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The length of the shadow of a pole incli...

The length of the shadow of a pole inclined at `10^@` to the vertical towards the sun is 2.05 metres, when theelevation of the sun is `38^@`. Then, find the length of the pole.

A

`(2.05 sin 42^@)/(sin 38^@)`

B

`(2.05 sin 42^@)/(cos 42^@)`

C

`(2.05 sin 38^@)/(sin 42^@)`

D

`(2.05 sin 42^@)/(sin 38^@)`

Text Solution

Verified by Experts

Let PO be the tower and OS be its shadow .

In triangle SOP , using sine rule ,
`(l)/(sin38^@)=(2.05)/(sin(angleSPO))` ...(1)
` angle SPO=180^@-38^@-90^@-10^@=42^@`
So from (1), `l=(2.05sin38^@)/(sin42^@)`
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