The sine rule

The trigonometric ratios are handy for real-life situations where right-angled triangles are involves, but what about triangles that are not right-angled? We use the sine rule. Use this resource to learn what the sine rule is and how it is applied.

The sine rule shows that the ratio of the length of a side, to the sine of its opposite angle, will be the same for all three sides.

Consider the triangle.

A triangle with sides a, b and c and opposite angles capital A, capital B and capital C, respectively.

Here:

the vertices \(A\), \(B\) and \(C\) have the angles \(A\), \(B\) and \(C\), respectively
\(a\), \(b\) and \(c\) are the side lengths opposite to the angles \(A\), \(B\) and \(C\), respectively.

The sine rule states that:

\[\frac{a}{\sin(A)} = \frac{b}{\sin(B)} = \frac{c}{\sin(C)}\]
or
\[\frac{\sin(A)}{a} = \frac{\sin(B)}{b} = \frac{\sin(C)}{c}\]

This rule can be used to find angles and sides in any triangle (not just a right-angled triangle) when given:

one side and any two angles; or
two sides and an angle opposite one of the given sides.

Example 1 – using the sine rule

Consider triangle \(PQR\).

A triangle with sides p, q and r where r equals 15 centimetres, and opposite angles capital P, capital Q and capital R, where angle P is 70 degrees and angle R is 30 degrees.

Find:

the side length \(p\)

the side length \(q\).

We have been given two angles and one side length, so we can use the sine rule.

\[\begin{align*} \frac{p}{\sin(P)} & = \frac{r}{\sin(R)}\\
\frac{p}{\sin(70^{\circ})} & = \frac{15}{\sin(30^{\circ})}\\
p & = \frac{15\times(\sin70^{\circ})}{\sin(30^{\circ})}\\
& = 28.2\textrm{ cm}\
\end{align*}\]

The side length \(p\) is \(28.2\textrm{ cm}\).

We can find the missing angle easily if we remember that the sum of the three interior angles of a triangle is \(180^{\circ}\).

\[\begin{align*} \angle Q & = 180^{\circ}-\left(70^{\circ}+30^{\circ}\right)\\
& =80^{\circ}
\end{align*}\]

Now that we have angle \(Q\), we can use the sine rule to find side length \(q\).

\[\begin{align*} \frac{q}{\sin(Q)} & = \frac{r}{\sin(R)}\\
\frac{q}{\sin(80^{\circ})} & = \frac{15}{\sin(30^{\circ})}\\
q & = \frac{15\times\sin(80^{\circ})}{\sin(30^{\circ})}\\
& = 29.6\textrm{ cm}
\end{align*}\]

The side length \(q\) is \(29.6\textrm{ cm}\).

Consider triangle \(ABC\).

Find:

angle \(C\)

angle \(A\)

side length \(a\).

We have been given two side lengths and one angle, so we can use the sine rule.

\[\begin{align*} \frac{\sin(B)}{b} & = \frac{\sin(C)}{c}\\
\frac{\sin(126^{\circ})}{20} & = \frac{\sin(C)}{12}\\
\frac{\sin(126^{\circ})}{20}\times12 & = \sin(C)\\
\sin(C) & = 0.485\\
C & = \sin^{-1}(0.485)\\
& = 29^{\circ}
\end{align*}\]

Angle \(C\) is \(29^{\circ}\).

We can find the angle \(A\) easily if we remember that the sum of the three interior angles of a triangle is \(180^{\circ}\).

\[\begin{align*} \angle A & = 180^{\circ}-\left(126^{\circ}+29^{\circ}\right)\\
& = 25^{\circ}
\end{align*}\]

Angle \(A\) is \(25^{\circ}\).

Now that we have angle \(A\), we can use the sine rule to find side length \(a\).

\[\begin{align*} \frac{a}{\sin(A)} & = \frac{b}{\sin(B)}\\
\frac{a}{\sin(25^{\circ})} & = \frac{20}{\sin(126^{\circ})}\\
a & = \frac{20}{\sin(126^{\circ})}\times\sin(25^{\circ})\\
& = 10.4\textrm{ m}
\end{align*}\]

The side length \(a\) is \(10.4\textrm{ m}\).

The ambiguous case

The ambiguous case of the sine rule occurs when you're given two side lengths and a non-included angle. This is an angle that is not between the two known side lengths.

The ambiguous case results in either two valid solutions or no solution. There is an easy way to determine which of these is at play:

If the original angle and the calculated angle add up to less than \(180^{\circ}\), there are two valid solutions.
If the original angle and the calculate angle add up to more than \(180^{\circ}\), there are no valid solutions.

Example 3 – using the sine rule

Consider triangle \(ABC\) with angle \(A=30^{\circ}\), side \(c=12\textrm{ cm}\) and side \(a=8\textrm{ cm}\). Find angle \(C\).

Two triangles that share corners A and B. Side AB is 12 centimetres long. The final corner of the triangles differs with one labelled C1 and the other, C2. Side BC2 is 8 centimetres long and side BC1 is 8 centimetres long. Sides AC1 and AC2 are different lengths. In both cases, angle A is 30 degrees, but for triangle ABC2, the angle B is acute. In triangle ABC1, angle B is obtuse.

When we use the sine rule:

\[\begin{align*} \frac{\sin(A)}{a} & = \frac{\sin(C)}{c}\\
\frac{\sin(30^{\circ})}{8} & = \frac{\sin(C)}{12}\\
\frac{\sin(30^{\circ})}{8}\times12 & = \sin(C)\\
\sin(C) & = \frac{1}{16}\times12\\
& = 0.75\\
C & = \sin^{-1}(0.75)\\
& = 48.6^{\circ}
\end{align*}\]

To determine whether the ambiguous case applies, we can add up the original angle given and the new calculated angle.

\[\begin{align*} A+C & = 30^{\circ}+48.6^{\circ}\\
& = 78.6^{\circ}\\
& < 180^{\circ}
\end{align*}\]

The original angle and the calculated angle add up to less than \(180^{\circ}\), so there are two valid solutions.

To find the other solution, we just need to minus the calculated angle from \(180^{\circ}\).

\[\begin{align*} C & =180^{\circ}-48.6^{\circ}\\
& = 131.4^{\circ}
\end{align*}\]

If \(C=48.6^{\circ}\), the triangle is \(ABC_{2}\). If \(C=131.4^{\circ}\), the triangle is \(ABC_{1}\).

Your turn – using the sine rule

Find the unknown sides for the following triangles.

Find the unknown sides and angles for the following triangles.

Given triangle \(ABC\), where angle \(A=35\), side \(a=16\) and side \(c=21\), find the magnitude of angles \(B\) and \(C\) and the length of side \(b\).

1. \(b=18.1\), \(c=19.9\)
2. \(q=53.5\), \(r=41.2\)
3. \(t=9.1\), \(r=20.2\)
1. \(L=35.4^{\circ}\), \(N=64.6^{\circ}\), \(n=7.8\)
2. \(B=41.7^{\circ}\), \(A=73.3^{\circ}\), \(a=6.8\)
\(C=131.2^{\circ}\), \(B=13.8^{\circ}\), \(b=6.7\) or \(C=48.8\), \(B=96.2\), \(b=27.7\)

Images on this page by RMIT, licensed under CC BY-NC 4.0

Learning Lab homepage