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Question

From the top of a lighthouse, the angles of depression of two ships on the opposite sides of it are observed to be α and β. If the height of the lighthouse be h metres and the line joining the ships passes through the foot of the lighthouse, show that the distance between the ships is h(tan α+tan β)tan α tan β metres. [3 MARKS]

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Solution

Let AB be the lighthouse and C and D be the positions of the two ships. Then, AB = h metres.

Clearly, ACB=α and ADB=β



Let AC = x metres and

AD = y metres

From right ΔCAB, we have

ACAB=cot αxh=cot αx=h cot α ........... (i)

From right ΔDAB, we have

ADAB=cot βyh=cot βy=h cot β ........... (ii)

Adding the corresponding siddes of (i) and (ii), we get

x+y=h(cot α+cot β)=h(1tan α+1tan β)

x+y=h(tan α+tan β)tan α tan β

Hence, the distance between the ships is h(tan α+tan β)tan α tan β m

Alternative Method,

In ΔDCB

tan α=hy

y=htan α [1 MARK]

In ΔDCA

tan β=hx

x=htan β [1 MARK]

The distance between the two ships =htan α+htan β

=h(tan β+tan αtan βtan α)m [1 MARK]


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