## University Calculus: Early Transcendentals (3rd Edition)

$$a=2\sqrt3-2$$
$$c = 2 \hspace{1cm}A=\frac{\pi}{4}\hspace{1cm}B=\frac{\pi}{3}$$ - From Exercise 61, the law of sines: $$\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}$$ To find $a$, we still need to find $C$. Notice that we already have 2 angles of the triangle here, which means $$C=\pi-A-B$$ $$C=\pi-\frac{\pi}{4}-\frac{\pi}{3}$$ $$C=\frac{12\pi-3\pi-4\pi}{12}$$ $$C=\frac{5\pi}{12}$$ So $$\sin C=\sin\frac{5\pi}{12}=\sin\Big(\frac{\pi}{6}+\frac{\pi}{4}\Big)=\sin\frac{\pi}{6}\cos\frac{\pi}{4}+\cos\frac{\pi}{6}\sin\frac{\pi}{4}$$ $$\sin C=\frac{1}{2}\frac{\sqrt2}{2}+\frac{\sqrt3}{2}\frac{\sqrt2}{2}=\frac{\sqrt2}{4}+\frac{\sqrt6}{4}=\frac{\sqrt2+\sqrt6}{4}$$ - Now to find $a$, we use the law of sines: $$\frac{\sin A}{a}=\frac{\sin C}{c}$$ $$a=\frac{c\sin A}{\sin C}$$ $$a=\frac{2\sin\frac{\pi}{4}}{\frac{\sqrt2+\sqrt6}{4}}=\frac{2\times\frac{\sqrt2}{2}}{\frac{\sqrt2+\sqrt6}{4}}=\frac{\sqrt2}{\frac{\sqrt2+\sqrt6}{4}}=\frac{4\sqrt2}{\sqrt2+\sqrt6}$$ - Multiply both numerator and denominator by $(\sqrt2-\sqrt6)$: $$a=\frac{4\sqrt2(\sqrt2-\sqrt6)}{(\sqrt2+\sqrt6)(\sqrt2-\sqrt6)}=\frac{4\times2-4\sqrt{12}}{2-6}=\frac{8-8\sqrt3}{-4}$$ $$a=2\sqrt3-2$$