Find the angles using the given conditions 
I got stuck in last step. What should I do next? Please help me
 A: $$
\text{Let } a = \frac{\alpha}{2}, b = \frac{\beta}{2}, c = \frac{\gamma}{2} \\
\sin{\frac{\alpha-\beta}{2}}+\sin{\frac{\alpha-\gamma}{2}}+\sin{\frac{3\alpha}{2}} = \frac{3}{2} \\
\sin\left(a-b\right)+\sin\left(a-c\right)+\sin{3a} = \frac{3}{2} 
\\
\sin\left(a-b\right)+\sin\left(a-c\right)+\sin\left(a-b+a-c+a-a+a+b+c\right) = \frac{3}{2}\\
\sin\left(a-b\right)+\sin\left(a-c\right)+\sin\left(a-b+a-c+90^{\circ}\right) = \frac{3}{2}\\
\sin\left(a-b\right)+\sin\left(a-c\right)+\cos\left(a-b+a-c\right) = \frac{3}{2}\\
\text{So, } a-b=30^{\circ},a-c=30^{\circ} \text{(messy proof below)}
$$
Do the rest and you should get $\alpha=100^{\circ}, \beta=40^{\circ}, \gamma=40^{\circ}$

Proof that $\sin x +\sin y + \cos(x+y) = \frac{3}{2} \implies x = y= 30^{\circ}$
It actually turns out that $\frac{3}{2}$ is the maximum of that two dimensional function. Finding the partial derivatives wrt $x$ and equating it to zero,
$$
\cos x = \sin(x+y) \\
\sin\left(\frac{\pi}{2}-x\right) = \sin(x+y) \\
y = \frac{\pi}{2}-2x
$$
Note that in general, $\left(\frac{\pi}{2}-x\right) = n\pi+(-1)^n(x+y)$. But for the range($0\leq x,y< 90^{\circ}$) we are working here this is the only possible condition. Similarly, the other partial derivative gives,
$$
x = \frac{\pi}{2}-2y
$$
Solving we get that $\sin x +\sin y + \cos(x+y) = \frac{3}{2}, 0\leq x,y< 90^{\circ}$ only if $x = y = 30^{\circ}$. 
I know this is not the most elegant proof, but it works and if you know a better way, please comment/answer.
