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

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