There is a triangle, sides = a,b,c and its angle = $\alpha, \beta, \gamma$

Question

$b=2\sqrt3$

$\sin(\alpha + \beta) + \cos(2\alpha-\beta) = 2$

Find a

I tried $\frac{\sin\alpha}{a}=\frac{\sin\beta}{2\sqrt3}$

Or, Using identities $\sin(\alpha + \beta) = \sin\alpha\cdot\cos\beta + \cos\alpha\cdot\sin\beta$

But i can't solve it. What formula should i use?

Answer 1

Hint

For $x,\sin x,\cos x\le1$

$\implies\sin(\alpha+\beta)+\cos(2\alpha-\beta)\le2$

So, we need $\sin(\alpha+\beta)=1=\cos(2\alpha-\beta)$

Also, $0<\alpha,\beta<\pi$

$\implies0<\alpha+\beta<\pi$

$0-\pi<2\alpha-\beta<2\pi-0\implies2\alpha-\beta$ has to be $0$