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Question Number 170323 by cortano1 last updated on 20/May/22

Answered by mr W last updated on 20/May/22

$$\frac{AC}{\sin 18°}=\frac{AD}{\sin 12°}...\left(i\right)$$$$\frac{\sin (\alpha +18°)}{DB}=\frac{\sin \alpha }{AD}...\left(ii\right)$$$$\left(i\right)\times \left(ii\right):$$$$\frac{\sin (\alpha +18°)}{\sin 18°}=\frac{\sin \alpha }{\sin 12°}$$$$\frac{\sin \alpha \cos 18°+\cos \alpha \sin 18°}{\sin 18°}=\frac{\sin \alpha }{\sin 12°}$$$$\frac{1}{\tan 18°}+\frac{1}{\tan \alpha }=\frac{1}{\sin 12°}$$$$\tan \alpha =\frac{1}{\frac{1}{\sin 12°}-\frac{1}{\tan 18°}}=\frac{1}{\sqrt{3}}$$$$\Rightarrow \alpha =30°$$

Commented by learner22 last updated on 21/May/22

$$\mathit{N}\mathit{i}\mathit{c}\mathit{e}$$

Commented by cortano1 last updated on 22/May/22

$$howtoshow\frac{1}{\sin 12°}-\frac{1}{\tan 18°}=\sqrt{3}?$$

Commented by Tawa11 last updated on 08/Oct/22

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

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