Question

If $r>0,-\pi \le \theta \le \pi$ and r, $\theta$ satisfy $r\sin \theta =3$ and $r=4(1+\sin \theta )$ then the number of possible solutions of the pair $(r,\theta )$ is

• A. 2
• B. 4
• C. 0
• D. infinite

Answer 1

Answer: (A)

2

$rsin\theta =3$ and $r=4(1+sin\theta )$
using these equation,
$4(1+sin\theta )sin\theta =3$
$\Rightarrow 4si{n}^2\theta +4sin\theta -3=0$
$\Rightarrow (2sin\theta -1)(2sin\theta +3)=0$
but $-1\le sin\theta \le 1$,so
$\Rightarrow sin\theta =1/2$
Therefore solution in $-\pi \le \theta \le \pi$ is,
$\theta =\frac{\pi }{6},\frac{5\pi }{6}$
and $r=6$
Thus, number of possible pair $(r,\theta )$ is 2.
Hence, option 'A' is correct.