NZ Level 7 (NZC) Level 2 (NCEA)
Sine Rule (Ambiguous Case)
Lesson

The sine rule applies to all triangles, like this one:

We can write it as follows:

$\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}$sinAa=sinBb=sinCc

This rule relates each side in a triangle with the sine ratio of the angle opposite to it. For example, if we know the side lengths $a$a and $b$b, and we know the size of angle $A$A, we can take the first equality and rearrange it to obtain

$\sin B=\frac{b\sin A}{a}$sinB=bsinAa,

which lets us determine the sine of $B$B. To find $B$B itself, we need to take the inverse:

$B=\sin^{-1}\left(\frac{b\sin A}{a}\right)$B=sin1(bsinAa).

We can then substitute the values of $a$a, $b$b, and $A$A to find a value for $B$B. But in this case, there is more to consider! To see when and why, we are going to dive a little deeper into the geometry of the situation.

### The ambiguous case

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

 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=sin1(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:

Sine rule (ambiguous case)

If you are trying to find an angle using the sine rule (with two known sides and a known angle), use the following rule to find one value of $B$B:

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

$B=\sin^{-1}\left(\frac{b\sin A}{a}\right)$B=sin1(bsinAa)

If the side opposite the known angle is the shorter side, you are in the ambiguous case. Subtract the first value of $B$B that you found from $180^\circ$180° to find the second solution.

#### Worked example

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=sin1(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=18073.9=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 1

Find the value of $x$x using the sine rule, noting that $x$x is obtuse.

Round your answer to two decimal places.

##### Question 2

Consider $\triangle ABC$ABC below:

1. Find $x$x, noting that $x$x is acute.

Round your answer to the nearest degree.

2. Now find $\angle ADB$ADB to the nearest whole degree, given that$\angle ADB>\angle ACB$ADB>ACB.

##### Question 3

$\triangle ABC$ABC consists of angles $A$A, $B$B and $C$C which appear opposite sides $a$a, $b$b and $c$c respectively where $\angle CAB=36^\circ$CAB=36°, $a=7$a=7 and $b=10$b=10.

1. Select the most appropriate option to complete the sentence below:

The triangle:

Can be either acute or obtuse.

A

Must be an obtuse triangle.

B

Must be an acute triangle.

C

Does not exist.

D

Must be a right-angled triangle.

E

Can be either acute or obtuse.

A

Must be an obtuse triangle.

B

Must be an acute triangle.

C

Does not exist.

D

Must be a right-angled triangle.

E

### Outcomes

#### M7-4

Apply trigonometric relationships, including the sine and cosine rules, in two and three dimensions

#### 91259

Apply trigonometric relationships in solving problems