Question

A given chord $AB$ of a given circle subtends an angle $\theta$ at a point $C$ on the circumference, then the triangle $ABC$ 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 $A=\frac{\pi }{2}-\frac{\theta }{2},B=\frac{\pi }{2}-\frac{\theta }{2}$

From the figure;
Area of triangle($\Delta c\times h/2$)=$c\times h/2$, where c is constant and $h$ is variable.
${\Delta }_{max}$ = $c\times h_{max}/2$.
h_{max} clearly from the figure is when angles A and B are same.
So if angle $C$ $=$ $\theta$, $A=B=$ $\pi /2-\theta /2$ for maximum area of triangle $ABC$.