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Question Number 172730 by cortano1 last updated on 30/Jun/22

Answered by mr W last updated on 30/Jun/22

Commented by mr W last updated on 30/Jun/22

$$\frac{\sin (x+45)}{AD}=\frac{\sin x}{BD}$$$$\frac{AD}{\sin (45-15)}=\frac{DC}{\sin 15}$$$$\frac{\sin (x+45)}{\sin (45-15)}=\frac{\sin x}{\sin 15}$$$$\frac{1+\frac{1}{\tan x}}{2\sqrt{2}\cos 15}=1$$$$1+\frac{1}{\tan x}=2\sqrt{2}\cos 15=\frac{2\sqrt{2}(\sqrt{2}+\sqrt{6})}{4}=1+\sqrt{3}$$$$\tan x=\frac{1}{\sqrt{3}}$$$$\Rightarrow x=30°✓$$

Commented by Tawa11 last updated on 01/Jul/22

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

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