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Question Number 75058 by behi83417@gmail.com last updated on 07/Dec/19

$$\mathbf{in}\mathbf{triangle}:\mathbf{ABC}:$$$$\mathbf{a}=\sqrt{2},\mathbf{b}-\mathbf{c}=\frac{\sqrt{2}+1}{2},\stackrel{}{\mathbf{B}}-\stackrel{}{\mathbf{C}}=\frac{\mathit{\pi }}{2}$$$$\mathbf{find}:{\mathbf{h}}_{\mathbf{a}},{\mathbf{S}}_{\mathbf{ABC}},{\mathbf{d}}_{\mathbf{a}},\mathbf{R},\stackrel{}{\mathbf{A}}.$$

Commented by mr W last updated on 07/Dec/19

$$b-c=\frac{1}{2}isnotpossiblesir!$$$$itmustbe1<b-c<\sqrt{2}!$$

Commented by behi83417@gmail.com last updated on 07/Dec/19

$$\mathrm{thank}\mathrm{you}\mathrm{very}\mathrm{much}\mathrm{dear}\mathrm{master}.$$$$\mathrm{now}\mathrm{it}\mathrm{is}\mathrm{fixed}.$$

Answered by mr W last updated on 07/Dec/19

$$let\mathrm{\angle }C=\theta$$$$\Rightarrow \mathrm{\angle }B=\frac{\pi }{2}+\theta$$$$\Rightarrow \mathrm{\angle }A=\pi -\theta -(\frac{\pi }{2}+\theta )=\frac{\pi }{2}-2\theta$$$$\frac{b}{\sin (\frac{\pi }{2}+\theta )}=\frac{c}{\sin \theta }=\frac{\sqrt{2}}{\sin (\frac{\pi }{2}-2\theta )}$$$$\Rightarrow b=\frac{\sqrt{2}\cos \theta }{\cos 2\theta }$$$$\Rightarrow c=\frac{\sqrt{2}\sin \theta }{\cos 2\theta }$$$$b-c=\frac{\sqrt{2}(\cos \theta -\sin \theta )}{\cos 2\theta }=\delta$$$$\frac{2(1-\sin 2\theta )}{{\cos }^{2}2\theta }={\delta }^2$$$$\frac{1}{1+\sin 2\theta }=\frac{{\delta }^2}{2}$$$$\sin 2\theta =\frac{2}{{\delta }^2}-1\Rightarrow 1<\delta <\sqrt{2}$$$$\Rightarrow \mathit{\theta }=\frac{1}{2}{\mathbf{sin}}^{-1}(\frac{2}{{\mathit{\delta }}^2}-1)$$$$with\delta =\frac{1+\sqrt{2}}{2}weget$$$$\theta =\frac{1}{2}{\sin }^{-1}(23-16\sqrt{2})\approx 10.9475°$$$$therestisclear.$$

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