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Question Number 163413 by mr W last updated on 06/Jan/22

Commented by mr W last updated on 06/Jan/22

$$AB=AC=DC=1,say$$$$BC=2\cos 40$$$$BD=2\cos 40-1$$$$ED=\frac{BD}{2\cos 40}=\frac{2\cos 40-1}{2\cos 40}$$$$\frac{\sin x}{DC}=\frac{\sin (40-x)}{ED}$$$$\frac{2\cos 40-1}{2\cos 40}=\frac{\sin (40-x)}{\sin x}$$$$\frac{2\cos 40-1}{2\cos 40}=\frac{\sin 40}{\tan x}-\cos 40$$$$\frac{2\cos 40-1+2{\cos }^{2}40}{2\cos 40}=\frac{\sin 40}{\tan x}$$$$\frac{2\cos 40+\cos 80}{2\cos 40\sin 40}=\frac{1}{\tan x}$$$$\tan x=\frac{\cos 10}{2\cos 40+\sin 10}=\frac{1}{\sqrt{3}}$$$$\Rightarrow x=30°$$

Commented by amin96 last updated on 06/Jan/22

$$greatsir$$

Commented by Tawa11 last updated on 06/Jan/22

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

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