NSWMaths Standard 2Syllabus dot point

How is the cosine rule used to find missing sides and angles in non-right-angled triangles?

Use the cosine rule to find a side given two sides and the included angle, or an angle given three sides

A focused answer to the HSC Maths Standard 2 dot point on the cosine rule. Both forms of the rule, when to use it (SAS or SSS), the side-finding and angle-finding versions, and worked navigation and engineering examples.

Generated by Claude OpusReviewed by Better Tuition Academy7 min answerUpdated 2026-05-20

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What this dot point is asking

NESA wants you to apply the cosine rule in two situations: finding the third side when you know two sides and the included angle (SAS), and finding an angle when you know all three sides (SSS). The rule is on the NESA reference sheet.

The answer

The cosine rule for sides (SAS)

For any triangle $ABC$ with sides $a$ , $b$ , $c$ opposite angles $A$ , $B$ , $C$ :

c^2 = a^2 + b^2 - 2 a b \cos C.

The variable $c$ is the side opposite the known angle $C$ . By symmetry:

a^2 = b^2 + c^2 - 2 b c \cos A, \quad b^2 = a^2 + c^2 - 2 a c \cos B.

The pattern: the unknown side squared equals the sum of squares of the other two sides, minus twice their product times the cosine of the included angle.

The cosine rule for angles (SSS)

Rearrange to find an angle from three sides:

\cos A = \frac{b^2 + c^2 - a^2}{2 b c}.

The angle $A$ is opposite the side $a$ . The other two angles follow similarly.

When to use it

SAS (two sides and the included angle): use cosine rule to find the third side.
SSS (three sides, no angles): use cosine rule to find any one angle. Then either use cosine rule again, or use the sine rule for the remaining angles.

For AAS or SSA, use the sine rule instead.

Identifying the included angle

The included angle is the angle between the two given sides. In a worded problem, look for "the angle at $B$ between $AB$ and $BC$ " or "an angle of $60\degree$ between the two roads".

Connection to Pythagoras

When $C = 90\degree$ , $\cos C = 0$ and the cosine rule reduces to $c^2 = a^2 + b^2$ , which is Pythagoras. The cosine rule is the generalisation of Pythagoras to non-right-angled triangles.

Worked example

Two roads meeting at an angle

Two roads leave a town at an angle of $80\degree$ . A car drives $30$ km along the first road and $40$ km along the second. How far apart are the cars now?

Triangle with two sides $30$ and $40$ km and included angle $80\degree$ . Distance between the cars is the third side $c$ .

$c^2 = 30^2 + 40^2 - 2 \times 30 \times 40 \times \cos 80\degree$ .

$\cos 80\degree \approx 0.1736$ .

$c^2 = 900 + 1600 - 2400 \times 0.1736 = 2500 - 416.74 = 2083.26$ .

$c \approx \sqrt{2083.26} \approx 45.64$ km.

Three sides, find an angle

A triangle has sides $5$ cm, $7$ cm and $10$ cm. Find the angle opposite the $7$ cm side.

Let $a = 7$ , $b = 5$ , $c = 10$ .

$\cos A = \frac{5^2 + 10^2 - 7^2}{2 \times 5 \times 10} = \frac{25 + 100 - 49}{100} = \frac{76}{100} = 0.76$ .

$A = \cos^{-1}(0.76) \approx 40.5\degree$ .

Coastal navigation (Australian context)

A yacht sails from Sydney Heads on a bearing of $130\degree$ for $25$ km, then changes course to bearing $200\degree$ for $40$ km. How far from the starting point is the yacht?

The interior angle at the bend is the supplement of the change of bearing: $200 - 130 = 70\degree$ , and the angle of the triangle at the bend is $180 - 70 = 110\degree$ .

$c^2 = 25^2 + 40^2 - 2 \times 25 \times 40 \times \cos 110\degree$ .

$\cos 110\degree \approx -0.3420$ .

$c^2 = 625 + 1600 - 2000 \times (-0.3420) = 2225 + 684 = 2909$ .

$c \approx \sqrt{2909} \approx 53.94$ km.

Note the cosine of an obtuse angle is negative, which is why the formula adds the product term rather than subtracting it.

Past exam questions, worked

Real questions from past NESA papers on this dot point, with our answer explainer.

2022 HSC Q203 marksIn triangle

ABC

a = 7

cm,

b = 9

cm and

C = 65\degree

. Find side

c

correct to one decimal place.

Show worked answer →

Use $c^2 = a^2 + b^2 - 2 a b \cos C$ .

$c^2 = 49 + 81 - 2 \times 7 \times 9 \times \cos 65\degree$ .

$\cos 65\degree \approx 0.4226$ .

$c^2 = 130 - 126 \times 0.4226 = 130 - 53.25 = 76.75$ .

$c = \sqrt{76.75} \approx 8.76$ .

Rounded to one decimal place: $c \approx 8.8$ cm.

Markers reward the cosine rule stated, correct substitution, intermediate computation kept to at least four decimal places, and final answer at the requested precision.

2023 HSC Q224 marksA triangular paddock has sides

80

60

m and

50

m. Find the largest angle of the paddock correct to the nearest degree.

Show worked answer →

The largest angle is opposite the longest side ( $80$ m). Use the cosine rule rearranged:

$\cos A = \frac{b^2 + c^2 - a^2}{2 b c}$ .

Let $a = 80$ , $b = 60$ , $c = 50$ .

$\cos A = \frac{60^2 + 50^2 - 80^2}{2 \times 60 \times 50} = \frac{3600 + 2500 - 6400}{6000} = \frac{-300}{6000} = -0.05$ .

$A = \cos^{-1}(-0.05) \approx 92.87\degree$ , round to $93\degree$ .

Markers reward identifying the largest angle as opposite the longest side, the rearranged cosine rule, and the angle rounded as requested. Half a mark if you find $A$ but do not justify which side it sits opposite.

What this dot point is asking

The answer

The cosine rule for sides (SAS)

The cosine rule for angles (SSS)

When to use it

Identifying the included angle

Connection to Pythagoras

Past exam questions, worked

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