Question

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​

Answer 1

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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