$ABC$ is a triangle with perimeter of $20$; length of the opposite side of $A$ is $4$, Find $\sin(\dfrac{A}{2}):\cos(\dfrac{B-C}{2})$
My attempt:
From $A+B+C=\pi$ gives $\sin(\dfrac{A}{2}):\cos(\dfrac{B-C}{2})=\cos(\dfrac{B+C}{2}):\cos(\dfrac{B-C}{2})$
Let opposite sides of the angles called $a,b,c$
From Law of Tangents $\dfrac{\tan(\frac{B+C}{2})}{\tan(\frac{B-C}{2})}=\dfrac{b+c}{b-c}$
gives $\dfrac{\cos(\frac{B+C}{2})}{\cos(\frac{B-C}{2})}=\dfrac{(b+c)\sin(\frac{B+C}{2})}{(b-c)\sin(\frac{B-C}{2})}$
There are quite a lot of unknowns but I still can't use length info that the problem given.
Let the lengths of the sides opposite to $B$ and $C$ be $b$ and $c$ respectively.
From Law of Sines we have $$\frac{4}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\Longrightarrow \sin A:\sin B:\sin C=4:b:c$$
Also note that $$\sin B+\sin C=2\sin\frac{B+C}{2}\cos\frac{B-C}{2}=2\cos\frac{A}{2}\cos\frac{B-C}{2}\\ \\\Longrightarrow\sin\frac{A}2(\sin B+\sin C)=\sin A\cos\frac{B-C}2$$
Therefore we have $$\sin \frac{A}2:\cos\frac{B-C}2=\sin A:(\sin B+\sin C)=4:(b+c)=4:16=\color{blue}{1:4}$$ which is the answer.