Question

Two sides of a triangle have lengths 'a' and 'b' and the angle between them is $\theta$. What value of $\theta$ will maximize the area of the triangle? Find the maximum area of the triangle also. [CBSE 2002 C]

Answer 1

$$\begin{gathered} \mathrm{As},\mathrm{the}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{triangle},A=\frac{1}{2}ab\sin \theta \\ \Rightarrow A\left(\theta \right)=\frac{1}{2}ab\sin \theta \\ \Rightarrow A\text{'}\left(\theta \right)=\frac{1}{2}ab\cos \theta \\ \mathrm{For}\mathrm{maxima}\mathrm{or}\mathrm{minima},A\text{'}\left(\theta \right)=0 \\ \Rightarrow \frac{1}{2}ab\cos \theta =0 \\ \Rightarrow \cos \theta =0 \\ \Rightarrow \theta =\frac{\mathrm{\pi }}{2} \\ \mathrm{Also},A\text{'}\text{'}\left(\theta \right)=-\frac{1}{2}ab\sin \theta \\ \mathrm{or},A\text{'}\text{'}\left(\frac{\mathrm{\pi }}{2}\right)=-\frac{1}{2}ab\sin \frac{\mathrm{\pi }}{2}=-\frac{1}{2}ab<0 \\ \mathrm{i}.\mathrm{e}.\theta =\frac{\mathrm{\pi }}{2}\mathrm{is}\mathrm{point}\mathrm{of}\mathrm{maxima} \\ \mathrm{Now}, \\ \mathrm{The}\mathrm{maximum}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{triangle}=\frac{1}{2}ab\sin \frac{\mathrm{\pi }}{2}=\frac{ab}{2} \end{gathered}$$