One side of a rectangular piece of paper is 6 cm, the adjacent sides being longer than 6cm. One corner of the paper is folded so that is sets on the opposite longer side. If the length of the crease is I cm and it makes an angle 9 with the long side as shown, then I is

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$\frac{3}{\sin θ \cos^{2} θ}$

$\frac{6}{\sin^{2} θ \cos θ}$

$\frac{3}{\sin θ \cos θ}$

$\frac{3}{\sin^{2} θ}$

answer is A.

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Detailed Solution

$\sin θ = \frac{x}{l}$
Also, $\cos 2 θ = \frac{6 - x}{x}$

$\begin{aligned} 1 + \cos 2 θ = \frac{6}{x} \\ 2 \cos^{2} θ = \frac{6}{l \sin θ} \end{aligned}$
$[substituting x = l \sin θ from Eq. (i)]$
$l = \frac{3}{\sin θ \cos^{2} θ}$

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