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In the given figure, $\Delta MEN$ is a right triangle, right-angled at E. If $\angle EMN = 60^\circ$ and ME = $5$ cm, find the length of MN. Also, find the value of $\sin^2 N$.
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Solution (a) In $\Delta MEN$,
$\frac{EM}{MN} = \cos 60^\circ$ (1/2 Mark)
$\frac{5}{MN} = \frac{1}{2} \Rightarrow MN = 10$ cm (1/2 Mark)
$\angle MNE = 30^\circ$ (1/2 Mark)
$\sin^2 N = \sin^2 30^\circ = (\frac{1}{2})^2 = \frac{1}{4}$ (1/2 Mark)
$\frac{EM}{MN} = \cos 60^\circ$ (1/2 Mark)
$\frac{5}{MN} = \frac{1}{2} \Rightarrow MN = 10$ cm (1/2 Mark)
$\angle MNE = 30^\circ$ (1/2 Mark)
$\sin^2 N = \sin^2 30^\circ = (\frac{1}{2})^2 = \frac{1}{4}$ (1/2 Mark)