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Question Number 207230 by mr W last updated on 10/May/24

Answered by mr W last updated on 10/May/24

Commented by mr W last updated on 10/May/24

$${\tan }^{-1}\frac{x}{y}+{\tan }^{-1}\frac{4-x}{y}=60°$$$$\frac{\frac{x}{y}+\frac{4-x}{y}}{1-\frac{x(4-x)}{y^2}}=\sqrt{3}$$$$\frac{4y}{y^2+x^2-4x}=\sqrt{3}$$$$\Rightarrow x^2-4x=\frac{4y}{\sqrt{3}}-y^2$$$${\tan }^{-1}\frac{x}{4-y}+{\tan }^{-1}\frac{4-x}{4-y}=105°$$$$\frac{\frac{x}{4-y}+\frac{4-x}{4-y}}{1-\frac{x(4-x)}{{(4-y)}^2}}=-\sqrt{3}-2$$$$\frac{4(4-y)}{y^2-8y+16+x^2-4x}=-\sqrt{3}-2$$$$\Rightarrow x^2-4x=-\frac{4(4-y)}{\sqrt{3}+2}-y^2+8y-16$$$$\Rightarrow \frac{4y}{\sqrt{3}}-y^2=-\frac{4(4-y)}{\sqrt{3}+2}-y^2+8y-16$$$$\Rightarrow y=3$$$$\frac{green}{yellow}=\frac{y}{4-y}=\frac{3}{4-1}=3✓$$

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