The Cosine Rule
Every question with a fully worked solution.
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Theory
The cosine rule finds the third side of a triangle from two sides and the angle between them, or any angle from three sides. This Year 10A maths page covers both forms, obtuse angles, exact answers, equations in \(x\) and bearing problems.
The cosine rule links the three sides of any triangle to one of its angles. For angle \(C\), between sides \(a\) and \(b\), with \(c\) opposite: \(c^2 = a^2 + b^2 - 2ab\cos C\).
It is Pythagoras' theorem with a correction term. When \(C = 90^\circ\), \(\cos C = 0\) and it becomes \(c^2 = a^2 + b^2\). When \(C\) is obtuse, \(\cos C\) is negative, so \(c^2\) is more than \(a^2 + b^2\).
Use it when you know two sides and the included angle (the angle between them) or all three sides. If you know a side and its opposite angle, the sine rule is usually quicker.
Rearranged for an angle, the cosine rule gives one answer only, because \(\cos^{-1}\) of a negative number gives the obtuse angle directly. There is no ambiguous case.
To find a side:
To find an angle:
The same pattern works for every angle: \(a^2 = b^2 + c^2 - 2bc\cos A\) and \(b^2 = a^2 + c^2 - 2ac\cos B\).
How to use the cosine rule
- Label the angle you know or want as \(C\), and the side opposite it as \(c\).
- Substitute the two sides that form the angle as \(a\) and \(b\).
- Calculate with brackets around \(2ab\cos C\), take the square root for a side, or use \(\cos^{-1}\) for an angle.
- Check: the largest side is opposite the largest angle.
In problems
- Largest or smallest angle: it is opposite the longest or shortest side, so put that side in the \(c^2\) position.
- Parallelogram diagonals: adjacent angles add to \(180^\circ\). The longer diagonal is opposite the obtuse angle.
- Bearings: find the angle at the turning point from a back bearing, or the angle between two paths by subtracting bearings.
- Unknown in the sides: substitute the expressions, expand and solve the quadratic. Reject negative lengths; if both answers are positive, two triangles may fit.
- Speeds and time: write each distance as speed \(\times t\), then solve for \(t\).
- Several steps: after finding a side, use the unrounded value for the next angle or for a perimeter and cost.
Two sides and the angle between them: use the cosine rule with \(a\) opposite \(A\).
| \(a^2\) | \(=\) | \(b^2 + c^2 - 2bc\cos A\) |
| \(\) | \(=\) | \(9.6^2 + 14^2\) \(- 2 \times 9.6 \times 14\cos 118^\circ\) |
| \(\) | \(=\) | \(288.16 - 268.8\cos 118^\circ\) |
| \(\) | \(=\) | \(414.353\ldots\) |
| \(a\) | \(=\) | \(20.355\ldots\) |
| \(\therefore\ BC\) | \(\approx\) | \(\mathbf{20.4}\text{ m}\) |
\(\cos 118^\circ\) is negative, so the last term is added. \(BC\) is longer than both other sides, as it must be opposite an obtuse angle.
(i) the largest angle
(ii) the smallest angle.
(i) The largest angle \(\theta\) is opposite the longest side, \(13\) cm.
| \(\cos\theta\) | \(=\) | \(\dfrac{6^2 + 9^2 - 13^2}{2 \times 6 \times 9}\) |
| \(\) | \(=\) | \(\dfrac{36 + 81 - 169}{108} = -\dfrac{52}{108}\) |
| \(\theta\) | \(=\) | \(118.782\ldots^\circ \approx \mathbf{118.8^\circ}\) |
The cosine is negative, so \(\theta\) is obtuse.
(ii) The smallest angle \(\phi\) is opposite the shortest side, \(6\) cm.
| \(\cos\phi\) | \(=\) | \(\dfrac{9^2 + 13^2 - 6^2}{2 \times 9 \times 13}\) \(= \dfrac{214}{234}\) |
| \(\phi\) | \(=\) | \(23.860\ldots^\circ \approx \mathbf{23.9^\circ}\) |
The side opposite the angle you want is the one that is subtracted on top.
The known angle \(A\) is opposite \(BC = 13\), so \(13\) goes on the left. Use the exact value \(\cos 60^\circ = \dfrac{1}{2}\).
| \(13^2\) | \(=\) | \(x^2 + 7^2\) \(- 2 \times x \times 7 \times \dfrac{1}{2}\) |
| \(169\) | \(=\) | \(x^2 + 49 - 7x\) |
| \(0\) | \(=\) | \(x^2 - 7x - 120\) |
| \(0\) | \(=\) | \((x - 15)(x + 8)\) |
| \(x\) | \(=\) | \(15 \text{ or } x = -8\) |
A length cannot be negative, so \(x = \mathbf{15}\).
Check: \(15^2 + 7^2 - 2 \times 15 \times 7 \times \dfrac{1}{2}\) \(= 225 + 49 - 105\) \(= 169 = 13^2\).
In \(1.5\) hours: \(PA = 18 \times 1.5 = 27\) km and \(PB = 24 \times 1.5 = 36\) km. The angle between the paths is \(157^\circ - 62^\circ = 95^\circ\).
| \(d^2\) | \(=\) | \(27^2 + 36^2\) \(- 2 \times 27 \times 36\cos 95^\circ\) |
| \(\) | \(=\) | \(2025 - 1944\cos 95^\circ\) |
| \(\) | \(=\) | \(2194.430\ldots\) |
| \(d\) | \(=\) | \(46.844\ldots\) |
| \(\) | \(\approx\) | \(\mathbf{46.8}\text{ km}\) |
If the time is unknown, write the sides as \(18t\) and \(24t\); the cosine rule then gives an equation for \(t^2\).
Common pitfalls
Frequently asked questions
When do you use the cosine rule instead of the sine rule?
Use the cosine rule when you know two sides and the angle between them, or all three sides. Use the sine rule when you know a side and the angle opposite it.
How do you find an angle using the cosine rule?
Use \(\cos C = \dfrac{a^2 + b^2 - c^2}{2ab}\), where \(c\) is the side opposite the angle you want. Then use \(\cos^{-1}\). A negative value of \(\cos C\) means the angle is obtuse.
Is the cosine rule the same as Pythagoras' theorem?
It is an extension of it. When the angle is 90 degrees, cos 90 degrees is 0, and the cosine rule becomes c squared equals a squared plus b squared.
Why does the cosine rule not have an ambiguous case?
Between 0 and 180 degrees, each value of cosine belongs to exactly one angle. Positive cosine gives an acute angle and negative cosine gives an obtuse angle, so inverse cos always gives the right one.
How do you find the largest angle of a triangle?
The largest angle is opposite the longest side. Put the longest side in the c squared position of the cosine rule and solve for the angle.