For a given constant $b > 10,$ there are two possible triangles $ABC$ satisfying $AB = 10,$ $AC = b,$ and $\sin B = \frac{3}{5}.$ Find the positive difference between the lengths of side $\overline{BC}$ in these two triangles.
I'm not sure how to set up a system of equations to solve for the side lengths of BC in both triangles. Can I solve this in another way other than using algebra? How do I incorporate trig functions in this problem?
This question is from a mock contest made specifically for a contest math class I am taking, I do not have a public link to cite it with.
Apply the sine rule
\begin{align} BC & = \frac{\sin A}{\sin B}b = \frac{\sin(B +C)}{\sin B}b = b(\cos C+ \cot B\sin C)\tag 1 \end{align} where $\cot B= \pm \frac43$ from $\sin B=\frac35$ and
$$\sin C= \frac{10}b \sin B=\frac6b, \>\>\>\>\> \cos C= \frac1b \sqrt{b^2-36} $$
Substitute above into (1) to get $BC_{1,2} = \sqrt{b^2-36} \pm 8$, which yields the difference $$BC_1-BC_2 = 16$$