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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 but after 10 minutes the angle of depression changes to 55. The speed of enemy ship is equal to [tan 55=1.42, tan 35=0.7]


A

7.9 m / min

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B

8.5 m / min

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C

7.2 m / min

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D

6.6 m / min

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Solution

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

tan55=CEAE

1.42 = 100AE

AE = 1001.42=70.42 m

In ADF

tan35=FDAF

0.7=100AF

AF = 1000.7=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 = EFTime Taken=72.4310=7.2m/min


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