As observed from the top of a lighthouse,the | vedclass

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(C) Let the height of the lighthouse be $h$. Let the distance of ship $Q$ from the foot of the lighthouse be $x$ and the distance of ship $P$ be $y$.F

Vedclass · Accepted Answer

The distance of $P$ is more than $Q$.

Vedclass · Answer

(C) Let the height of the lighthouse be $h$. Let the distance of ship $Q$ from the foot of the lighthouse be $x$ and the distance of ship $P$ be $y$.From the geometry of the problem,the angle of elevation of the top of the lighthouse from the ships is equal to the angle of depression.For ship $Q$: $	an(50^{\circ}) = \frac{h}{x} \implies x = \frac{h}{	an(50^{\circ})}$.For ship $P$: $	an(35^{\circ}) = \frac{h}{y} \implies y = \frac{h}{	an(35^{\circ})}$.Since $35^{\circ} Therefore,$\frac{h}{	an(35^{\circ})} > \frac{h}{	an(50^{\circ})}$,which means $y > x$.Thus,the distance of ship $P$ from the lighthouse is greater than the distance of ship $Q$.

As observed from the top of a lighthouse,the angles of depression of two ships $P$ and $Q$ anchored in the sea on the same side are found to be $35^{\circ}$ and $50^{\circ}$ respectively. Then,from the lighthouse:

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