Question

A security guard standing on the top of a 100 m high lighthouse sees an enemy ship coming towards it. It was initially at an angle of depression of ${35}^{\circ }$ but after 10 minutes the angle of depression changes to ${55}^{\circ }$. The speed of enemy ship is equal to $[tan{55}^{\circ }=1.42,tan{35}^{\circ }=0.7]$

• A. 7.9 m / min
• B. 8.5 m / min
• C. 7.2 m / min
• D. 6.6 m / min

Answer 1

The correct option is C

7.2 m / min

Let AB be the height of the lighthouse.

Distance travelled by the ship in 10 minutes = CD

In $△$ ACE

$\tan {55}^{\circ }=\frac{CE}{AE}$

1.42 = $\frac{100}{AE}$

AE = $\frac{100}{1.42}=70.42$ m

In $△$ ADF

$\tan {35}^{\circ }=\frac{FD}{AF}$

$0.7=\frac{100}{AF}$

AF = $\frac{100}{0.7}=142.85$ m

From the figure we can see that the distance travelled by enemy ship is EF

EF = AF - AE

EF = 142.85 - 70.42

EF = 72.43 m

Speed of enemy ship = $\frac{EF}{TimeTaken}=\frac{72.43}{10}=7.2m/min$