Two ships are sailing in the sea on the two sides of a lighthouse. The angle of elevation of the top of the lighthouse is observed from the ships are 30° and 45° respectively. If the lighthouse is 100 m high, the distance between the two ships is:
A. 173 m
B. 200 m
C. 273 m
D. 300 m
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Solution
Given :-
- The angle of elevation of the top of the lighthouse is observed from the ships are 30° and 45° respectively.
- If the lighthouse is 100 m high
Find :-
- Distance between both ship .
Explantion
Let AB be the Lighthouse and C & D be the position of the ship .
Then, AB = 100 m, < BCA = 30° , < BDA = 45° .
Now, take ∆ BCA,
==> tan 30° = AB/AC
==> 1/√3 = 100/AC
==> √3 = AC/100
==> AC = √3 × 100
==> AC =100√3 m
Again,
Take ∆ BDA ,
==> tan 45° = AB/AD
==> 1 = 100/AD
==> AD = 100 m
So, Now calculate CD
==> CD = CA + DA
==> CD = 100√3 + 100
==> CD = 100(√3 + 1)
==> CD = 100( 1.73 + 1)
==> CD = 100 × 2.73
==> CD = 273 m
Hence
- Distance between both ship be 273 m
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