15 √3 m

Next, we can see from the diagram, that PQ = PD + QD

∴ PQ = 45√3 + 15 √3 = 60 √3 m

∴ PQ = 60 x 1.73 = 103.8 m

Therefore, distance between two ships is 103.8 m.

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