# $\frac{\sin^2 A}{1+2 \cos^2 A}=\frac{3}{19},90^\circ <A < 180^\circ$

Given that $$\frac{\sin^2 A}{1+2 \cos^2 A}=\frac{3}{19},90^\circ <A < 180^\circ$$

find the value of $$\frac{\sin A}{1+2 \cos A}$$

without using a calculator.

I've no idea how to solve it. Can anyone give me some hints? Thanks in advance.

$$19\sin^2 A=3(1+2(1-\sin^2 A))$$
Find $\sin A$ first.