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Question Number 187810 by Rupesh123 last updated on 22/Feb/23

Commented by a.lgnaoui last updated on 23/Feb/23

$$\mathrm{E})30$$

Commented by mr W last updated on 23/Feb/23

$$wrong!$$

Answered by mr W last updated on 23/Feb/23

Commented by mr W last updated on 23/Feb/23

$$sayR=2$$$$ED=\frac{4}{\sqrt{5}}\times 1=\frac{4}{\sqrt{5}}$$$$FD=\sqrt{2}$$$$\tan \beta =\frac{1}{2}$$$$\alpha =\frac{\pi }{4}-\beta$$$$\sin \alpha =\frac{1}{\sqrt{2}}(\cos \beta -\sin \alpha )=\frac{1}{\sqrt{2}}(\frac{2}{\sqrt{5}}-\frac{1}{\sqrt{5}})=\frac{1}{\sqrt{10}}$$$$\cos \alpha =\frac{3}{\sqrt{10}}$$$$\frac{\sin (x+\alpha )}{\sin x}=\frac{ED}{FD}=\frac{4}{\sqrt{10}}$$$$\cos \alpha +\frac{\sin \alpha }{\tan x}=\frac{4}{\sqrt{10}}$$$$\frac{3}{\sqrt{10}}+\frac{1}{\tan x\sqrt{10}}=\frac{4}{\sqrt{10}}$$$$\frac{1}{\tan x}=1$$$$\Rightarrow \tan x=1\Rightarrow x=45°✓$$

Commented by Rupesh123 last updated on 23/Feb/23

Perfect ��

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