Question

Find the values of $\theta$ lying between $0$ and $2\pi$ satisfying the equation $r\sin \theta =\sqrt{3}$ and $r+4\sin \theta =2(\sqrt{3}+1)$

• A. $\frac{\pi }{3},\frac{\pi }{6}$
• B. $\frac{2\pi }{3},\frac{5\pi }{6}$
• C. $\frac{\pi }{6},\frac{\pi }{3},\frac{2\pi }{3}$ and $\frac{5\pi }{6}$
• D. $Noneofthese$

Answer 1

The correct option is A $\frac{\pi }{3},\frac{\pi }{6}$

Given

$rsin\theta =\sqrt{3}⟹r=\frac{\sqrt{3}}{sin\theta }$

Then $\frac{\sqrt{3}}{sin\theta }+4sin\theta =2(\sqrt{3}+1)$

$⟹\sqrt{3}+4si{n}^2\theta =2(\sqrt{3}+1)sin\theta$

$⟹4si{n}^2\theta -2\sqrt{3}sin\theta -2sin\theta +\sqrt{3}=0$

$⟹(2sin\theta -\sqrt{3})(2sin\theta -1)=0$

$⟹sin\theta =\frac{\sqrt{3}}{2},sin\theta =\frac{1}{2}$

$⟹\theta =\frac{\pi }{3},\frac{\pi }{6}$