The Sine Rule and the Ambiguous Case | IEB Grade 11 Maths

Q: When do I use the sine rule?

When you know a side and the angle opposite it, plus one more side or angle.

Q: Is the proof of the sine rule examinable in Grade 11?

Yes, in acute-angled triangles. Learn the construction of the height and how it gives two expressions for $h$.

Q: How do I know if there are two triangles?

If you are finding an angle opposite the longer of two given sides, check whether the obtuse option plus the given angle is less than 180°.

The sine rule solves triangles that are not right-angled when you know a side and its opposite angle. Grade 11 covers the proof, the calculations and the ambiguous case, where two different triangles fit the same information.

Proof of the sine rule

In $△ A B C$ , draw the height $h$ from $C$ perpendicular to $A B$ (or $A B$ extended) at $D$ .

In $△ A D C$ : $h = b sin A$ . In $△ B D C$ : $h = a sin B$ .

So $b sin A = a sin B$ , which gives

$\frac{a}{s i n A} = \frac{b}{s i n B}$

Drawing a height from $A$ in the same way gives $\frac{b}{sin B} = \frac{c}{sin C}$ .

For an obtuse angle at $A$ , $D$ lies on $B A$ extended, and $h = b sin (180° - A) = b sin A$ . The result is the same.

Finding sides and angles

$\frac{a}{s i n A} = \frac{b}{s i n B} = \frac{c}{s i n C}$

Use the sine rule when you know a side and its opposite angle.

Finding a side. $\hat{A} = 42°$ , $\hat{B} = 71°$ , $a = 15$ cm:

$b = \frac{15 s i n 71°}{s i n 42°} \approx 21, 20 cm$

Finding an angle. Put the angles on top: $\frac{sin B}{b} = \frac{sin A}{a}$ .

With $a = 12$ , $b = 9$ , $\hat{A} = 64°$ : $sin B = \frac{9 sin 64°}{12}$ , so $\hat{B} \approx 42, 38°$ . Since $b < a$ , $B < A$ , and only the acute answer is possible.

The ambiguous case

When you know two sides and an angle not between them, and you use the sine rule to find an angle, there may be two triangles: one with an acute angle and one with an obtuse angle, because $sin θ = sin (180° - θ)$ .

Example. $c = 10$ , $b = 7$ , $\hat{B} = 40°$ .

$sin C = \frac{10 s i n 40°}{7} \approx 0, 918 \Rightarrow \hat{C} \approx 66, 67° or 113, 33°$

Both are valid, since $40° + 113, 33° < 180°$ . So there are two possible triangles, with $\hat{A} \approx 73, 33°$ or $\hat{A} \approx 26, 67°$ .

If the obtuse angle plus the given angle is 180° or more, only the acute triangle exists.

Key terms

Sine rule: $\frac{a}{s i n A} = \frac{b}{s i n B} = \frac{c}{s i n C}$ .
Ambiguous case: When two triangles fit the given information.
Opposite side: The side across the triangle from an angle, named with the same letter in lower case.

Common mistakes

Using the sine rule without a known side and opposite angle pair.
Missing the obtuse option in the ambiguous case.
Having the calculator in radian mode.
Rounding too early and carrying the error forward.

Practice questions

Reveal steps one at a time, then tick the marks you earned.

Question 1

4 marks total

routine

In $△ A B C$ , $\hat{A} = 42°$ , $\hat{B} = 71°$ and $a = 15$ cm. Calculate $b$ .

Worked solution

Hidden - try the question first

Solution

$\frac{b}{sin 71°} = \frac{15}{sin 42°}$
$b = \frac{15 sin 71°}{sin 42°}$
$b \approx 21, 20$ cm

Question 2

5 marks total

routine

In $△ A B C$ , $a = 12$ cm, $b = 9$ cm and $\hat{A} = 64°$ . Calculate $\hat{B}$ and explain why there is only one answer.

Worked solution

Hidden - try the question first

Solution

$\frac{b}{sin 71°} = \frac{15}{sin 42°}$
$b = \frac{15 sin 71°}{sin 42°}$
$\hat{B} \approx 42, 38°$
$b < a$ , so $\hat{B} < \hat{A}$
The obtuse option $137, 62°$ is impossible

Question 3

6 marks total

complex

Prove the sine rule for an acute-angled triangle $A B C$ : $\frac{a}{sin A} = \frac{b}{sin B}$ .

Worked solution

Hidden - try the question first

Solution

Construction: draw $C D ⊥ A B$ , with $C D = h$
In $△ A D C$ : $sin A = \frac{h}{b}$ , so $h = b sin A$
In $b \approx 21, 20$ : $\hat{B} \approx 42, 38°$ , so $h = a sin B$
$b sin A = a sin B \Rightarrow \frac{a}{sin A} = \frac{b}{sin B}$

Question 4

9 marks total

problem-solving

In $△ A B C$ , $c = 10$ cm, $b = 7$ cm and $\hat{B} = 40°$ .

(a) Calculate the two possible sizes of $\hat{C}$ . (3)

(b) Explain why both are possible. (1)

Worked solution

Hidden - try the question first

Solution

(a) $C D ⊥ A B$
$\hat{C} \approx 66, 67°$
or $\hat{C} \approx 113, 33°$
(b) $\hat{B} \approx 42, 38°$ , so the obtuse triangle also exists
(c) $\hat{A} \approx 73, 33°$ or $\hat{A} \approx 26, 67°$
$a = \frac{7 sin A}{sin 40°}$
$a = \frac{7 sin 73 , 33°}{sin 40°} \approx 10, 43$ cm
or $a = \frac{7 sin 26 , 67°}{sin 40°}$
$△ A D C$ cm

FAQs

When do I use the sine rule?

When you know a side and the angle opposite it, plus one more side or angle.

Is the proof of the sine rule examinable in Grade 11?

Yes, in acute-angled triangles. Learn the construction of the height and how it gives two expressions for $h$ .

How do I know if there are two triangles?

If you are finding an angle opposite the longer of two given sides, check whether the obtuse option plus the given angle is less than 180°.

The sine rule

Proof of the sine rule

Finding sides and angles

The ambiguous case

Key terms

Common mistakes

Practice questions

FAQs

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