Solving Triangles - Part 2

The area of the triangle is $\frac{1}{2}bc \sin A=\frac{1}{2}ab \sin C=\frac{1}{2}ac \sin B$ from which we have that

$\sin A=\frac{a}{c} \sin C$ and $\sin B=\frac{b}{c} \sin C$ . Given that $c=2$ and $C=\frac{\pi}{3}$ we have $\sin A=\frac{\sqrt{3}a}{4}$ and $\sin B=\frac{\sqrt{3}b}{4}$ . We need to find $ab$ to use the formula above since we know $\sin C$ . The first equation comes from the cosines law

$c^{2}=a^{2}+b^{2}-2ab\cos C$ , this is $\boxed{a^{2}+b^{2}=ab+4} \,\ (1)$ .

Now simplify the given expression

$2 \sin 2A=\sin C-\sin (2B+C)\\ 2 \sin 2A=2\cos (\frac{C+2B+C}{2}) \sin (\frac{C-2B-C}{2}) \\ \sin 2A=\cos (B+C)\sin(-B)$

Notice that $A+B+C=\pi \implies B+C=\pi -A \implies \sin (B+C)=\sin (\pi-A)=\sin A$ . Therefore $\cos(B+C)=\pm \sqrt{1-\sin ^{2} A}\\$ and $\cos^{2}(B+C)=\cos^{2}A$ . Then squaring both sides of the equation above gives

$\sin^{2} 2A=(-\sin B)^{2}\cos^{2} A\\ 4\sin^{2} A \cos^{2} A=\sin^{2} B\cos^{2} A\\ \cos^{2} A(4\sin^{2} A-\sin^{2} B)=0\\ (\cos^{2} A)(2\sin A-\sin B)(2\sin A+\sin B)=0\\\\ \cos A=0 \,\ or \,\ 2\sin A=\sin B \,\ or \,\ 2\sin A=-\sin B$

In the first case A would be the right angle and the equation $(1)$ would give us that $a^{2}+b^{2}=c^{2}=4$ but $a^{2}+b^{2}=ab+4$ therefore $ab=0 \implies the \,\ area \,\ is \,\ 0$ The third case, though, would give us the negative value of $a$ or $b$ . So we have that

$2\sin A=\sin B\\ 2 \times \frac{\sqrt{3}a}{4}=\frac{\sqrt{3}b}{4} \implies \boxed{b=2a} \,\ (2)$

We can finally combine $(1)$ and $(2)$

$a^{2}+4a^{2}=a\times 2a+4 \implies a=\frac{2}{\sqrt{3}}$ and $b= \frac{4}{\sqrt{2}}$ .

The area of the triangle is $\frac{1}{2}ab \sin C=\frac{1}{2}\times \frac{2}{\sqrt{3}}\times \frac{4}{\sqrt{3}}\times \frac{\sqrt{3}}{2}=\boxed{\frac{2}{\sqrt{3}}}$ . The final answer is therefore $2\times 3^{2}=18$ . :)

$\begin{aligned} 2\sin 2A + \sin (2B+C) & = \sin C \\ 2\sin 2A + \sin \left(2 \left(\pi - A - C\right)+C\right) & = \sin C & \small \color{#3D99F6} \text{Given that }C = \frac \pi 3 \\ 2\sin 2A + \sin \left(\frac 53 \pi - 2A \right) & = \frac {\sqrt 3}2 \\ 2\sin 2A - \sin \left(2A + \frac \pi 3 \right) & = \frac {\sqrt 3}2 \\ 2\sin 2A - \frac 12 \sin 2A - \frac {\sqrt 3}2\cos 2A & = \frac {\sqrt 3}2 \\ \frac 32 \sin 2A - \frac {\sqrt 3}2\cos 2A & = \frac {\sqrt 3}2 & \small \color{#3D99F6} \text{Divide by } \sqrt 3 \text{ throughout.} \\ \frac {\sqrt 3}2 \sin 2A - \frac 12\cos 2A & = \frac 12 \\ \sin \left(2A - \frac \pi 6\right) & = \frac 12 \\ 2A - \frac \pi 6 & = \frac \pi 6 \\ \implies A & = \frac \pi 6 \end{aligned}$

$\implies B = \frac \pi 2$ , $a = \frac 2{\sqrt 3}$ and area of $\triangle ABC$ is $\frac 12 a c = \frac 2{\sqrt 3}$ . Therefore $pq^2 = \boxed {18}$ .