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Question Number 126509 by Algoritm last updated on 21/Dec/20

Answered by mr W last updated on 21/Dec/20

Commented by talminator2856791 last updated on 22/Dec/20

$$\mathrm{so}\mathrm{then}\mathrm{the}\mathrm{answer}?$$

Commented by mr W last updated on 22/Dec/20

$$\alpha =\frac{5\times 180}{7}=\frac{900}{7}$$$$DG=1+2\times 1\times \cos (180-\frac{900}{7})=1+2\cos \frac{360}{7}$$$$\mathrm{\angle }DGH=3\times 180-60-2\times \frac{900}{7}-90=\frac{930}{7}$$$$\frac{HG}{\sin (\frac{900}{7}-90)}=\frac{DG}{\sin \frac{930}{7}}$$$$\Rightarrow HG=\frac{\sin \frac{270}{7}}{\cos \frac{300}{7}}\times (1+2\cos \frac{360}{7})$$$$BG=2\times 1\times \sin \frac{900}{2\times 7}=2\cos \frac{180}{7}$$$$\mathrm{\angle }BGH=\frac{900}{7}-60-\frac{1}{2}(180-\frac{900}{7})=\frac{300}{7}$$$$\frac{\sin (\frac{300}{7}+\mathrm{\angle }GBH)}{BG}=\frac{\sin \mathrm{\angle }GBH}{HG}$$$$\sin \frac{300}{7}\times \frac{1}{\tan \mathrm{\angle }GBH}+\cos \frac{300}{7}=\frac{2\cos \frac{180}{7}}{\frac{\sin \frac{270}{7}}{\cos \frac{300}{7}}\times (1+2\cos \frac{360}{7})}$$$$\frac{1}{\tan \mathrm{\angle }GBH}=(\frac{2\cos \frac{180}{7}}{\sin \frac{270}{7}\times (1+2\cos \frac{360}{7})}-1)\frac{1}{\tan \frac{300}{7}}$$$$\Rightarrow \mathrm{\angle }GBH={\tan }^{-1}\frac{\tan \frac{300}{7}}{\frac{2\cos \frac{180}{7}}{\sin \frac{270}{7}\times (1+2\cos \frac{360}{7})}-1}$$$$\Rightarrow \mathrm{\angle }GBH={\tan }^{-1}\frac{\tan \frac{300}{7}}{\frac{2\cos \frac{180}{7}}{\sin \frac{270}{7}\times (1+2\cos \frac{360}{7})}-1}$$$$x=\frac{900}{7}-\frac{1}{2}(180-\frac{900}{2})-\mathrm{\angle }GBH$$$$x=\frac{720}{7}-{\tan }^{-1}\frac{\tan \frac{300}{7}}{\frac{2\cos \frac{180}{7}}{\sin \frac{270}{7}\times (1+2\cos \frac{360}{7})}-1}$$$$=30°$$

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