Question Number 57700 by Nameed last updated on 10/Apr/19
$$\mathrm{R}\left(\mathrm{1}\:−\:\mathrm{cos}\theta\right)\:=\:\mathrm{0}.\mathrm{5} \\ $$$$\mathrm{Rsin}\theta\:=\:\mathrm{4} \\ $$$$\mathrm{R}\:=\:? \\ $$$$\theta\:=\:? \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 10/Apr/19
$${R}×\mathrm{2}{sin}\frac{\theta}{\mathrm{2}}{cos}\frac{\theta}{\mathrm{2}}=\mathrm{4} \\ $$$${R}\left(\mathrm{2}{sin}^{\mathrm{2}} \frac{\theta}{\mathrm{2}}\right)=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\frac{\mathrm{2}{Rsin}^{\mathrm{2}} \frac{\theta}{\mathrm{2}}}{\mathrm{2}{Rsin}\frac{\theta}{\mathrm{2}}{cos}\frac{\theta}{\mathrm{2}}}=\frac{\mathrm{1}}{\mathrm{2}×\mathrm{4}} \\ $$$${tan}\frac{\theta}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{8}}\rightarrow\theta=\mathrm{2}{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{8}}\right) \\ $$$${sin}\frac{\theta}{\mathrm{2}}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{8}^{\mathrm{2}} +\mathrm{1}^{\mathrm{2}} }}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{65}}} \\ $$$${R}×\mathrm{2}{sin}^{\mathrm{2}} \frac{\theta}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${R}×\mathrm{2}×\frac{\mathrm{1}}{\mathrm{65}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${R}=\frac{\mathrm{65}}{\mathrm{4}} \\ $$
Commented by math1967 last updated on 10/Apr/19
$${tan}\frac{\theta}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{8}}\:\therefore\:\theta=\mathrm{2tan}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{8}}\:\: \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 10/Apr/19
$${thank}\:{you}\:{dada}…{vote}\:{ar}\:{ki}\:{khabar}…{kolkata}\:{te} \\ $$
Commented by math1967 last updated on 11/Apr/19
$${Jor}\:{prochar}\:{cholche} \\ $$