Question

A Person Standing on a boat at point A sees a submarine at point B, making an angle of depression of ${60}^{\circ }$. After sometime, the submarine travels 100 m and moves to point C. Now, the angle of depression becomes 30${}^{\circ }$. Find the distance the boat has to travel to reach a point that is exactly above the submarine. $[Tan{30}^{\circ }=0.57,tan{60}^{\circ }=1.73]$

• A. 86.2 m
• B. 96.4 m
• C. 82.4 m
• D. 78.6 m

Answer 1

The correct option is A

86.2 m

In the given figure, the person is standing on the boat at A and the submarine moves from B to C.

The boat has to travel distance AD to reach D, a point which is exactly above the submarine.

In $△ADCtan{30}^{\circ }=\frac{CD}{AD}0.57=\frac{CD}{AD}CD=0.57AD\left(i\right)$

In $△ADBtan{60}^{\circ }=\frac{BD}{AD}1.73=\frac{BD}{AD}BD=1.73AD\left(ii\right)$

Now, we can observe that $CB=BD-CD100=1.73AD-0.57AD100=1.16ADAD=86.2m$

Hence, the boat has to travel 86.2 m to reach a point which is exactly above the submarine.