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Question Number 154343 by Lekhraj last updated on 17/Sep/21

Answered by qaz last updated on 17/Sep/21

$$\frac{\sin \mathrm{x}}{\bullet \mathrm{O}}=\frac{\sin \mathrm{\angle }\mathrm{M}\bullet \mathrm{C}}{\mathrm{OC}},\frac{\sin 30°}{\bullet \mathrm{O}}=\frac{\sin 105°}{\mathrm{OA}}$$$$\because \mathrm{OA}=\mathrm{OC}$$$$\therefore \sin \mathrm{x}=\frac{\sin 30°}{\sin 105°}\cdot \sin \mathrm{\angle }\mathrm{M}\bullet \mathrm{C}=\frac{\sqrt{2}}{\sqrt{3}+1}\sin \mathrm{\angle }\mathrm{M}\bullet \mathrm{C}$$$$\because \mathrm{\angle }\mathrm{M}\bullet \mathrm{C}=180°-135°-\mathrm{x}=45°-\mathrm{x}$$$$\therefore \sin \mathrm{x}=\frac{\sqrt{2}}{\sqrt{3}+1}\sin (45°-\mathrm{x})=\frac{1}{\sqrt{3}+1}(\cos \mathrm{x}-\sin \mathrm{x})$$$$\Rightarrow \tan \mathrm{x}=2-\sqrt{3}$$$$\Rightarrow \mathrm{x}={\tan }^{-1}(2-\sqrt{3})$$

Commented by Lekhraj last updated on 17/Sep/21

$$\mathrm{Thank}\mathrm{you}.\mathrm{You}\mathrm{does}\mathrm{not}\mathrm{mark}\mathrm{in}\mathrm{the}\mathrm{figure}\mathrm{which}\mathrm{point}\mathrm{is}\mathrm{the}\mathrm{blue}\mathrm{dot}$$$$\mathrm{and}\mathrm{which}\mathrm{point}\mathrm{is}\mathrm{o}.$$

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