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Question Number 141812 by mathsuji last updated on 23/May/21

Commented by mr W last updated on 23/May/21

$$checkyourquestionanddiagram!$$$$\mathrm{\angle }BEC=\alpha ?$$$$butaccordingtodiagram$$$$\mathrm{\angle }BEC=180°$$

Commented by mathsuji last updated on 23/May/21

$$Sorrysir,BED=\alpha$$

Answered by mr W last updated on 24/May/21

Commented by mr W last updated on 24/May/21

$$given:a,b,\alpha$$$$letp=EB,q=EA,\mathrm{\angle }EAB=\beta$$$$b^2=p^2+q^2-2pq\cos \alpha$$$$a^2=p^2+q^2+2pq\cos \alpha$$$$if\alpha =90°:p=q=\frac{\sqrt{a^2+b^2}}{2}$$$$otherwise:$$$$\Rightarrow pq=\frac{a^2-b^2}{4\cos \alpha }=\mu ,say$$$$\Rightarrow p^2+q^2=\frac{a^2+b^2}{2}=\lambda ,say$$$$\Rightarrow p^4+q^2{p}^2=\lambda p^2$$$$\Rightarrow p^4-\lambda p^2+{\mu }^2=0$$$$\Rightarrow p^2=\frac{\lambda \pm \sqrt{{\lambda }^2-4{\mu }^2}}{2}$$$$\Rightarrow p=\sqrt{\frac{\lambda +\sqrt{{\lambda }^2-4{\mu }^2}}{2}}$$$$\Rightarrow q=\sqrt{\frac{\lambda -\sqrt{{\lambda }^2-4{\mu }^2}}{2}}$$$$\frac{\sin \beta }{p}=\frac{\sin \alpha }{a}$$$$\Rightarrow \sin \beta =\frac{p\sin \alpha }{a}$$$$\Rightarrow \beta ={\sin }^{-1}\left(\frac{p\sin \alpha }{a}\right)$$$$EF=q\sin \beta =\frac{pq\sin \alpha }{a}=\frac{(a^2-b^2)\tan \alpha }{4a}$$$$\frac{EH}{\sin \frac{\beta }{2}}=\frac{q}{\sin \alpha }$$$$\Rightarrow EH=\frac{q\sin \frac{\beta }{2}}{\sin \alpha }$$$$A_{\mathrm{\Delta }EFH}=\frac{EF\times EH\times \sin (\frac{\pi }{2}-\alpha +\beta )}{2}$$$$\Rightarrow A_{\mathrm{\Delta }EFH}=\frac{(a^2-b^2)q\sin \frac{\beta }{2}\cos (\alpha -\beta )}{8a\cos \alpha }$$$${FH}^2={EF}^2+{EH}^2-2\times EF\times EH\times \cos (\frac{\pi }{2}-\alpha +\beta )$$$$\Rightarrow EH=\sqrt{{\left[\frac{(a^2-b^2)\tan \alpha }{4a}\right]}^2+{\left[\frac{q\sin \frac{\beta }{2}}{\sin \alpha }\right]}^2-\frac{(a^2-b^2)q\sin \frac{\beta }{2}\sin (\alpha -\beta )}{2a\cos \alpha }}$$

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