The Sine Rule
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Theory
The sine rule solves triangles that are not right-angled: each side divided by the sine of its opposite angle gives the same number. This Year 10A maths page shows how to find a side, find an angle, decide whether an obtuse answer is possible, handle the ambiguous case, and use the rule in elevation and bearing problems.
Label a triangle with capital letters \(A\), \(B\), \(C\) for the angles and small letters \(a\), \(b\), \(c\) for the sides, so that side \(a\) is opposite angle \(A\).
The sine rule says that the ratio of a side to the sine of its opposite angle is the same for all three pairs. It works in every triangle, including obtuse-angled ones.
Use the sine rule when you know a side and its opposite angle, plus one more side or angle. If you know two angles, find the third first: the angles add to \(180^\circ\).
When you solve for an angle, \(\sin\theta = k\) has an acute answer and an obtuse answer, \(180^\circ - \theta\). Check whether the obtuse one fits. When it does, two triangles match the data: this is the ambiguous case.
To find a side:
To find an angle:
How to use the sine rule
- Label the triangle so each side sits opposite its angle. Find a third angle from the angle sum if you can.
- Write two of the ratios: the complete pair (side and opposite angle both known) and the pair with the unknown.
- Solve. For a side, put sides on top; for an angle, put sines on top and use \(\sin^{-1}\).
- Check any angle for an obtuse second answer, and check the largest side is opposite the largest angle.
In problems
- Elevation from two points: the angle at the near point is \(180^\circ\) minus its elevation, and the angle at the top is the difference of the elevations (Example 4).
- Bearings: draw north lines, then use bearings to find each angle of the triangle.
- Slopes: a vertical object on a slope of \(\alpha\) makes \(90^\circ + \alpha\) with the downhill slope.
- Quadrilaterals: split along a diagonal, find the diagonal in one triangle and use it, unrounded, in the other.
- Isosceles shortcut: two equal angles mean the sides opposite them are equal, with no sine rule needed.
- Algebraic sides: if \(a = x + 2\) and \(b = x\), cross-multiply \((x + 2)\sin B = x\sin A\), collect the \(x\) terms and factorise.
\(YZ\) is opposite \(X\), \(XZ\) is opposite \(Y\) and \(XY\) is opposite \(Z\). Find the third angle first.
| \(\angle Z\) | \(=\) | \(180^\circ - 52^\circ - 73^\circ = 55^\circ\) |
| \(\dfrac{XZ}{\sin 73^\circ}\) | \(=\) | \(\dfrac{11.5}{\sin 52^\circ}\) |
| \(XZ\) | \(=\) | \(\dfrac{11.5\sin 73^\circ}{\sin 52^\circ} = 13.956\ldots\) |
| \(\) | \(\approx\) | \(\mathbf{14.0}\text{ cm}\) |
| \(XY\) | \(=\) | \(\dfrac{11.5\sin 55^\circ}{\sin 52^\circ} = 11.954\ldots\) |
| \(\) | \(\approx\) | \(\mathbf{12.0}\text{ cm}\) |
Check: the longest side \(XZ\) is opposite the largest angle, \(73^\circ\).
To find an angle, use the sine rule upside down, with the sines on top.
| \(\dfrac{\sin C}{6.2}\) | \(=\) | \(\dfrac{\sin 71^\circ}{8.4}\) |
| \(\sin C\) | \(=\) | \(\dfrac{6.2\sin 71^\circ}{8.4} = 0.6978\ldots\) |
| \(C\) | \(=\) | \(44.257\ldots^\circ\) or \(180^\circ - 44.257\ldots^\circ = 135.742\ldots^\circ\) |
Test the obtuse value: \(135.742\ldots^\circ + 71^\circ \gt 180^\circ\), so it cannot be an angle of this triangle. (Also, \(c < b\), so \(C\) must be smaller than \(B\).)
| \(C\) | \(\approx\) | \(\mathbf{44.3^\circ}\) |
The given angle is opposite the shorter of the two given sides, so two triangles may fit.
| \(\sin B\) | \(=\) | \(\dfrac{11\sin 32^\circ}{7} = 0.8327\ldots\) |
| \(B\) | \(=\) | \(56.380\ldots^\circ \text{ or } 123.619\ldots^\circ\) |
Both are possible, because \(32^\circ + 123.619\ldots^\circ < 180^\circ\). So \(B \approx \mathbf{56.4^\circ}\) or \(\mathbf{123.6^\circ}\).
For each one, find \(C\), then \(c\):
| \(B \text{ acute: } C\) | \(=\) | \(91.619\ldots^\circ\) |
| \(c\) | \(=\) | \(\dfrac{7\sin 91.619\ldots^\circ}{\sin 32^\circ} \approx \mathbf{13.2}\text{ cm}\) |
| \(B \text{ obtuse: } C\) | \(=\) | \(24.380\ldots^\circ\) |
| \(c\) | \(=\) | \(\dfrac{7\sin 24.380\ldots^\circ}{\sin 32^\circ} \approx \mathbf{5.5}\text{ cm}\) |
Triangle \(ABT\) is not right-angled, but it has two known angles and a known side.
| \(\angle ABT\) | \(=\) | \(180^\circ - 47^\circ = 133^\circ\) |
| \(\angle ATB\) | \(=\) | \(180^\circ - 29^\circ - 133^\circ = 18^\circ\) |
| \(\dfrac{BT}{\sin 29^\circ}\) | \(=\) | \(\dfrac{80}{\sin 18^\circ}\) |
| \(BT\) | \(=\) | \(125.510\ldots \approx \mathbf{125.5}\text{ m}\) |
Now triangle \(BGT\) is right-angled, with \(BT\) as the hypotenuse:
| \(h\) | \(=\) | \(125.510\ldots\sin 47^\circ = 91.792\ldots\) |
| \(\) | \(\approx\) | \(\mathbf{91.8}\text{ m}\) |
The angle at \(T\) is the difference of the two elevations: \(47^\circ - 29^\circ = 18^\circ\) (exterior angle of a triangle).
Common pitfalls
Frequently asked questions
When do you use the sine rule?
Use the sine rule when you know a side and the angle opposite it, plus one other side or angle. If you know two sides and the angle between them, or all three sides, use the cosine rule instead.
How do you find an angle using the sine rule?
Put the sines on top: \(\dfrac{\sin A}{a} = \dfrac{\sin B}{b}\). Substitute, make the sine of the unknown angle the subject, then use \(\sin^{-1}\). Check whether the obtuse angle \(180^\circ - \theta\) also fits.
What is the ambiguous case of the sine rule?
When you are given two sides and an angle opposite the shorter one, the shorter side can swing into two positions. Two different triangles fit, with one acute and one obtuse angle that add to 180 degrees.
Does the sine rule work for right-angled triangles?
Yes. With a 90 degree angle, sin 90 degrees is 1, and the sine rule reduces to the ordinary sine ratio. SOH CAH TOA is usually quicker, though.
How do you know if there are two possible triangles?
Find the acute angle from inverse sine, then subtract it from 180 degrees. If that obtuse angle plus the given angle is less than 180 degrees, both triangles exist.