Question

An isosceles triangle of vertical angle $2\theta$ is inscribed in a circle of radius a . Show that the area of the triangle is maximum when $\theta =\frac{\pi }{6}$ .

Answer 1

$BD=\sqrt{a^2-x^2}$

Area $=\frac{1}{2}.BC.BD$

$A^{\prime }=1.\sqrt{a^2-x^2}+(a-x).\frac{1}{2}\frac{(a-2x)}{\sqrt{a^2-x^2}}$

$=\frac{a^2-x^2+(a+x)(-x)}{\sqrt{a^2-x^2}}$

$=\frac{a^2-ax-2x^2}{\sqrt{a^2-x^2}}=0$

$\frac{+-}{x=a/2}\Rightarrow$ maximum

$BD=\sqrt{a^2-\frac{a^2}{4}}=\frac{\sqrt{3}a}{2}$

$2x^2+ax-a^2=0$

$x=-a,a/2$

$x=a/2$

$\tan \angle BOD=\frac{BD}{OD}=\frac{\sqrt{2}a/2}{a/2}=\sqrt{3}$

$\angle BOD=\pi /3$

$\angle BOC=2\pi /3$

$\angle BAC=\frac{\angle BOC}{2}$

$=\pi /3=2\theta$

$2\theta =\pi /3$

$\theta =\pi /6$