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Question Number 72934 by Maclaurin Stickker last updated on 05/Nov/19

Answered by mr W last updated on 05/Nov/19

Commented by mr W last updated on 05/Nov/19

$$\beta +\gamma =45°$$$$CE=\frac{CA}{\cos \gamma }=\frac{b}{\cos \gamma }=\frac{b}{\cos (45-\beta )}$$$$BD=\frac{AB}{\cos \beta }=\frac{c}{\cos \beta }$$$$k=\frac{CE\times BD}{{BC}^2}=\frac{1}{\cos (45-\beta )\cos \beta }\times \frac{bc}{a^2}$$$$\frac{b}{a}=\sin 2\beta$$$$\frac{c}{a}=\cos 2\beta$$$$k=\frac{\sin 2\beta \cos 2\beta }{\cos (45-\beta )\cos \beta }$$$$k=\frac{2\sqrt{2}\sin \beta ({\cos }^2\beta -{\sin }^2\beta )}{\cos \beta +\sin \beta }$$$$k=2\sqrt{2}\sin \beta (\cos \beta -\sin \beta )$$$$k=\sqrt{2}(\sin 2\beta -1+\cos 2\beta )$$$$\frac{k}{\sqrt{2}}+1=\sin 2\beta +\cos 2\beta$$$$\frac{k}{\sqrt{2}}+1=\sin B+\cos B$$$$\frac{k}{\sqrt{2}}+1=\sqrt{2}\sin (B+45)$$$$\frac{k+\sqrt{2}}{2}=\sin (B+45)$$$$\Rightarrow \mathrm{\angle }B={\sin }^{-1}\frac{k+\sqrt{2}}{2}-45°$$$$$$$$0<\mathrm{\angle }B<90°$$$$\frac{\sqrt{2}}{2}<\frac{k+\sqrt{2}}{2}⩽1$$$$\sqrt{2}<k+\sqrt{2}⩽2$$$$\Rightarrow 0<k⩽2-\sqrt{2}$$

Commented by Maclaurin Stickker last updated on 06/Nov/19

$$perfect$$

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