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Question Number 210667 by mr W last updated on 15/Aug/24

Answered by ajfour last updated on 16/Aug/24

$$letCbeoriginandA(p,q)$$$$q=\frac{p}{\sqrt{3}}$$$$q=p-a$$$$0-q=m(2a-p)$$$$\Rightarrow m=-\frac{q}{2a-p}=-\frac{q}{p-2q}=-\frac{1}{\sqrt{3}-2}$$$$\tan {\theta }_{?}=-(2+\sqrt{3})$$$${\theta }_{?}=\pi -{\tan }^{-1}(2+\sqrt{3})$$

Commented by mr W last updated on 16/Aug/24

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Answered by A5T last updated on 15/Aug/24

Commented by A5T last updated on 15/Aug/24

$$\frac{sin30°}{y}=\frac{sin(150°-?)}{2x}\Rightarrow sin(150°-?)=\frac{x}{y}$$$$\frac{sin45°}{y}=\frac{sin(135°-?)}{x}\Rightarrow sin(135°-?)=\frac{x}{y}\times sin45°$$$$\Rightarrow \frac{sin(135°-?)}{sin(150°-?)}=sin45°=\frac{sin(45°+?)}{sin(30°+?)}$$$$\Rightarrow ?=105°$$

Commented by mr W last updated on 15/Aug/24

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Answered by mr W last updated on 15/Aug/24

Commented by mr W last updated on 15/Aug/24

$$?=60+45=105°$$

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