Question

A chord AB of a circle subtends an angle $\theta$ at a point C on the circumference, $△ABC$ has the maximum area when

• A. $A=\frac{\pi }{2}-\frac{\theta }{2},B=\frac{\pi }{2}-\frac{\theta }{2}$
• B. $A=\frac{\pi }{4}-\frac{\theta }{2},B=\frac{3\pi }{4}-\frac{3\theta }{2}$
• C. $A=\frac{\pi }{6}+\theta ,B=\frac{5\pi }{6}+2\theta$
• D. None of these

Answer 1

The correct option is A $A=\frac{\pi }{2}-\frac{\theta }{2},B=\frac{\pi }{2}-\frac{\theta }{2}$

The base AB of triangle ABC is fixed.
To maximize the area we have to maximize the height.
Maximum height will be achieved when perpendicular from point C passes through centre, then the triangle will be isosceles triangle.
$\Rightarrow A=B=\frac{\pi }{2}-\frac{\theta }{2}$