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Question Number 172628 by mnjuly1970 last updated on 29/Jun/22

Commented by mr W last updated on 29/Jun/22

$$maybe\frac{2}{\tan \gamma }=\frac{1}{\tan \beta }-\frac{1}{\tan \alpha }?$$

Commented by mnjuly1970 last updated on 29/Jun/22

$$\mathrm{yes}\mathrm{sir}\mathrm{W}...\mathrm{thanks}\mathrm{alot}$$

Answered by mr W last updated on 29/Jun/22

$$\frac{BM}{\sin \alpha }=\frac{AM}{\sin (\gamma +\alpha )}...\left(i\right)$$$$\frac{\sin \beta }{MC}=\frac{\sin (\gamma -\beta )}{AM}...\left(ii\right)$$$$\left(i\right)\times \left(ii\right):$$$$\frac{\sin \beta }{\sin \alpha }=\frac{\sin (\gamma -\beta )}{\sin (\gamma +\alpha )}=\frac{\sin \gamma \cos \beta -\cos \gamma \sin \beta }{\sin \gamma \cos \alpha +\cos \gamma \sin \alpha }$$$$1=\frac{\frac{\tan \gamma }{\tan \beta }-1}{\frac{\tan \gamma }{\tan \alpha }+1}$$$$\Rightarrow \frac{2}{\tan \gamma }=\frac{1}{\tan \beta }-\frac{1}{\tan \alpha }$$

Commented by Tawa11 last updated on 30/Jun/22

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

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