\(A = 28^\circ, b = 14, c = 10\)
Since we do not have a side-sine pair, we must use the Law of Cosines.
This problem is of the SAS type: we are given two sides and the included angle (the angle between the two given sides).
First, we will solve for the side opposite the given angle :
By the Law of Cosines, \(a^2=b^2+c^2-2ab\cos A=196+100-280\cos28^\circ\approx48.77\), so \(a=6.98\).
Next we will use the Law of Sines to solve the angle opposite the smaller of the two given sides. It's important that we solve for the angle opposite the smaller of the two sides, since we know that it will not be obtuse. The angle opposite the largest side of a triangle may be an obtuse angle, that is, it may be greater than \(90^\circ\). It is best to find obtuse angles last by subtracting the other two angles from \(180^\circ\).
\(\dfrac{\sin C}{10}=\dfrac{\sin28^\circ}{6.98}\), so \(\sin A=\dfrac{10\sin28^\circ}{6.98}=0.6722\). Therefore, \(C=\arcsin(0.6722)=42.2^\circ\).
\(B = 180^\circ - 28^\circ - 42.2^\circ = 109.8^\circ\)
| \(a=6.98\) |
| \(C=42.2^\circ\) |
| \(B=109.8^\circ\) |