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 $\theta$ = $\frac{\mathrm{\pi }}{6}$.
[NCERT EXEMPLAR]

Answer 1

Let ABC be an isosceles triangle inscribed in the circle with radius a such that AB = AC.

$$\begin{gathered} \mathrm{AD}=\mathrm{AO}+\mathrm{OD}=a+a\cos 2\theta =a\left(1+\cos 2\theta \right)\mathrm{and} \\ \mathrm{BC}=2\mathrm{BD}=2a\sin 2\theta \\ \mathrm{As},\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{triangle}\mathrm{AC},A=\frac{1}{2}\mathrm{BC}\times \mathrm{AD} \\ \Rightarrow A\left(\theta \right)=\frac{1}{2}\times 2a\sin 2\theta \times a\left(1+\cos 2\theta \right) \\ =a^2\sin 2\theta \left(1+\cos 2\theta \right) \\ =a^2\sin 2\theta +a^2\sin 2\theta \cos 2\theta \\ \Rightarrow A\left(\theta \right)=a^2\sin 2\theta +\frac{a^2\sin 4\theta }{2} \\ \Rightarrow A\text{'}\left(\theta \right)=2a^2\cos 2\theta +\frac{4a^2\cos 4\mathrm{\theta }}{2} \\ \Rightarrow A\text{'}\left(\theta \right)=2a^2\cos 2\theta +2a^2\cos 4\theta \\ \Rightarrow A\text{'}\left(\theta \right)=2a^2\left(\cos 2\theta +\cos 4\theta \right) \\ \mathrm{For}\mathrm{maxima}\mathrm{or}\mathrm{minima},A\text{'}\left(\theta \right)=0 \\ \Rightarrow 2a^2\left(\cos 2\theta +\cos 4\theta \right)=0 \\ \Rightarrow \cos 2\theta +\cos 4\theta =0 \\ \Rightarrow \cos 2\theta =-\cos 4\theta \\ \Rightarrow \cos 2\theta =\cos \left(\mathrm{\pi }-4\theta \right) \\ \Rightarrow 2\theta =\mathrm{\pi }-4\theta \\ \Rightarrow 6\theta =\mathrm{\pi } \\ \Rightarrow \theta =\frac{\mathrm{\pi }}{6} \\ \mathrm{Also},A\text{'}\text{'}\left(\theta \right)=2a^2\left(-\sin 2\theta -\sin 4\mathrm{\theta }\right)=-2a^2\left(\sin 2\theta +\sin 4\theta \right)<0\mathrm{at}\theta =\frac{\mathrm{\pi }}{6}. \\ \mathrm{So},\mathrm{the}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{triangle}\mathrm{is}\mathrm{maximum}\mathrm{at}\theta =\frac{\mathrm{\pi }}{6}. \end{gathered}$$