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Year 12 Maths Standard 2 (2027) Trigonometry

The Cosine Rule

20 practice questions 0 video lessons Theory + worked examples

Master the cosine rule for NSW Year 12 Mathematics Standard 2. The cosine rule, \(c^2=a^2+b^2-2ab\cos C\), works in any triangle: use it to find a side from two sides and the included angle, or rearrange it to \(\cos C=\tfrac{a^2+b^2-c^2}{2ab}\) to find an angle from all three sides.

You will learn to label a triangle correctly, choose the right form of the rule, handle obtuse angles where the cosine is negative, and apply it to real problems — distances, navigation legs and surveying triangular blocks of land — a core Standard 2 trigonometry skill.

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Theory

The cosine rule finds a missing side or angle in any triangle. This Year 12 Standard 2 (NSW) guide shows how to use \(c^2=a^2+b^2-2ab\cos C\) to find a side from two sides and the included angle, and its rearrangement \(\cos C=\tfrac{a^2+b^2-c^2}{2ab}\) to find an angle from three sides — including obtuse angles.

The cosine rule works in any triangle — including non-right-angled ones — by linking the three side lengths to one angle. It is the tool to reach for when right-angled trigonometry (SOH CAH TOA) does not apply.

Label the triangle so that side \(a\) is opposite angle \(A\), side \(b\) opposite \(B\) and side \(c\) opposite \(C\). The rule then has two everyday uses: find a side from two sides and the included angle between them (SAS), or find an angle from all three sides (SSS).

Because the rule uses \(\cos C\) directly, it handles obtuse angles automatically: a negative cosine simply makes the opposite side longer, or signals an obtuse angle when you solve for one. This is a core Year 12 Standard 2 (NSW) skill for distances, navigation and surveying.

Triangle ABC with sides a, b, c opposite vertices A, B, CA general (non-right-angled) triangle ABC. The lower-case side a is opposite vertex A, side b is opposite B and side c is opposite C. A B C a b c
Label the triangle so each lower-case side sits opposite the matching upper-case angle: \(a\) opposite \(A\), \(b\) opposite \(B\), \(c\) opposite \(C\).
Two sides a and b with included angle C and the opposite side cTwo sides a and b meet at the included angle C; the side c lies opposite the angle C. The cosine rule links c to a, b and angle C. C B A a b c C
Two sides \(a\) and \(b\) meet at the included angle \(C\); the cosine rule finds the opposite side \(c\) (or, rearranged, the angle \(C\)).

To find a side from two sides and the included angle:

\[c^2 = a^2 + b^2 - 2ab\cos C\]
c2=a2+b22abcosC

Rearranged, to find an angle from the three sides:

\[\cos C = \dfrac{a^2 + b^2 - c^2}{2ab}\]
cosC=a2+b2c22ab
Which letter is which? \(C\) is the angle you are working with; \(c\) is the side opposite it; \(a\) and \(b\) are the other two sides. For an obtuse angle \(\cos C<0\), so \(-2ab\cos C\) becomes a positive amount added on.

Finding a side (two sides + included angle)

  1. Name the unknown side \(c\) and the angle opposite it \(C\); the two known sides are \(a\) and \(b\).
  2. Substitute into \(c^2 = a^2 + b^2 - 2ab\cos C\).
  3. Evaluate the right-hand side (keep the sign of \(\cos C\)).
  4. Square root to get \(c\), then round as asked.

Finding an angle (three sides)

  1. Choose the angle \(C\) and identify the side \(c\) opposite it.
  2. Substitute into \(\cos C = \dfrac{a^2 + b^2 - c^2}{2ab}\).
  3. Apply \(\cos^{-1}\) to the whole fraction (degree mode). A negative value gives an obtuse angle.
Example 1 — Find a side (SAS)
In triangle \(ABC\), \(AB=10\) cm, \(AC=14\) cm and the included angle \(A=58^\circ\). Find \(BC\), to \(2\) decimal places.
Solution

Two sides and the included angle are known, so use \(c^2=a^2+b^2-2ab\cos C\).

Triangle with sides 10 cm and 14 cm and an included angle of 58 degreesTriangle ABC with AB = 10 cm, AC = 14 cm, the included angle A = 58 degrees and the opposite side BC labelled x. A B C 10 cm 14 cm x 58°
\(BC^2\)\(=\)\(10^2 + 14^2 - 2(10)(14)\cos 58^\circ\)
\(\)\(=\)\(296 - 280\cos 58^\circ\)
\(\)\(=\)\(147.62\)
\(BC\)\(=\)\(\sqrt{147.62}\)
\(\therefore\ BC\)\(\approx\)\(12.15\text{ cm}\)
BC12.15
Example 2 — Obtuse included angle
Two yachts leave a marina \(M\): one sails \(16\) km, the other \(22\) km, with a \(124^\circ\) angle between their courses. How far apart are they, to \(2\) decimal places?
Solution

The angle is obtuse, so \(\cos 124^\circ<0\) and the last term is added on.

Two courses of 16 km and 22 km from a marina with a 124 degree angle between themTriangle with the marina M at the obtuse vertex, two courses MA = 16 km and MB = 22 km, the included angle 124 degrees and the distance AB labelled d. M A B 16 km 22 km d 124°
\(d^2\)\(=\)\(16^2 + 22^2 - 2(16)(22)\cos 124^\circ\)
\(\)\(=\)\(740 - 704\cos 124^\circ\)
\(\)\(=\)\(740 + 393.67\)
\(d\)\(=\)\(\sqrt{1133.67}\)
\(\therefore\ d\)\(\approx\)\(33.67\text{ km}\)
d33.67
Example 3 — Find an angle (SSS)
A triangle has sides \(9\) cm, \(12\) cm and \(16\) cm. Find the largest angle \(\theta\), to the nearest degree.
Solution

The largest angle is opposite the longest side (\(16\)), so let \(c=16\).

Triangle with sides 16 cm, 12 cm and 9 cm, angle theta opposite the 16 cm sideTriangle with sides 16 cm, 12 cm and 9 cm; the angle theta is opposite the longest (16 cm) side. B C A 16 cm 12 cm 9 cm θ
\(\cos\theta\)\(=\)\(\dfrac{9^2 + 12^2 - 16^2}{2(9)(12)}\)
\(\)\(=\)\(\dfrac{-31}{216} = -0.1435\)
\(\theta\)\(=\)\(\cos^{-1}(-0.1435)\)
\(\therefore\ \theta\)\(\approx\)\(98^\circ\)
θ98°
Example 4 — Applied (three sides)
A triangular paddock has sides \(45\) m, \(60\) m and \(80\) m. Find the smallest angle \(\theta\), to the nearest degree.
Solution

The smallest angle is opposite the shortest side (\(45\)), so let \(c=45\).

Triangular paddock with sides 80 m, 60 m and 45 m, angle theta opposite the 45 m sideTriangular paddock with sides 80 m, 60 m and 45 m; the angle theta is opposite the shortest (45 m) side. B C A 80 m 60 m 45 m θ
\(\cos\theta\)\(=\)\(\dfrac{60^2 + 80^2 - 45^2}{2(60)(80)}\)
\(\)\(=\)\(\dfrac{7975}{9600} = 0.8307\)
\(\theta\)\(=\)\(\cos^{-1}(0.8307)\)
\(\therefore\ \theta\)\(\approx\)\(34^\circ\)
θ34°

Common pitfalls

Use the included angle. For \(c^2=a^2+b^2-2ab\cos C\) the angle \(C\) must sit between the two known sides, and \(c\) is the side opposite it. Using a non-included angle gives the wrong answer.
Keep the sign for obtuse angles. When \(C>90^\circ\), \(\cos C\) is negative, so \(-2ab\cos C\) is added on. Dropping the minus sign makes the side too short.
Finish the right way. Take the square root for a side; for an angle apply \(\cos^{-1}\) to the whole fraction, with the calculator in degree mode. A negative cosine is fine — it just means an obtuse angle.

Frequently asked questions

When do I use the cosine rule instead of the sine rule?

Use the cosine rule when you know two sides and the included angle (to find the third side), or when you know all three sides (to find an angle). Use the sine rule when you have a matching side-and-opposite-angle pair. If the triangle is right-angled, plain SOH CAH TOA is usually quicker.

What is the included angle?

The included angle is the angle formed between the two sides you already know — the angle at the vertex where those two sides meet. In c squared equals a squared plus b squared minus 2ab cos C, the side c is the one opposite that included angle C.

How do I rearrange the cosine rule to find an angle?

Start from c squared equals a squared plus b squared minus 2ab cos C and make cos C the subject: cos C equals (a squared plus b squared minus c squared) divided by 2ab. Then take the inverse cosine of that value. Put the side opposite the angle you want as c.

What happens when the angle is obtuse?

The cosine of an obtuse angle is negative, so the term minus 2ab cos C becomes a positive amount that is added on, making the opposite side longer. When you solve for an angle and the cosine comes out negative, the inverse cosine gives an obtuse angle — that is correct, not an error.

How do I find the largest or smallest angle in a triangle?

The largest angle is always opposite the longest side, and the smallest angle is opposite the shortest side. Put that side as c in cos C equals (a squared plus b squared minus c squared) over 2ab, then take the inverse cosine to get the angle.