Question

Let $P(\sin \theta ,\cos \theta )$, where $0\le \theta \le 2\pi$ be a point and let $OAB$ be a triangle with vertices $(0,0),(\frac{\sqrt{3}}{2},0)$ and $(0,\frac{\sqrt{3}}{2})$. Find $\theta$ if $P$ lies inside the $△OAB$.

Answer 1

In $\Delta OAB$

Given that ,point

$\left(O\right)=(0,0)$

$A=(\frac{\sqrt{3}}{2},0)$

$B=(0,\frac{\sqrt{3}}{2})$

And point $P=(\sin \theta ,\cos \theta )$

Then equation of line $AB$ is $\frac{x}{\frac{\sqrt{3}}{2}}+\frac{y}{\frac{\sqrt{3}}{2}}=1$

$x+y=\frac{\sqrt{3}}{2}$ …… (1)

Now, point $O$ and $P$

Line O

$0+0-\frac{\sqrt{3}}{2}<0$

$=-\frac{\sqrt{3}}{2}<0$

Line.P$\sin \theta +\cos \theta <\frac{\sqrt{3}}{2}$

Squaring both side and we get,

${(\sin \theta +\cos \theta )}^2<{\left(\frac{\sqrt{3}}{\sqrt{2}}\right)}^2$

$\Rightarrow {\sin }^2\theta +{\cos }^2\theta +2\sin \theta \cos \theta <\frac{3}{2}$

$\Rightarrow 1+2\sin \theta \cos \theta <\frac{3}{2}$

$\Rightarrow \sin 2\theta <\frac{3}{2}-1$

$\Rightarrow \sin 2\theta <\frac{1}{2}$

$\Rightarrow \sin 2\theta <\sin \frac{\pi }{6}0\le \theta \le 2\pi$

$\Rightarrow 2\theta \in (0,\frac{\pi }{6})\cup (\frac{\pi }{2}-\frac{\pi }{6},\pi +\frac{\pi }{6})\cup (\frac{3\pi }{2}-\frac{\pi }{6},2\pi )0\le 2\theta \le 4\pi$

$\Rightarrow 2\theta \in (0,\frac{\pi }{6})\cup (\frac{2\pi }{6},\frac{7\pi }{6})\cup (\frac{4\pi }{3},2\pi )$

$\Rightarrow \theta \in (0,\frac{\pi }{12})\cup (\frac{2\pi }{12},\frac{7\pi }{12})\cup (\frac{4\pi }{12},\frac{2\pi }{2})$

Hence, this is the answer.