Question and Answers Forum

All Questions      Topic List

Mensuration Questions

Previous in All Question      Next in All Question      

Previous in Mensuration      Next in Mensuration      

Question Number 185978 by Rupesh123 last updated on 30/Jan/23

Answered by ajfour last updated on 30/Jan/23

$$x=?=2\left(2\sin \theta \right)$$$$Y=\frac{1}{2}\left(2^2\right)\left(2\theta \right)-\frac{1}{2}\left(2\right)\left(2\sin 2\theta \right)$$$$G=\frac{1}{2}\left(2^2\right)(\frac{\pi }{2}-2\theta )$$$$-\frac{1}{2}\left(2\cos 2\theta \right)\left(2\sin 2\theta \right)$$$$Y=2G\Rightarrow$$$$12\theta =2\pi +\left(2\sin 2\theta \right)(1-2\cos 2\theta )$$$$\Rightarrow 6\theta =\pi +\sin 2\theta (1-2\cos 2\theta )$$$$\Rightarrow \theta =\frac{\pi }{6}$$$$x=4\sin \theta =4\sin \frac{\pi }{6}=2$$

Answered by mr W last updated on 30/Jan/23

Commented by mr W last updated on 30/Jan/23

$$?=2$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com