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Question Number 70887 by naka3546 last updated on 09/Oct/19

Commented by ajfour last updated on 09/Oct/19

Commented by ajfour last updated on 09/Oct/19

$$s\sin \alpha =r\sin \gamma$$$$b^2={(s\sin \alpha +r\cos \delta )}^2+{(s\cos \alpha -r\sin \delta )}^2$$$$r\sin \delta =a\sin \beta$$$$b^2={(r\sin \delta +a\sin \beta )}^2+{(a\cos \beta -r\cos \delta )}^2$$$$\delta +\gamma =\frac{\pi }{2}$$$$\frac{s\cos \alpha -r\sin \delta }{s\sin \alpha +r\cos \delta }=\frac{r\sin \delta +a\sin \beta }{a\cos \beta -r\cos \delta }$$$$.........$$

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