The sine rule

The trigonometric ratios are handy for real-life situations where right-angled triangles are involved, 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.

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 \(PQR\).

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 \(29.6\textrm{ cm}\).

Example 2

Consider triangle \(ABC\).

Find:

1. angle \(C\)
2. angle \(A\)
3. 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 \(C\).

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 \(ABC_{1}\).

Exercise – using the sine rule

1. Find the unknown sides for the following triangles.

1. 

2. 

3. 

2. Find the unknown sides and angles for the following triangles.

1. 

2. 

3. 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 \(b\).

The sine rule in context

In astronomy, astronomers are observing a distant star. They measure angles from two separate observatories on Earth to determine how far away the star is. From Observatory Alpha, the angle to the star is \(60^{\circ}\). From Observatory Beta, the angle to the star is \(45^{\circ}\).

If the distance between observatories is \(0.000896\) astronomical units (AU), how far away is the star from Observatory Alpha?

Solution

Let the angle from Observatory Alpha to the star be \(\alpha\) with a distance of \(a\). The sine rule connects lengths and their opposite angles. We need to find the angle opposite to the \(0.000896\textrm{ AU}\). Let's call this \(\gamma\).
\[\begin{align*} \gamma & = 180 - \alpha - \beta\\
& = 180-60-45\\
& = 75^{\circ}
\end{align*}\]

Now, we can use the sine rule to find the distance \(a\).
\[\begin{align*} \frac{a}{\sin(\alpha)} & = \frac{d}{\sin(\gamma)}\\
\frac{a}{\sin(60^{\circ})} & = \frac{0.000896}{\sin(75^{\circ})}\\
\frac{a}{\frac{\sqrt{3}}{2}} & = 0.000928\\
a & = 0.00107
\end{align*}\]

The distance between Observatory Alpha and the star is \(0.00107\textrm{ AU}\).

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