Trigonometric relationship in a triangle If in a triangle $ABC$, $$3\sin^2B+4\sin A\sin B+16\sin^2A-8\sin B-20\sin A+9=0$$ find the angles of the triangle. I am unable to manipulate the given expression. Thanks.

The equation can be rearranged as
$$(16\sin A+2\sin B-10)^2+44(\sin B-1)^2=0\ .$$
The only way this can happen is if
$$16\sin A+2\sin B-10=0\ ,\quad \sin B-1=0\ ;$$
these equations are easily solved to give
$$\sin A=\frac12\ ,\quad \sin B=1\ ,$$
and since $A,B$ are angles in a triangle we have
$$A=\frac\pi2\ ,\quad B=\frac\pi6\ .$$
