Question and Answers Forum

All Questions      Topic List

Geometry Questions

Previous in All Question      Next in All Question      

Previous in Geometry      Next in Geometry      

Question Number 172484 by mnjuly1970 last updated on 27/Jun/22

Answered by mr W last updated on 28/Jun/22

$$let\theta =\mathrm{\angle }ADC$$$$\frac{10}{\sin \theta }=\frac{5}{\sin \alpha }=\frac{AD}{\sin (\theta +\alpha )}$$$$\frac{6}{\sin 2\alpha }=\frac{AD}{\sin (\theta -2\alpha )}$$$$\sin \theta =2\sin \alpha$$$$AD=\frac{5\sin (\theta +\alpha )}{\sin \alpha }=5(\frac{\sin \theta }{\tan \alpha }+\cos \theta )$$$$AD=\frac{6\sin (\theta -2\alpha )}{\sin 2\alpha }=6(\frac{\sin \theta }{\tan 2\alpha }-\cos \theta )$$$$5(\frac{\sin \theta }{\tan \alpha }+\cos \theta )=6(\frac{\sin \theta }{\tan 2\alpha }-\cos \theta )$$$$11\cos \theta =2\sin \alpha (\frac{6}{\tan 2\alpha }-\frac{5}{\tan \alpha })$$$$11\cos \theta =-2\left(\frac{3-{\cos }^2\alpha }{\cos \alpha }\right)$$$$121{\cos }^2\theta =4\left(\frac{9-{6\cos }^2\alpha +{\cos }^4\alpha }{{\cos }^2\alpha }\right)$$$$121(1-4{\sin }^2\alpha )=4\left(\frac{9-{6\cos }^2\alpha +{\cos }^4\alpha }{{\cos }^2\alpha }\right)$$$$121(4{\cos }^2\alpha -3)=4\left(\frac{9-{6\cos }^2\alpha +{\cos }^4\alpha }{{\cos }^2\alpha }\right)$$$$40\times 4{\cos }^4\alpha -113{\cos }^2\alpha -12=0$$$${\cos }^2\alpha =\frac{113+\sqrt{{113}^2+4\times 40\times 4\times 12}}{2\times 40\times 4}=\frac{4}{5}$$$$\cos \alpha =\frac{2}{\sqrt{5}}\Rightarrow \sin \alpha =\frac{1}{\sqrt{5}}$$$$\sin \theta =2\sin \alpha =\frac{2}{\sqrt{5}}\Rightarrow \cos \theta =-\frac{1}{\sqrt{5}}$$$$\sin \mathrm{\angle }C=\sin (\pi -\alpha -\theta )=\sin (\alpha +\theta )$$$$=\sin \alpha \cos \theta +\cos \alpha \sin \theta$$$$=-\frac{1}{\sqrt{5}}\times \frac{1}{\sqrt{5}}+\frac{2}{\sqrt{5}}\times \frac{2}{\sqrt{5}}=\frac{3}{5}$$$$Area=\frac{10\times 11}{2}\times \frac{3}{5}=33$$

Commented by Tawa11 last updated on 28/Jun/22

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

Terms of Service

Privacy Policy

Contact: info@tinkutara.com