Question

From the top of a cliff of $92m$ height, the angle of depression of a buoy is ${20}^{\circ }$. Calculate to the nearest meter, the distance of the buoy from the foot of the cliff. {Use $\tan {20}^o=0.36$}

Answer 1

Let AB be the height of the Cliff and C be the buoy.

In triangle ABC,

$\begin{array}{c}\tan {20}^{\circ }=\frac{AB}{BC}=\frac{92}{x}\\ \\ \Rightarrow 0.36=\frac{92}{x}\\ \\ x=255.55m\end{array}$

Therefore, the distance of buoy from foot $=$ 255.55 m