In a non-right angled triangle, the cosine rule states that –

- b
^{2}= a^{2}+ c^{2}– 2ac cosB - a
^{2}= b^{2}+ c^{2}– 2bc cosA - c
^{2}= a^{2}+ b^{2}– 2ab cosC

We’ll deduct the proof of the first of the above rules below. In the triangle ABC, draw the perpendicular AD to BC, and we get two right angled triangles ABD and ADC.

In ABD, applying the Pythagoras Theorem, c^{2} = h^{2} + x^{2}, and

sinB = and cosB =

Therefore h = c sinB, and x = c cosB

h^{2} = c^{2} – x^{2} – **Equation 1**

From ADC, we can obtain

h^{2} = b^{2} – (a – x)^{2}

h^{2} = b^{2} – a^{2} + 2ax – x^{2} – **Equation 2**

Using equations 1 and 2,

h^{2} = c^{2} – x^{2} = b^{2} – a^{2} + 2ax – x^{2}

c^{2} – x^{2} = b^{2} – a^{2} + 2ax – x^{2}

c^{2} = b^{2} – a^{2} + 2ax^{
}

Substituting x = c cosB,

c^{2} = b^{2} – a^{2} + 2ac cosB

-b^{2} = -c^{2} – a^{2} + 2ac cosB

Therefore b^{2} = a^{2} + c^{2} – 2ac cosB

Similarly we can prove that a^{2} = b^{2} + c^{2} – 2bc cosA, and c^{2} = a^{2} + b^{2} – 2ab cosC.

**Remember**: We use cosine rule when all three sides and an included angle is given or is involved.

Now, let us look at some examples of using cosine rule

**Example 1**: In the triangle ABC, a = 5 cm, c = 6 cm, and angle B = 40°. What is x?

Applying the cosine rule b^{2} = a^{2} + c^{2} – 2ac cosB

x^{2} = 5^{2} + 6^{2} – 2 x 5 x 6 x cos40°

= 25 + 36 – 60 cos40°

= 15.03733

x = 3.88 cm

**Example 2**: In the triangle ABC, a = 6 cm, b = 5 cm, c = 3 cm. What is angle A?

Applying the cosine rule a^{2} = b^{2} + c^{2} – 2bc cosA, and rearranging the formula, we get

cosA =

=

=

=

A =

= 97.66255566

= 97°40′ (to the nearest minute)

## Summary of Cosine Rule

- a
^{2}= b^{2}+ c^{2}– 2bc cosA or cosA = - b
^{2}= a^{2}+ c^{2}– 2ac cosB or cosB = - c
^{2}= a^{2}+ b^{2}– 2ab cosC or cosC =