Question

An isosceles triangle with vertical angle $2\theta$ is inscribed in a circle of radius $a$. The area of the triangle will be maximum when $\theta =$

• A. $\pi /6$
• B. $\pi /4$
• C. $\pi /3$
• D. $\pi /2$

Answer 1

The correct option is A $\pi /6$

Let equal side of an isoceles triangle be a.
$a/2=rcos\theta$
Substituting in equation (1)
Area$=1/2\left(2rcos\theta \right)\left(2rcos\theta \right)sin2\theta$ (1)
A=$4r^2{cos}^3\theta sin\theta$
Differentiate wrt to $\theta$ and equate it to zero
${cos}^2\theta ({cos}^2\theta -3{sin}^2\theta )=0$
${tan}^2\theta =1/3$
$\theta =\pi /6$