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Question Number 16364 by Tinkutara last updated on 21/Jun/17

Answered by ajfour last updated on 21/Jun/17

Commented by ajfour last updated on 21/Jun/17

$$In△BCG,applyingthesinerule:$$$$firsttheangles,$$$$\mathrm{\angle }CBG=\frac{\pi }{2}-\frac{B}{2},\mathrm{\angle }BGC=\frac{B}{2}+\frac{C}{2}$$$$\mathrm{\angle }CGD=C/2\Rightarrow CG=\frac{r_1}{\cos \left(C/2\right)}$$$$\frac{CG}{\sin \mathrm{\angle }CBG}=\frac{BC}{\sin \mathrm{\angle }BGC}$$$$\frac{\left[r_1/\cos \left(C/2\right)\right]}{\sin (\pi /2-B/2)}=\frac{a}{\sin (B/2+C/2)}$$$$\frac{r_1}{\cos \frac{B}{2}\cos \frac{C}{2}}=\frac{a}{\sin (\frac{\pi }{2}-\frac{A}{2})}$$$$\Rightarrow r_1=\frac{a\cos \frac{B}{2}\cos \frac{C}{2}}{\cos \frac{A}{2}}.$$$$similarlyfor{r}_2,and{r}_3.$$

Commented by Tinkutara last updated on 21/Jun/17

$$\mathrm{Thanks}\mathrm{Sir}!$$

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