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Question Number 169088 by cortano1 last updated on 24/Apr/22

Answered by mr W last updated on 24/Apr/22

$$a=18$$$$b=15$$$$c=12$$$$s=(18+15+12)/2=22.5$$$$\mathrm{\Delta }=\sqrt{22.5\times 4.5\times 7.5\times 10.5}=\frac{135\sqrt{7}}{4}$$$$\sin B=\frac{2\mathrm{\Delta }}{ac}=\frac{2\times 135\sqrt{7}}{4\times 18\times 12}=\frac{5\sqrt{7}}{16}\Rightarrow \cos B=\frac{9}{16}$$$$\sin C=\frac{2\mathrm{\Delta }}{ab}=\frac{2\times 135\sqrt{7}}{4\times 18\times 15}=\frac{\sqrt{7}}{4}\Rightarrow \cos C=\frac{3}{4}$$$$\tan \frac{B}{2}=\frac{\frac{5\sqrt{7}}{16}}{1+\frac{9}{16}}=\frac{\sqrt{7}}{5}$$$$\tan \frac{C}{2}=\frac{\frac{\sqrt{7}}{4}}{1+\frac{3}{4}}=\frac{\sqrt{7}}{7}$$$${(18-\frac{3}{\tan \frac{B}{2}}-\frac{R}{\tan \frac{C}{2}})}^2={(3+R)}^2-{(3-R)}^2$$$${(18-\frac{15\sqrt{7}}{7}-\sqrt{7}R)}^2=12R$$$${(18-\frac{15\sqrt{7}}{7})}^2-18(2\sqrt{7}-1)R+7R^2=0$$$$R=\frac{1}{7}[9(2\sqrt{7}-1)-\sqrt{81{(2\sqrt{7}-1)}^2-7{(18-\frac{15\sqrt{7}}{7})}^2}]$$$$=\frac{9(2\sqrt{7}-1)-6\sqrt{2(3\sqrt{7}-2)}}{7}\approx 2.564$$

Commented by mr W last updated on 24/Apr/22

Commented by Tawa11 last updated on 24/Apr/22

$$\mathrm{Great}\mathrm{sir}$$

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