We can use the law of sines:
\(\begin{array}{rcl} \frac{\sin A}{a}&\,=\,&\frac{\sin B}{b}\\~\\ \frac{\sin(50°)}{3}&\,=\,&\frac{\sin B}{2}\\~\\ 2\cdot\frac{\sin(50°)}{3}&\,=\,&\sin B\\~\\ \frac{2}{3}\sin(50°)&=\,&\sin B \end{array} \\ \ \\~\\ B\approx30.71°\quad\text{or}\quad B\approx149.29° \)
There are infinitely many values of B that produce the same sin value, but we only want to consider those
that are in the interval (0, 180°) because no angle in a triangle can have a measure outside that interval.
This is how we find those two solutions:
\(B\,=\,\arcsin(\frac{2}{3}\sin(50°))\qquad\text{or}\qquad B\,=\,180°-\arcsin(\frac{2}{3}\sin(50°))\)
But in this case, B ≈ 149.29° can't be a solution because we already have an angle that is 50°, and
149.29° + 50° > 180°
So we know B ≈ 30.71° is the only solution for this problem. So there is only one possible triangle.
Now we can find the measure of angle C .
C = 180° - A - B ≈ 180° - 50° - 30.71° ≈ 99.29°
All we are missing is the length of side c. We can find it using the law of cosines.
c2 = a2 + b2 - 2ab cos C
c2 ≈ 32 + 22 - 2(3)(2) cos( 99.29° )
c2 ≈ 13 - 12 cos( 99.29° ) c is a length, so take the positive sqrt of both sides
c ≈ √[ 13 - 12 cos( 99.29° ) ] Plug this into a calculator
c ≈ 3.865
Now we have found:
B ≈ 30.71°
C ≈ 99.29°
c ≈ 3.865
