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 $\pi - C$.

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 $x \in [-1,1]$?
