Question

Let $P(\sin \theta ,\cos \theta )(0\le \theta \le 2\pi )$ be a point inside the triangle with vertices $(0,0),(\sqrt{\frac{3}{2}},0)$ and $(0,\sqrt{\frac{3}{2}})$. Then,

• A. $0<\theta <\frac{\pi }{12}$
• B. $\frac{5\pi }{12}<\theta <\frac{\pi }{2}$
• C. $0<\theta <\frac{5\pi }{12}$
• D. $\frac{5\pi }{12}<\theta <\pi$

Answer 1

Answer: (A), (B)

The correct options are
A $0<\theta <\frac{\pi }{12}$
B $\frac{5\pi }{12}<\theta <\frac{\pi }{2}$

Equations of lines along $OA,OB$ and $AB$ are $y=0,x=0$ and $x+y=\sqrt{\frac{3}{2}}$ respectively.
Now $P$ and $B$ will lie on the same side of $y=0$ if $\cos \theta >0$.

Similarly, $P$ and $A$ will lie on the same side of $x=0$ if $\sin \theta >0$ and $P$ and $O$ will lie on the same side of $x+y=\sqrt{\frac{3}{2}}$ if $\sin \theta +\cos \theta <\sqrt{\frac{3}{2}}$.

Hence, $P$ will lie inside the $\Delta$ $ABC$ if $\sin \theta >0,\cos \theta >0$ and $\sin \theta +\cos \theta <\sqrt{\frac{3}{2}}.$ Now,
$\sin \theta +\cos \theta <\sqrt{\frac{3}{2}}$
$\Rightarrow \sin (\theta +\frac{\pi }{4})<\sqrt{\frac{3}{4}}$
Since $\sin \theta >0$ and $\cos \theta >0$

So $0<\theta <\frac{\pi }{12}$ or $\frac{5\pi }{12}<\theta <\frac{\pi }{2}$