There are different kinds of triangle: a right, acute and obtuse triangle. Solving for the sides of a right triangle, we commonly used the Pythagorean theorem. How do you solve for the side and angle of a non-right triangle?

### Sine rule

Sine rule determines the relations of the lengths of the sides to its opposite angles. A triangle labelled with capital A, B, and C are the angles of the triangle. a , b, and c are the sides opposite to the angle. Angle A is opposite to side a. According to the Sine rule:

The equations for the sine rule to find the sides of a triangle are:

\(\frac{a}{sinA}=\frac{b}{sinB}=\frac{c}{sinC}\)

It can also be written this way:

\(\frac{sinA}{a}=\frac{sinB}{b}=\frac{sin}{sinC}\)

Note: Sine Rule is used if the angle and the sides opposite to the angle are given.

Notice that the given measure of an angle and its opposite side, and the angle and the opposite side is unknown.

Let’s say that A = 30, a = 10, C = 80 and c = ?

Using sine law, \(\frac{a}{sinA}=\frac{c}{sinC}\) Substitute the given to find the missing side.

\(\frac{a}{sinA}=\frac{c}{sinC}\)

\(\frac{10}{sin30}=\frac{c}{sin80}\) Multiply both sides by sin 80.

\(c=\frac{10sin80}{sin30}\)

c = 19.7

We can also solve for the remaining sides and angles. The sum of all angles of a triangle are 180.

A + B + C = 180 Substitute the angles

30 + B + 80 = 180

B = 180 – 30 – 80

B = 70

Since angle B = 70, we can proceed to get the value of its opposite side.

\(\frac{10}{sin30}=\frac{b}{sin70}\)

\(b=\frac{10sin70}{sin30}\)

b = 18.79

The measure of all angles are A = 30, B = 70 and C=80 and its opposite sides are a = 10, b = 18.79 and c = 19.7.

### Cosine Rule

Cosine rule is used when the given in a triangle are the three sides or two adjacent sides and the angle in between the two sides.

\(a^{2}=b^{2}+c^{2}-2bc\) \(cosA\)

\(b^{2}=a^{2}+c^{2}-2ac\) \(cosB\)

\(c^{2}=a^{2}+b^{2}-2ab\) \(cosC\)

It can be written this way:

\(cosA=\frac{b^{2}+c^{2}-a^{2}}{2bc}\)

\(cosB=\frac{a^{2}+c^{2}-b^{2}}{2ac}\)

\(cosC=\frac{a^{2}+b^{2}-c^{2}}{2ab}\)

Note: Cosine rule is used when the given are the three sides of a triangle or two adjacent sides and the angle.

Example: Given the below triangle, solve for the missing sides and angles. Label the sides and angles of a triangle A = 100, b = 12, c = 10. Let’s find the sides a using cosine rule:

\(a^{2}=b^{2}+c^{2}-2bc\) \(cosA\)

\(a^{2}=b^{2}+c^{2}-2bc\) \(cosA\)

\(a^{2}=12^{2}+10^{2}-2(12)(10)\) \(cos100\)

\(a^{2}=144+100-240\) \(cos100\)

\(a = \sqrt{244-240 cos 100}\)

a = 16.90

Solve for the angles

\(cosB=\frac{a^{2}+c^{2}-b^{2}}{2ac}\)

\(cosB=\frac{(16.90)^{2}+(10)^{2}-(12)^{2}}{2(16.90)(10)}\)

\(cos B =\frac{285.61+100-144}{338}\)

B = 44.37

For angle C

A + B + C = 180

100 + 44.37 + C = 180

C = 180 – 100 – 44.37

C = 35.63

Therefore the measure of the sides are a = 16.90, b = 12 and c = 10, and the angles are A = 100, B = 44.37 and C = 35.63.