Question

From a cliff $150$m above the shore line the angle of depression of a ship is ${19}^{\circ }{30}^{\prime }$.Find the distance from the ship to a point on the shore directly below the observer.

Answer 1

Let $OB$ be the cliff of height $150m$ and $A$ be the position of the ship.

The angle of depression of the ship is ${19}^{\circ }{30}^{\prime }$

Then $\angle BAC={19}^{\circ }{30}^{\prime }$

In right angled triangle $△AOB$

$\tan {19}^{\circ }{30}^{\prime }=\frac{OB}{OA}=\frac{150}{OA}$

$\Rightarrow OA=\frac{150}{\tan {19}^{\circ }{30}^{\prime }}$

$\Rightarrow OA=150\cot {19}^{\circ }{30}^{\prime }$

$\Rightarrow OA=150\tan ({90}^{\circ }-{19}^{\circ }{30}^{\prime })$

$\Rightarrow OA=150\tan {70}^{\circ }{30}^{\prime }$

$\Rightarrow OA=150\times 2.8239=423.59m$