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Question Number 117480 by I want to learn more last updated on 12/Oct/20

Commented by Tawa11 last updated on 15/Sep/21

$$\mathrm{nice}$$

Answered by mr W last updated on 12/Oct/20

Commented by mr W last updated on 12/Oct/20

$$letBC=a$$$$\cos \beta =\frac{21}{a}$$$$\cos \gamma =\frac{35}{a}$$$$\frac{AC}{\sin \beta }=\frac{BC}{\sin (\beta +\gamma )}=\frac{a}{\sin \beta \cos \gamma +\cos \beta \sin \gamma }$$$$AC=\frac{a\sin \beta }{\sin \beta \cos \gamma +\cos \beta \sin \gamma }$$$$DC=AC\cos \gamma =\frac{a\sin \beta \cos \gamma }{\sin \beta \cos \gamma +\cos \beta \sin \gamma }$$$$DG\tan \beta =DC$$$$15\tan \beta =\frac{a\sin \beta \cos \gamma }{\sin \beta \cos \gamma +\cos \beta \sin \gamma }$$$$15=\frac{a\cos \beta \cos \gamma }{\sin \beta \cos \gamma +\cos \beta \sin \gamma }$$$$\tan \beta +\tan \gamma =\frac{a}{15}$$$$\sqrt{{\left(\frac{a}{21}\right)}^2-1}+\sqrt{{\left(\frac{a}{35}\right)}^2-1}=\frac{a}{15}$$$${\left(\frac{a}{21}\right)}^2-1={\left(\frac{a}{15}\right)}^2+{\left(\frac{a}{35}\right)}^2-1-2\left(\frac{a}{15}\right)\sqrt{{\left(\frac{a}{35}\right)}^2-1}$$$$a^2[\frac{1}{{35}^2}-{(\frac{1}{{15}^2}+\frac{1}{{35}^2}-\frac{1}{{21}^2})}^2{\left(\frac{15}{2}\right)}^2]=1$$$$\Rightarrow a=\frac{1}{\sqrt{\frac{1}{{35}^2}-{(\frac{1}{{15}^2}+\frac{1}{{35}^2}-\frac{1}{{21}^2})}^2{\left(\frac{15}{2}\right)}^2}}=\frac{98}{\sqrt{3}}$$$$\Rightarrow \beta ={\cos }^{-1}\frac{21}{a}={\cos }^{-1}\frac{3\sqrt{3}}{14}$$$$\Rightarrow \gamma ={\cos }^{-1}\frac{35}{a}={\cos }^{-1}\frac{5\sqrt{3}}{14}$$$$\Rightarrow \alpha =\pi -{\cos }^{-1}\frac{3\sqrt{3}}{14}-{\cos }^{-1}\frac{5\sqrt{3}}{14}=\frac{\pi }{3}=60°$$

Commented by I want to learn more last updated on 12/Oct/20

$$\mathrm{I}\mathrm{appreciate}\mathrm{sir}.$$

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