\(a = 10, b = 4, B = 20^\circ\)
\(\dfrac{\sin A}{10}=\dfrac{\sin 20^\circ}{4}\) so \(\sin A=\dfrac{10\sin20^\circ}{4}=0.8550\)
There are two angles between \(0^\circ\) and \(180^\circ\) whose sine is \(0.8550\): \(58.8^\circ\) and \(121.2^\circ\).
Since the larger of the two angles, when added to the given angle \(20^\circ\) is only \(141.2^\circ\), which is still less than \(180^\circ\), the larger angle is a possible solution. There will be two different solutions to this triangle.
Solution #1
\(A = 58.8^\circ, C = 180^\circ - 2^\circ - 58.8^\circ = 101.2^\circ\).
Then \(\dfrac{c}{\sin101.2^\circ}=\dfrac{4}{\sin20^\circ}\) so \(c=\dfrac{4\sin101.2^\circ}{\sin20^\circ}=11.4\).
So one solution is \(A = 58.8^\circ, C = 101.2^\circ, c=11.4\)
Solution #2
\(A = 121.2^\circ, C = 180^\circ - 20^\circ - 121.2^\circ = 38.8^\circ\)
Then \(\dfrac{c}{\sin38.8^\circ}=\dfrac{4}{\sin20^\circ}\) so \(c=\dfrac{4\sin38.8^\circ}{\sin20^\circ}=7.3\)
So a second solution is \(A = 121.2^\circ, C = 38.8^\circ, c=7.3\)