If none of the angles of a triangle is a right angle, the triangle is called oblique triangle. An oblique triangle is labeled such that side a is opposite to angle A, side b is opposite to angle B and side c is opposite to angle C. To solve triangle means to find its lengths and angles. The law of sine is follows:
\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} \\ \\
Example:
Solve for triangle with A = 40^\circ, \: B =20^\circ, \: a =2
From sine law, \frac{a}{\sin A} = \frac{b}{\sin B} \\
or, \frac{2}{\sin 40} = \frac{b}{\sin 20} \\
or, b = \frac{2 \sin 20}{\sin 40} = 1.1 \\
C = 180^\circ - (40^\circ + 20^\circ) = 120^\circ \\\frac{a}{\sin A} = \frac{c}{\sin C} \\ \\
\frac{2}{\sin 40} = \frac{c}{\sin 120} \\or, c = \frac{2 \sin 120}{\sin 40} = 2.7
Example:
Solve for triangle with b= 4, \: c= 6, \: B =20^\circ \\ \\
From sine law, \frac{b}{\sin B} = \frac{c}{\sin C} \\ \\
or, \frac{4}{\sin 20} = \frac{6}{\sin C} \\ \\
or, \sin C = \frac{6 \sin 20}{4} = 0.51 = \sin 30.9
or, C_1 = 30.9^\circ
Again, \sin C = \sin 30.9 = \sin (180-30.9)
So, C_2 = 149.1^\circ
A_1 = 180 - (30.9 + 20) = 129.1^\circ \\A_2 = 180 - (149.1 + 20) = 10.9^\circ \\ \\
\frac{a_1}{\sin A_1} = \frac{b}{\sin B} \\or, \frac{a_1}{\sin 129.1} = \frac{4}{\sin 20} \\
or, a_1 = \frac{4 \sin 129.1}{\sin 20} = 9.1 \\
a_2 = \frac{b \sin A_2}{\sin B} = \frac{4 \sin 10.9}{\sin 20} = 2.2