Cosine Rule — Finding an Angle

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The Cosine Rule — Finding an Angle

Three sides, one angle

If you know all three sides, rearrange the Cosine Rule for the angle. Put the side facing the wanted angle as $a$: $a^2 = b^2 + c^2 - 2bc\cos A$, then solve for $\cos A$.

Worked example — all three sides

A triangle has $a = 7$, $b = 5$, $c = 6$. Find $A$ (the angle facing the side of length $7$).

$7^2 = 5^2 + 6^2 - 2(5)(6)\cos A \Rightarrow 49 = 61 - 60\cos A$

$60\cos A = 12 \Rightarrow \cos A = \tfrac{1}{5}$

$A = 78.4 = 78^\circ$

$A = 78^\circ$

Worked example — an obtuse angle

A triangle has $a = 6$, $b = 3$, $c = 5$. Find $A$.

$6^2 = 3^2 + 5^2 - 2(3)(5)\cos A \Rightarrow 36 = 34 - 30\cos A$

$30\cos A = -2 \Rightarrow \cos A = -\tfrac{2}{30}$

A negative cosine means an obtuse angle: $A = 93.8 = 94^\circ$

$A = 94^\circ$

Worked example — finding a named angle

In $\triangle PQR$, $|PQ| = 8$, $|QR| = 7$, $|PR| = 9$. Find $|\angle PQR|$.

The side facing $Q$ is $|PR| = 9$

$9^2 = 8^2 + 7^2 - 2(8)(7)\cos Q \Rightarrow 81 = 113 - 112\cos Q$

$112\cos Q = 32 \Rightarrow \cos Q = \tfrac{32}{112} \Rightarrow Q = 73^\circ$

$|\angle PQR| = 73^\circ$

You try

A triangle has $a = 10$, $b = 8$, $c = 7$. Find $A$.

$10^2 = 8^2 + 7^2 - 2(8)(7)\cos A$.

$100 = 113 - 112\cos A \Rightarrow \cos A = \tfrac{13}{112} \Rightarrow A = 83^\circ$

$A = 83^\circ$

You try

A triangle has $a = 15$, $b = 8$, $c = 10$. Find $A$.

$15^2 = 8^2 + 10^2 - 2(8)(10)\cos A$.

$225 = 164 - 160\cos A \Rightarrow \cos A = -\tfrac{61}{160} \Rightarrow A = 112^\circ$

$A = 112^\circ$

You try

In $\triangle XYZ$, $|XY| = 800$, $|XZ| = 700$, $|YZ| = 550$. Find $|\angle X|$ to one decimal place.

The side facing $X$ is $|YZ| = 550$: $550^2 = 700^2 + 800^2 - 2(700)(800)\cos X$.

$302500 = 1130000 - 1120000\cos X$

$\cos X = \dfrac{827500}{1120000} = 0.7388 \Rightarrow X = 42.4^\circ$

$|\angle X| = 42.4^\circ$

That’s Class 9.

Three sides → an angle with the Cosine Rule (watch for obtuse). Class 10: the area of a triangle.