Question

As observed from the top of a watch tower, 200 m above sea level, the angle of depression of a boat, sailing directly towards it, changes from ${30}^{\circ }$ to ${45}^{\circ }$. Determine the distance travelled by boat during the period of observation. (Assume $\surd$3 = 1.732)

• A. 146.4 m
• B. 130.8 m
• C. 100 m
• D. 50 m

Answer 1

The correct option is A

146.4 m

Let A and B be the two positions of the boat. Let `d' be the distance travelled by boat during the period of observation i.e. AB = `d' meters.

Let the observer be at O, the top of the watch tower PO. It is given that PO = 200 m and the angles of depression from O of A and B are ${30}^{\circ }$ and ${45}^{\circ }$ respectively.

$\therefore$ $\angle$OAP = 30${}^0$ and $\angle$OBP = 45${}^0$

In $\Delta$ OPB, we have

tan 45${}^0$ = $\frac{OP}{BP}$

$\Rightarrow 1=\frac{200}{BP}$

$\Rightarrow$ BP = 200 m In $\Delta$ OPA, we have -

$\Rightarrow$ tan 30${}^0$ = $\frac{OP}{AP}$ $\Rightarrow \frac{1}{\sqrt{3}}=\frac{200}{d+BP}$

$\Rightarrow d+BP=200\sqrt{3}$

$\Rightarrow d+200=200\sqrt{3}$

$\Rightarrow d=200\left(\sqrt{3}-1\right)=200\left(1.732-1\right)=146.4m$

Hence, the distance travelled by the boat from A to B is 146.4 m.