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.
 A: Let $X = \sin A$ and $Y = \sin B$. Then we have
$$16X^2 + 4XY + 3Y^2 - 20X - 8Y + 9 = 0$$
Rearranging the terms to solve for Y, we get
$$3Y^2 + (4X-8)Y + (16X^2 - 20X + 9) = 0$$
The discriminant for this quadratic equation becomes
$$(4X-8)^2 - 4*3*(16X^2-20X+9) = 16X^2 - 64X + 64 - 192X^2 + 240X - 108 = -176X^2 + 176X - 44 = -44(4X^2 - 4X + 1) = -44 (2X-1)^2$$
This must equal $0$ (or $Y$ will not be a real number), which means that $$X = \frac{1}{2}$$.
Plugging this back into the original equation, we get
$$3Y^2 - 6Y + 3 = 3 (Y-1)^2 = 0$$
which means that $$Y = 1$$
Since $Y = \sin B$, we now know that $B$ must be $90$ degrees. Because $X = \sin A$, we know that $A$ is either $30$ degrees or $150$ degrees, but only the former makes sense because the sum of a triangle's internal angles cannot exceed $180$ degrees.
As a final answer, we get that $ABC$ is a $30-60-90$ triangle, with $B$ being the right angle and $A$ being $30$ degrees.
A: 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
$$A=\frac\pi2\ ,\quad B=\frac\pi6\ .$$
