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Question Number 162481 by mnjuly1970 last updated on 29/Dec/21

Answered by mr W last updated on 29/Dec/21

$$lengthofbase=h$$$$\tan \alpha =\frac{b}{h}$$$$\tan 2\alpha =\frac{b+a}{h}$$$$\tan 3\alpha =\frac{b+a+x}{h}$$$$\tan 2\alpha =\frac{2\tan \alpha }{1-{\tan }^2\alpha }$$$$\frac{b+a}{h}=\frac{2\times \frac{b}{h}}{1-\frac{b^2}{h^2}}$$$$b+a=\frac{2{bh}^2}{h^2-b^2}$$$$\Rightarrow h^2=\frac{(a+b)b^2}{a-b}$$$$\tan 3\alpha =\frac{\tan 2\alpha +\tan \alpha }{1-\tan 2\alpha \tan \alpha }$$$$x=\frac{(a+2b)h^2}{h^2-(a+b)b}-(a+b)$$$$x=\frac{(a+2b)}{1-(a+b)b\times \frac{(a-b)}{(a+b)b^2}}-(a+b)$$$$x=\frac{(a+2b)b}{2b-a}-(a+b)$$$$x=\frac{a^2}{2b-a}✓$$

Commented by mnjuly1970 last updated on 29/Dec/21

$$grateful$$

Commented by Tawa11 last updated on 30/Dec/21

$$\mathrm{Great}\mathrm{sir}$$

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