Question

If $A\left(0 ,0\right), \, B\left(\theta , cos \theta \right)$ and $C\left(\left(sin\right)^{3} \theta , 0\right)$ are the vertices of a triangle $A B C,$ then the value of $\theta$ for which the triangle has the maximum area is (where $\left.\theta \in\left(0, \frac{\pi}{2}\right)\right)$

• A. $\frac{\pi }{6}$
• B. $\frac{\pi }{4}$
• C. $\frac{\pi }{3}$
• D. $\frac{\pi }{2}$

Answer 1

Answer: (C)

Area of triangle $ABC,$ $\triangle =\begin{vmatrix} \frac{1}{2} \begin{vmatrix} 0 & 0 & 1 \\ \theta & cos \theta & 1 \\ sin^{3}⁡\theta & 0 & 1 \end{vmatrix} \end{vmatrix}$
$\Rightarrow \triangle =\frac{1}{2}sin^{3} \theta cos⁡\theta $
$\frac{d \triangle }{d \theta }=\frac{3 sin^{2} \theta cos^{2} ⁡ \theta - sin^{4} ⁡ \theta }{2}=0$
$\Rightarrow \left(sin\right)^{2} \theta \left(3 \left(cos\right)^{2} ⁡ \theta - \left(sin\right)^{2} ⁡ \theta \right)=0$
$\Rightarrow sin \theta =0$ or $tan^{2} \theta =3$
$\theta =0$ or $\theta =\frac{\pi }{3}$
$\left(\frac{d^{2} \triangle }{d \left(\theta \right)^{2}}\right) < 0for\theta =\frac{\pi }{3}$