In a triangle $a:b:c =4:5:6$, then $3A+B$ equals to?

Question

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 $\pi - C$.

Answer 1

$$\cos{A}=\frac{5^2+6^2-4^2}{2\cdot5\cdot6}=\frac{3}{4}.$$ 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!

Answer 2

HINT:

We have $\dfrac a4=\dfrac b5=\dfrac c6=k$(say)

$\implies a=4k$ etc.

Use cosine formula

$$\cos A=\dfrac{b^2+c^2-a^2}{2bc}=\cdots=\dfrac{45}{60}>\dfrac12\implies0<A<60^\circ$$

and $$\cos B=\dfrac9{16}$$

$$\cos3A=-\dfrac9{16}$$

Now use How do I prove that $\arccos(x) + \arccos(-x)=\pi$ when $x \in [-1,1]$?