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