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 123843 by mnjuly1970 last updated on 28/Nov/20

Commented by mnjuly1970 last updated on 28/Nov/20

$$pleasesolve\Uparrow \Uparrow$$

Answered by mr W last updated on 28/Nov/20

Commented by mr W last updated on 28/Nov/20

$$sayradius=R$$$$areaofsegment=\frac{semicircle}{3}=\frac{circle}{6}$$$$\frac{R^2}{2}(\pi -2\beta -\sin 2\beta )=\frac{\pi R^2}{6}$$$$2\beta +\sin 2\beta =\frac{2\pi }{3}$$$$\Rightarrow \beta =0.5863(=33.6°)$$$$$$$$PQ=2R\cos \beta$$$$\frac{1}{2}\times 2R\cos \beta \times PS\times \sin \beta =\frac{\pi R^2}{6}$$$$\Rightarrow PS=\frac{\pi R}{3\sin 2\beta }$$$$\frac{\sin (\alpha -\beta )}{\sin \alpha }=\frac{PS}{PQ}=\frac{\pi R}{3\sin 2\beta \times 2R\cos \beta }$$$$\cos \beta -\frac{\sin \beta }{\tan \alpha }=\frac{\pi }{6\sin 2\beta \cos \beta }$$$$\tan \alpha =\frac{\sin \beta }{\cos \beta -\frac{\pi }{6\sin 2\beta \cos \beta }}$$$$\tan \alpha =\frac{\sin 2\beta }{1+\cos 2\beta -\frac{\pi }{3\sin 2\beta }}$$$$\alpha ={\tan }^{-1}\left(\frac{\sin 2\beta }{1+\cos 2\beta -\frac{\pi }{3\sin 2\beta }}\right)=74.7°$$

Commented by mnjuly1970 last updated on 29/Nov/20

$$thankyousomuch$$$$masterW....$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com