In a triangle $a:b:c =4:5:6$, then $3A+B$ equals to? In the above question $a,b,c$ are sides of triangle and $A,B,C$ are angles. The correct answer is $\pi$ but I am getting $\pi - C$.

$$\cos{A}=\frac{5^2+6^2-4^2}{2\cdot5\cdot6}=\frac{3}{4}.$$
Thus, $$\sin{A}=\sqrt{1-\frac{9}{16}}=\frac{\sqrt{7}}{4}<\frac{1}{\sqrt2},$$
which says that $A<45^{\circ}$, which gives that $2A$ is an acute angle
and $$\sin2A=2\cdot\frac{\sqrt7}{4}\cdot\frac{3}{4}=\frac{3\sqrt7}{8}.$$
In another hand, $$\cos{C}=\frac{4^2+5^2-6^2}{2\cdot4\cdot5}=\frac{1}{8},$$
which gives $$\sin{C}=\sqrt{1-\frac{1}{64}}=\frac{3\sqrt7}{8}.$$
Thus, $C=2A$ and
$$3A+B=2A+A+B=C+A+B=180^{\circ}$$
and we are done!
