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Question Number 207946 by cherokeesay last updated on 31/May/24

Answered by mr W last updated on 01/Jun/24

Commented by mr W last updated on 01/Jun/24

$$AC=\frac{x}{\cos \theta }$$$$BC=x\tan \theta$$$$CD=x(1-\tan \theta )$$$$ED=x(1-\tan \theta )\tan \theta$$$$CE=\frac{x(1-\tan \theta )}{\cos \theta }$$$$EG=x-x(1-\tan \theta )\tan \theta =x(1-\tan \theta +{\tan }^2\theta )$$$$EF=\frac{x}{\cos \theta }-\frac{x(1-\tan \theta )}{\cos \theta }=\frac{x\tan \theta }{\cos \theta }$$$$5=\frac{1}{2}\times \frac{x}{\cos \theta }\times \frac{x(1-\tan \theta )}{\cos \theta }$$$$\Rightarrow 10=x^2(1-\tan \theta )(1+{\tan }^2\theta )...\left(i\right)$$$$3=\frac{\sin (\frac{\pi }{2}-\theta )}{2}\times x(1-\tan \theta +{\tan }^2\theta )\times \frac{x\tan \theta }{\cos \theta }$$$$\Rightarrow 6=x^2(1-\tan \theta +{\tan }^2\theta )\tan \theta ...\left(ii\right)$$$$\left(i\right)/\left(ii\right):$$$$\frac{5}{3}=\frac{(1-\tan \theta )(1+{\tan }^2\theta )}{(1-\tan \theta +{\tan }^2\theta )\tan \theta }$$$$lett=\tan \theta$$$$8t^3-8t^2+8t-3=0$$$$(2t-1)(4t^2-2t+3)=0$$$$\Rightarrow 2t-1=0\Rightarrow t=\frac{1}{2}$$$$10=x^2(1-\frac{1}{2})(1+\frac{1}{4})$$$$\Rightarrow x^2=16\Rightarrow x=4✓$$

Commented by mnjuly1970 last updated on 01/Jun/24

$$\underset{―}{\underbrace{⋚}}$$

Commented by Tawa11 last updated on 21/Jun/24

$$\mathrm{Weldone}\mathrm{sir}$$

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