Question

A person is standing on the deck of a ship which is 10 m above the water level. He observes that the angle of elevation of the top of a cliff is ${45}^{\circ }$ and the angle of depression of its base is ${30}^{\circ }$. Claculate the distance of the cliff from the ship and the height of the cliff. $(\sqrt{3}=1.732)$

Answer 1

$tan45=\frac{x}{d}andtan30=\frac{y}{d}$.
$y=10.\to d=\frac{10}{tan30}=10\times \sqrt{3}=17.32m$
$x=dtan45=d=17.32$.
Height of cliff = x + y = 17.32 + 10 = 27. 32 m.