5.02 Sine rule and the ambiguous case

The sine rule

If we have a right-angled triangle, we can use trigonometric ratios to relate the sides and angles:

Here:

$\sin(A)$sin(A)

$=$=

$\frac{a}{c}$ac​

$\sin(B)$sin(B)

$=$=

$\frac{b}{c}$bc​

 

But what happens when we have a different kind of triangle with possible internal angles greater than $90^\circ$90°?

Let's start with a non-right-angled triangle, like the acute triangle or the obtuse triangle pictured here.  This derivation applies to both kinds of triangles.

Note the special way the sides are labelled.  Side $a$a, is opposite Angle $A$A, and side $b$b is opposite Angle $B$B and side $c$c is opposite Angle $C$C. Both triangles have the same perpendicular height, marked $h$h.  

$c$c is the hypotenuse of the right-angled triangle $DBA$DBA, and $a$a is the hypotenuse of the right-angled triangle $DBC$DBC.

In triangle $DBA$DBA:

$\sin(A)$sin(A)

$=$=

$\frac{h}{c}$hc​

Which we can rearrange to be:

$h$h

$=$=

$c\sin(A)$csin(A)

In triangle $DBC$DBC:

$\sin(C)$sin(C)

$=$=

$\frac{h}{a}$ha​

Which we can rearrange to be:

$h$h

$=$=

$a\sin(C)$asin(C)

 

Of course length $h$h is the same in both triangles $DBA$DBAand $DBC$DBC so we can equate these two statements:

$c\sin(A)$csin(A)

$=$=

$a\sin(c)$asin(c)

Which we can rearrange like this:

$\frac{a}{\sin(A)}$asin(A)​

$=$=

$\frac{c}{\sin(C)}$csin(C)​

 

In a similar way, and by constructing perpendicular lines from the other vertices we can also show:

$\frac{a}{\sin(A)}$asin(A)​

$=$=

$\frac{b}{\sin(B)}$bsin(B)​

And:

$\frac{b}{\sin(B)}$bsin(B)​

$=$=

$\frac{c}{\sin(C)}$csin(C)​

 

The following demonstration illustrates the rule. It shows that regardless of where you drag the vertices, all three ratios change by the same amount. Also, notice the special case of when the triangle is right-angled. Suppose $C$C is a right angle. Then $\sin C=1$sinC=1 and the formula becomes:

$\frac{\sin A}{a}=\frac{1}{c}$sinAa​=1c​ which becomes $\sin A=\frac{a}{c}$sinA=ac​ and

$\frac{\sin B}{b}=\frac{1}{c}$sinBb​=1c​ which becomes $\sin B=\frac{b}{c}$sinB=bc​.

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These results then combine to create the Sine Rule. Note that sine can be applied to angles of any magnitude. 

Finding a side length using the sine rule

Suppose we had the angles $A$A and $B$B and the length $b$b and we wanted to find the length $a$a. Using the form of the sine rule with numerator lengths $\frac{a}{\sin A}=\frac{b}{\sin B}$asinA​=bsinB​, we can make $a$a the subject 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 $2$2 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$2 d.p.)

 

 

 

Finding an angle using the sine rule

Suppose we had the side lengths $a$a and $b$b, and the angle $B$B, and we want to find the angle $A$A. Using the form of the sine rule with numerator sines $\frac{\sin A}{a}=\frac{\sin B}{b}$sinAa​=sinBb​, we first multiply both sides by $a$a. This gives $\sin A=\frac{a\sin B}{b}$sinA=asinBb​. We then take the inverse sine of both sides 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 $1$1 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$1 d.p.)

Practice questions

Question 1

Question 2

Sine rule - the ambiguous case

In right-angled triangle trigonometry the angles we dealt with were restricted to between $0^\circ$0° and $90^\circ$90°. As shown in the obtuse triangle above, the sine rule can be used for triangles with internal angles greater than $90^\circ$90°.

The trigonometric ratios - $\sin\theta$sinθ, $\cos\theta$cosθ, $\tan\theta$tanθ can be applied to angles of any magnitude. 

To make sense of this consider the unit circle applet below. This is a circle with a radius of $1$1unit centred at the origin on the Cartesian plane. A radius to any point $P$P on the circle is free to move and makes an angle $\theta$θ with the positive horizontal axis. By convention, the angle is measured anticlockwise from the positive horizontal axis. The angle can have any size, positive or negative, depending on how far the point has moved around the circle.

Using the applet below, change the angle between $0^\circ$0° and $180^\circ$180° and take note that $\sin\theta$sinθ is always positive.

The sine of the angle is defined to be the $y$y-coordinate of point $P$P. For an acute angle $\theta$θ, this makes sense as we have always thought of $\sin\theta=\frac{\text{opposite}}{\text{adjacent}}$sinθ=oppositeadjacent​. In the triangle above, this gives us $\sin\theta=\frac{y\text{-coordinate of }P}{1}$sinθ=y-coordinate of P1​. 

Investigate further and you will find that supplementary angles (those that add to $180^\circ$180°) share the same sine ratio. For example try  angles $30^\circ$30° and $150^\circ$150° and notice that sine of these angles is $0.5$0.5.

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The fact that there are two possible angles that give the same sine ratio introduces ambiguity when dealing with non-right-angled triangles.  There is the possibility of obtaining acute and obtuse triangle solutions for the same sine ratio. We will investigate the geometry of this situation further below.

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.

The second case is called the ambiguous case.

We can also see this algebraically by looking at the equation we used before:

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

This equation can have more than one valid solution (though your calculator will only ever give you one!). Luckily for us, the two values of $B$B that are produced (red and green angles) always add to $180^\circ$180°, as we saw in the applet. 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,

$\frac{\sin A}{a}$sinAa​

$=$=

$\frac{\sin B}{b}$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:

$\frac{\sin43^\circ}{11}$sin43°11​

$=$=

$\frac{\sin B}{15.5}$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:

$B$B	$=$=	$\sin^{-1}(0.961)$sin−1(0.961)
$B$B	$=$=	$73.9^\circ$73.9°

Since we are in the ambiguous case, we then subtract this value from $180^\circ$180° to find the other solution:

$B$B	$=$=	$180-73.9$180−73.9
$B$B	$=$=	$106.1^\circ$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 3

Question 4

Question 5