Question

The maximum area of a triangle formed by the points $(0,0)$ , $(\sin \theta ,0)$ and $(0,\cos \theta )$ is obtained when $\theta$ equals

• A. $\pi /2$
• B. $\pi /3$
• C. $\pi /4$
• D. None

Answer 1

Answer: (C)

$\pi /4$

The area of triangle formed by points $(x_1,y_1)$ , $(x_2,y_2)$ and $(x_3,y_3)$ is $|\frac{1}{2}\left(x_1\right(y_1-y_2)+x_2(y_2-y_3)+x_3(y_3-y_1)|$

So the area of triangle with vertices $(0,0)$ , $(\sin \theta ,0)$ and $(0,\cos \theta )$ is $|\frac{1}{2}\left(0\right(0-\cos \theta )+\sin \theta (\cos \theta -0)+0(0-0)|$

$\Rightarrow |\frac{1}{2}\sin \theta \cos \theta |=\frac{1}{4}|\sin \left(2\theta \right)|$

The area is maximum when $\sin \left(2\theta \right)$ is maximum.

The maximum value of $\sin \left(2\theta \right)$ is $1$ and occurs at $2\theta =\frac{\pi }{2}$

$\Rightarrow \theta =\frac{\pi }{4}$

Therefore option $C$ is correct.