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Question Number 57700 by Nameed last updated on 10/Apr/19

R(1 − cosθ) = 0.5  Rsinθ = 4  R = ?  θ = ?

$$\mathrm{R}\left(\mathrm{1}\:−\:\mathrm{cos}\theta\right)\:=\:\mathrm{0}.\mathrm{5} \\ $$$$\mathrm{Rsin}\theta\:=\:\mathrm{4} \\ $$$$\mathrm{R}\:=\:? \\ $$$$\theta\:=\:? \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 10/Apr/19

R×2sin(θ/2)cos(θ/2)=4  R(2sin^2 (θ/2))=(1/2)  ((2Rsin^2 (θ/2))/(2Rsin(θ/2)cos(θ/2)))=(1/(2×4))  tan(θ/2)=(1/8)→θ=2tan^(−1) ((1/8))  sin(θ/2)=(1/(√(8^2 +1^2 )))=(1/(√(65)))  R×2sin^2 (θ/2)=(1/2)  R×2×(1/(65))=(1/2)  R=((65)/4)

$${R}×\mathrm{2}{sin}\frac{\theta}{\mathrm{2}}{cos}\frac{\theta}{\mathrm{2}}=\mathrm{4} \\ $$$${R}\left(\mathrm{2}{sin}^{\mathrm{2}} \frac{\theta}{\mathrm{2}}\right)=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\frac{\mathrm{2}{Rsin}^{\mathrm{2}} \frac{\theta}{\mathrm{2}}}{\mathrm{2}{Rsin}\frac{\theta}{\mathrm{2}}{cos}\frac{\theta}{\mathrm{2}}}=\frac{\mathrm{1}}{\mathrm{2}×\mathrm{4}} \\ $$$${tan}\frac{\theta}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{8}}\rightarrow\theta=\mathrm{2}{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{8}}\right) \\ $$$${sin}\frac{\theta}{\mathrm{2}}=\frac{\mathrm{1}}{\sqrt{\mathrm{8}^{\mathrm{2}} +\mathrm{1}^{\mathrm{2}} }}=\frac{\mathrm{1}}{\sqrt{\mathrm{65}}} \\ $$$${R}×\mathrm{2}{sin}^{\mathrm{2}} \frac{\theta}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${R}×\mathrm{2}×\frac{\mathrm{1}}{\mathrm{65}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${R}=\frac{\mathrm{65}}{\mathrm{4}} \\ $$

Commented by math1967 last updated on 10/Apr/19

tan(θ/2)=(1/8) ∴ θ=2tan^(−1) (1/8)

$${tan}\frac{\theta}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{8}}\:\therefore\:\theta=\mathrm{2tan}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{8}}\:\: \\ $$

Commented by tanmay.chaudhury50@gmail.com last updated on 10/Apr/19

thank you dada...vote ar ki khabar...kolkata te

$${thank}\:{you}\:{dada}...{vote}\:{ar}\:{ki}\:{khabar}...{kolkata}\:{te} \\ $$

Commented by math1967 last updated on 11/Apr/19

Jor prochar cholche

$${Jor}\:{prochar}\:{cholche} \\ $$

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