Question

Two triangles can be formed with the given information. Use the Law of Sines to solve the triangles.
A = 56°, a = 12, b = 14

Answer 1

I'm having trouble solving for c, I've already been able to solve for both B1 and B2 as well as C2 and C1.

Answer 2

Eyeball this
http://hotmath.com/hotmath_help/topics/law-of-sines.html
and take a stab at it.

Answer 3

I'm assuming side \(a\) is the side opposite angle \(A\), and likewise for \(b/B\) and \(c/C\).

The law of sines gives you
\[\begin{align*}\frac{\sin A}{a}&=\frac{\sin B}{b}\\
\frac{\sin56^\circ}{12}&=\frac{\sin B}{14}
\end{align*}\]
Solve for angle \(B\).

The interior angles of a triangle add up to 180°, which means
\[A+B+C=180^\circ\]
You have \(A\) already, you'll have \(B\) from the law of sines equation, so solve for \(C\).

Another application of law of sines to find the remaining side \(c\):
\[\begin{align*}\frac{\sin A}{a}&=\frac{\sin C}{c}\\
\frac{\sin56^\circ}{12}&=\frac{\sin C}{c}
\end{align*}\]

Answer 4

I've been using that same formula as said in your link, but get a completely different answer.
\[12\div \sin56 = c \div \sin48.7\]

Answer 5

Show your work, what you got?

Answer 6

A = 56
B1 = 75.3
C1 = 48.7
B2 = 104.7
C2 = 19.3
a = 12
b = 14
@agent0smith

Answer 7

The answers are:
B = 75.3°, C = 48.7°, c = 10.9; B = 104.7°, C = 19.3°, c = 4.8
B = 14.7°, C = 109.3°, c = 13.7; B = 165.3°, C = 70.7°, c = 13.7
B = 75.3°, C = 48.7°, c = 13.2; B = 104.7°, C = 19.3°, c = 13.2
B = 14.7°, C = 109.3°, c = 10.5; B = 165.3°, C = 70.7°, c = 10.5