Given a triangle $\triangle ABC$, which fulfills the equations $6\cos{A}+2\cos{B}=7$ and $6\sin{A}-2\sin{B}=\sqrt{3}$, find the value of $\angle C$

Question

Given a triangle $\triangle ABC$, which fulfills the equations $6\cos{A}+2\cos{B}=7$ and $6\sin{A}-2\sin{B}=\sqrt{3}$, find the value of $\angle C$.

I though of solving it as follows:

$36\cos^2{A}+24\cos{B}\cos{A}+4\cos^2{B}=49$

$36\sin^2{A}-24\sin{B}\sin{A}+4\sin^2{B}=3$

Hence $12=24(\cos{A}\cos{B}-\sin{B}\sin{A})$

And I got stuck here. I then tried putting in different values for $\angle A, \angle B$ to find a pair that works, but I couldn't find one. Could you please explain to me how to solve the question?

Answer 1

Hint

You have done the hard work!

Use $\cos(A+B)$ formula

Finally use

$$\cos(A+B)=\cos(\pi-C)=?$$

Answer 2

$6\cos{A}+2\cos{B}=7$ and $6\sin{A}-2\sin{B}=\sqrt{3}$ Squaring and adding these two equations, we get $$36(\sin^2 A+\cos^2 A)+4(\cos^2 B+\sin^2 b)+24(\cos A \cos B-\sin A \sin B)=52$$ $$\implies 36+4+24\sin (A+B)=52 \implies \cos(A+B)=1/2=-\cos C \implies C= 2\pi/3$$