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.

Options

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^@)`

Explanation

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^@)`