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Question Number 28070 by naka3546 last updated on 19/Jan/18

Answered by mrW2 last updated on 20/Jan/18

$$let\mathrm{\angle }QSR=\alpha$$$$\mathrm{\angle }PQS=\alpha -40$$$$\frac{PS}{\sin (\alpha -40)}=\frac{PQ}{\sin (180-\alpha )}=\frac{PQ}{\sin \alpha }$$$$\Rightarrow PS=\frac{\sin (\alpha -40)}{\sin \alpha }\times PQ$$$$\frac{QR}{\sin 40}=\frac{PQ}{\sin (90-\frac{40}{2})}=\frac{PQ}{\cos 20}$$$$\Rightarrow QR=\frac{\sin 40}{\cos 20}\times PQ=2\sin 20\times PQ$$$$PS=QR$$$$\Rightarrow \frac{\sin (\alpha -40)}{\sin \alpha }=2\sin 20$$$$\Rightarrow \sin \alpha \cos 40-\cos \alpha \sin 40=2\sin 20\sin \alpha$$$$\Rightarrow \tan \alpha \cos 40-\sin 40=2\sin 20\tan \alpha$$$$\Rightarrow \tan \alpha =\frac{\sin 40}{\cos 40-2\sin 20}=\frac{\sin 40}{\cos 40-\sqrt{2(1-\cos 40)}}$$$$\Rightarrow \alpha ={\tan }^{-1}\left[\frac{\sin 40}{\cos 40-\sqrt{2(1-\cos 40)}}\right]=82.7°$$$$\Rightarrow \beta =180-\alpha =97.3°$$

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