11.06 Sine rule

In right-angled triangles, trigonometric ratios are used to relate the sides and angles:

In this triangle, $\sin A=\frac{a}{c}$sinA=ac​ and $\sin B=\frac{b}{c}$sinB=bc​.

These trigonometric ratios only apply to right-angled triangles. They do not work for other triangles where there are no right angles.

However, there is a relationship between each of the three interior angles and their opposite side-lengths, defined by the sine rule.

 

The sine rule

Let's start by drawing a line segment from the vertex $C$C perpendicular to the edge $c$c. The length of this segment is labelled as $x$x.

Since $x$x is perpendicular to $c$c, the two line segments meet at right angles. This means that the triangle above has been divided into two right-angled triangles. Using basic trigonometry, the relationships for the sines of the angles $A$A and $B$B is given by

$\sin A=\frac{x}{b}$sinA=xb​ and $\sin B=\frac{x}{a}$sinB=xa​.

Multiply the first equation by $b$b and the second by $a$a. This rearranges the sine ratios above so that $x$x is the subject of both equations.

$x=b\sin A$x=bsinA and $x=a\sin B$x=asinB,

Equating these two equations eliminates the $x$x, creating the following relationship between angles $A$A & $B$B and side-lengths $a$a & $b$b.

$b\sin A=a\sin B$bsinA=asinB.

Dividing this last equation by the side lengths a & b gives the following important relationship:

$\frac{\sin A}{a}=\frac{\sin B}{b}$sinAa​=sinBb​.

This process can be repeated to find how these two angles relate to $c$c and $C$C, and this gives us the sine rule (sometimes called the law of sines).

 

Finding a side length using the sine rule

Suppose a triangle has known angles $A$A and $B$B and a known side length $b$b (opposite to angle $B$B). The sine rule can be used to calculate unknown side length $a$a (opposite to angle $A$A).  Using the form of the sine rule with numerator lengths $\frac{a}{\sin A}=\frac{b}{\sin B}$asinA​=bsinB​, this equation can be transposed to make $a$a the subject of the equation, by multiplying both sides by $\sin A$sinA. This gives

$a=\frac{b\sin A}{\sin B}$a=bsinAsinB​.

 

Worked example

Example 1

Solve: Find the length of $PQ$PQ to two decimal places.

Think: The side we want to find is opposite a known angle, and we also know a matching side and angle. This means we can use the sine rule.

Do:

$\frac{PQ}{\sin48^\circ}$PQsin48°​	$=$=	$\frac{18.3}{\sin27^\circ}$18.3sin27°​
$PQ$PQ	$=$=	$\frac{18.3\sin48^\circ}{\sin27^\circ}$18.3sin48°sin27°​
$PQ$PQ	$=$=	$29.96$29.96 (to 2 d.p.)

 

Finding an angle using the sine rule

Suppose a triangle has known side lengths $a$a and $b$b, and the angle $B$B (opposite side length $b$b). The sine rule can be used to calculate unknown angle $A$A (opposite side length $a$a). Using the form of the sine rule with numerator sines $\frac{\sin A}{a}=\frac{\sin B}{b}$sinAa​=sinBb​, this equation can be transposed to make $\sin A$sinA the subject of the equation, by multiplying both sides by $a$a. This gives

$\sin A=\frac{a\sin B}{b}$sinA=asinBb​

Finally, take the inverse sine of both sides of the equation to make $A$A the subject, which gives

$A=\sin^{-1}\left(\frac{a\sin B}{b}\right)$A=sin−1(asinBb​).

 

Worked example

Example 2

Solve: Find $\angle PRQ$∠PRQ to one decimal place.

Think: The angle we want to find is opposite a known side, and we also know a matching angle and side. This means we can use the sine rule.

Do:

$\frac{\sin R}{28}$sinR28​	$=$=	$\frac{\sin39^\circ}{41}$sin39°41​
$\sin R$sinR	$=$=	$\frac{28\times\sin39^\circ}{41}$28×sin39°41​
$R$R	$=$=	$\sin^{-1}\left(\frac{28\times\sin39}{41}\right)$sin−1(28×sin3941​)
$R$R	$=$=	$25.5^\circ$25.5° (to 1 d.p.)

 

Practice questions

Question 1

Question 2

Question 3

 

The ambiguous case

Sometimes when trying to find an unknown angle in a triangle, the sides of the triangle can be arranged in two different ways, to create either an obtuse angle or an acute angle. When this can occur, this is known as the ambiguous case of using the sine rule.

Let's start by investigating this applet. Video instructions can be found here: Watch video

Created with Geogebra

Set the value of the length of $a$a (blue side), and the length of $b$b (red side). Click "Show triangle" and move the point $A$A to change the value of $\angle CAB$∠CAB (blue angle). These represent the three known quantities - two lengths and an angle.

The angle $\angle ABC$∠ABC (red angle) is the value we find with the sine rule. You will sometimes be able to see a second triangle, with an additional point $B'$B′ and another possible solution $\angle AB'C$∠AB′C (green angle). What do you notice about the relationship between these two solutions?

There are two cases, and what separates one from the other is summarized in this table.

When the length of the (blue) side opposite the known angle is equal to or greater than the length of the other known (red) side, there is only one possible triangle, and only one possible value for the other angle.	When the length of the (blue) side opposite the known angle is less than the length of the other known (red) side, there are two possible triangles, and two possible values for the other angle. This "second triangle" is shown here on its own.

This can be seen algebraically by looking at the equation used before:

$B=\sin^{-1}\left(\frac{b\sin A}{a}\right)$B=sin−1(bsinAa​)

In the previous lesson, it was seen that identical (positive) trigonometric ratios can be used to find both an acute angle solution and an obtuse angle solution. The two angles that create the same sine ratio are always supplementary, meaning that their sum is $180^\circ$180°.

A calculator will only ever give the acute angle solution and not the obtuse angle. However, since the two values of $B$B that are produced (red and green angles) always add to $180^\circ$180°, this sum can be used to find the other obtuse angle. In summary:

 

Worked example

Example 3

Solve: Suppose we are looking at a triangle labelled like this:

We are given $A=43^\circ$A=43°, $a=11$a=11 cm and $b=15.5$b=15.5 cm. We wish to determine the angle $B=\angle ABC$B=∠ABC opposite the side of length $b$b.

Think: We do not know whether the triangle has an obtuse angle or not - the diagram above is not necessarily to scale. According to the sine rule,

$\frac{\sin A}{a}=\frac{\sin B}{b}$sinAa​=sinBb​.

We can use this rule to find $B$B since we know the values of $a$a, $b$b, and $A$A. We need to be mindful that we are in the ambiguous case, as the value of $a$a is smaller than the value of $b$b.

Do: Substituting in the values provided into the version of the sine rule above gives us this equation:

$\frac{\sin43^\circ}{11}=\frac{\sin B}{15.5}$sin43°11​=sinB15.5​.

Rearranging,

$\sin B$sinB	$=$=	$15.5\times\frac{\sin43^\circ}{11}$15.5×sin43°11​
	$=$=	$0.961$0.961

Using the inverse sine function, we calculate $B=\sin^{-1}(0.961)=73.9^\circ$B=sin−1(0.961)=73.9°.

Since we are in the ambiguous case, we then subtract this value from $180^\circ$180° to find the other solution, $B=180-73.9=106.1^\circ$B=180−73.9=106.1°.

This means that with the information we were given, there are two possible triangles that can be formed, one with $B=73.9^\circ$B=73.9° and one with $B=106.1^\circ$B=106.1°.

 

Practice questions

Question 4

Question 5

Question 6