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Question Number 165057 by MathsFan last updated on 25/Jan/22

Commented by MathsFan last updated on 25/Jan/22

$$howdoisolveforx?$$

Answered by mr W last updated on 25/Jan/22

$$saysidelengthofkare=1$$$$BD=BE=\sqrt{2}$$$$\frac{BE}{BC}=\frac{\sin (x+15)}{\sin x}$$$$\sqrt{2}=\frac{\sin x\cos 15+\cos x\sin 15}{\sin x}$$$$\sqrt{2}=\cos 15+\frac{\sin 15}{\tan x}$$$$\tan x=\frac{\sin 15}{\sqrt{2}-\cos 15}=\frac{\sin (45-30)}{\sqrt{2}-\cos (45-30)}$$$$=\frac{\sqrt{3}-1}{4-(\sqrt{3}+1)}=\frac{\sqrt{3}-1}{3-\sqrt{3}}=\frac{1}{\sqrt{3}}$$$$x=30°$$

Commented by MathsFan last updated on 25/Jan/22

$$thankverymuchsir$$

Commented by Tawa11 last updated on 25/Jan/22

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

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