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Question Number 213519 by RojaTaniya last updated on 07/Nov/24

Answered by mr W last updated on 07/Nov/24

$$\tan \alpha =\frac{1}{x}$$$$\tan (\alpha +\beta )=\frac{3}{x}$$$$\tan (\alpha +\alpha +\beta )=\frac{6}{x}$$$$\frac{6}{x}=\frac{\frac{1}{x}+\frac{3}{x}}{1-\frac{1}{x}\times \frac{3}{x}}=\frac{4x}{x^2-3}$$$$x^2=9\Rightarrow x=3$$

Commented by RojaTaniya last updated on 07/Nov/24

$$Sir,perferct.Thanks.$$

Answered by A5T last updated on 07/Nov/24

$$\left[ABC\right]=\frac{3x\sqrt{x^2+1}sin\alpha }{2}=\frac{\sqrt{x^2+9}\times \sqrt{x^2+36}sin\alpha }{2}$$$$\Rightarrow 9x^2(x^2+1)=(x^2+9)(x^2+36)$$$$\Rightarrow 9x^4+9x^2=x^4+45x^2+324\Rightarrow 2x^4-9x^2-81=0$$$$\Rightarrow (x^2-9)(2x^2+9)=0\Rightarrow x^2=9\Rightarrow x=3$$

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