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Question Number 61335 by necx1 last updated on 01/Jun/19

Answered by mr W last updated on 02/Jun/19

$$\cos \mathrm{\angle }A=\frac{5^2+7^2-8^2}{2\times 5\times 7}=\frac{1}{7}$$$$\Rightarrow \sin \mathrm{\angle }A=\frac{4\sqrt{3}}{7}$$$$\cos \mathrm{\angle }B=\frac{5^2+8^2-7^2}{2\times 5\times 8}=\frac{1}{2}$$$$\Rightarrow \mathrm{\angle }B=60°$$$$\Rightarrow \sin \mathrm{\angle }B=\frac{\sqrt{3}}{2}$$$$\cos \mathrm{\angle }C=\frac{8^2+7^2-5^2}{2\times 8\times 7}=\frac{11}{14}$$$$\Rightarrow \sin \mathrm{\angle }C=\frac{5\sqrt{3}}{14}$$$$EC=\sqrt{2^2+8^2-2\times 2\times 8\times \frac{1}{2}}=2\sqrt{13}$$$$AD=\sqrt{2^2+7^2-2\times 2\times 7\times \frac{11}{14}}=\sqrt{31}$$$$\frac{\sin \mathrm{\angle }BAD}{6}=\frac{\sin \mathrm{\angle }B}{\sqrt{31}}$$$$\Rightarrow \sin \mathrm{\angle }BAD=\frac{6\times \sqrt{3}}{2\times \sqrt{31}}=\frac{3\sqrt{93}}{31}$$$$\Rightarrow \tan \frac{\mathrm{\angle }BAD}{2}=\frac{3\sqrt{93}}{31[1+\sqrt{1-{\left(\frac{3\sqrt{93}}{31}\right)}^2}]}=\frac{3\sqrt{3}}{2+\sqrt{31}}$$$$\frac{\sin \mathrm{\angle }AEC}{7}=\frac{\sin \mathrm{\angle }A}{2\sqrt{13}}$$$$\Rightarrow \sin \mathrm{\angle }AEC=\frac{7\times 4\sqrt{3}}{7\times 2\sqrt{13}}=\frac{2\sqrt{39}}{13}$$$$\Rightarrow \tan \frac{\mathrm{\angle }AEC}{2}=\frac{2\sqrt{39}}{13[1+\sqrt{1-{\left(\frac{2\sqrt{39}}{13}\right)}^2}]}=\frac{2\sqrt{3}}{1+\sqrt{13}}$$$$\frac{r}{\tan \frac{\mathrm{\angle }BAD}{2}}+\frac{r}{\tan \frac{\mathrm{\angle }AEC}{2}}=3$$$$\Rightarrow r=\frac{3}{\frac{2+\sqrt{31}}{3\sqrt{3}}+\frac{1+\sqrt{13}}{2\sqrt{3}}}=\frac{18\sqrt{3}}{7+3\sqrt{13}+2\sqrt{31}}=1.077$$

Commented by tanmay last updated on 01/Jun/19

$$excellentsir...$$

Commented by Tawa1 last updated on 02/Jun/19

$$\mathrm{Sir},\mathrm{is}\mathrm{the}\mathrm{answer}:\frac{18\sqrt{3}}{7+3\sqrt{13}+2\sqrt{31}}=1.0768$$$$\mathrm{please}\mathrm{sir},\mathrm{check}\mathrm{if}\mathrm{there}\mathrm{is}\mathrm{a}\mathrm{mistake}\mathrm{in}\mathrm{your}\mathrm{work}.$$$$\mathrm{If}\mathrm{non},\mathrm{i}\mathrm{will}\mathrm{go}\mathrm{with}\mathrm{your}\mathrm{answer}.$$

Commented by mr W last updated on 02/Jun/19

$$youareright,thanks!$$$$ihadamistakeduetotypo.$$

Commented by Tawa1 last updated on 02/Jun/19

$$\mathrm{Thanks}\mathrm{so}\mathrm{much}\mathrm{sir}\mathrm{for}\mathrm{checking}.\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.$$

Commented by necx1 last updated on 04/Jun/19

$$excellentsir$$

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