Trigonometry – Problem 27: Find the maximum area of

Maximum Area Problem

In $\triangle ABC$, given $a = 1$ and $(a+b)\sin A\sin B + (c-b)\sin C = 0$. Find the maximum area $S$ of $\triangle ABC$.

By Law of Sines, $\sin A : \sin B : \sin C = a : b : c$:

$$(a+b)(a-b) + (c-b)c = 0$$

$$a^2 - b^2 + c^2 - bc = 0$$

$$b^2 + c^2 - a^2 = bc$$

By Law of Cosines:

$$\cos A = \frac{b^2 + c^2 - a^2}{2bc} = \frac{bc}{2bc} = \frac{1}{2}$$

So $A = \frac{\pi}{3}$.

From $a^2 = b^2 + c^2 - bc = 1$:

$$1 = b^2 + c^2 - bc \geq 2bc - bc = bc$$

So $bc \leq 1$ (equality when $b = c = 1$).

Maximum area:

$$S = \frac{1}{2}bc\sin A = \frac{\sqrt{3}}{4}bc \leq \frac{\sqrt{3}}{4}$$

$\frac{\sqrt{3}}{4}$