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This one comes from my module without proper explanation.Please suggest how to approach this problem ?

PS:How do you write degree (in symbol) in math-mode ?

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    @Debanjan: Let $P'$ be the mirror image of $P$ across the segment $QR$. The triangle $P'QR$ is the copy that is pasted "back to back" with $PQR$ along $QR$. Since $P'Q$ and $PQ$ are both perpendicular to $QR$, the two segments together form a larger segment $PP'$ of the triangle $PP'R$. Since $PR=2PQ$, we know that $P'R=2PQ$. Now, observe that $PP'=...$, so ... (I'm simply walking through the traditional construction used to determine sines and cosines of two fundamental angles, but avoiding the trigonometric terminology, which isn't needed here, and which may be unfamiliar to the Questioner.)2010-11-06

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As Djaian pointed out use the fact that $\sin\angle{R} = \frac{PQ}{PR}=\frac{PQ}{2PQ}=\frac{1}{2}$ So $\angle{R} = ?$

Also i hope you are aware of the fact that in a right angled triangle $\sin\theta =\frac{\text{Opposite Side}}{\text{Hypotenuse}}$