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Question Number 153977 by mr W last updated on 12/Sep/21

Answered by mr W last updated on 12/Sep/21

$$\frac{l}{a}=\frac{\sin (45+\alpha )}{\sin \alpha }=\frac{1}{\sqrt{2}}(\frac{1}{\tan \alpha }+1)$$$$\frac{1}{\tan \alpha }=\frac{\sqrt{2}l}{a}-1$$$$similarly$$$$\frac{1}{\tan \beta }=\frac{\sqrt{2}l}{b}-1$$$$\alpha +\beta =90°-\theta$$$$\tan (\alpha +\beta )=\frac{\tan \alpha +\tan \beta }{1-\tan \alpha \tan \beta }=\tan (90-\theta )=\frac{1}{\tan \theta }$$$$\frac{\frac{1}{\tan \alpha }+\frac{1}{\tan \beta }}{\frac{1}{\tan \alpha }\times \frac{1}{\tan \beta }-1}=\frac{1}{\tan \theta }$$$$\frac{\frac{\sqrt{2}l}{a}-1+\frac{\sqrt{2}l}{a}-1}{(\frac{\sqrt{2}l}{a}-1)\times (\frac{\sqrt{2}l}{a}-1)-1}=\frac{1}{\tan \theta }$$$$\frac{\frac{\sqrt{2}l}{a}-1+\frac{\sqrt{2}l}{b}-1}{(\frac{\sqrt{2}l}{a}-1)\times (\frac{\sqrt{2}l}{b}-1)-1}=\frac{1}{\tan \theta }$$$$2l^2-\sqrt{2}(a+b)(1+\tan \theta )l+2ab\tan \theta =0$$$$l=\frac{(a+b)(1+\tan \theta )+\sqrt{{(a+b)}^2{(1+\tan \theta )}^2-8ab\tan \theta }}{2\sqrt{2}}$$$$c=\sqrt{2}l-(a+b)$$$$\Rightarrow c=\frac{(a+b)(\tan \theta -1)+\sqrt{{(a+b)}^2{(1+\tan \theta )}^2-8ab\tan \theta }}{2}$$

Commented by Tawa11 last updated on 12/Sep/21

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

Commented by liberty last updated on 13/Sep/21

$$waw$$

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