Sine, Cosine and Area Rules | IEB Grade 12 Maths

Q: How do I know whether to use the sine rule or the cosine rule?

If you know a side and the angle opposite it, use the sine rule. If you know two sides and the angle between them, or all three sides, use the cosine rule.

Q: When does the ambiguous case happen?

When you use the sine rule to find an angle from two sides and an angle that is not between them. Check whether the obtuse option also fits.

Q: Do I need to prove the sine, cosine and area rules?

Yes. The proofs of the sine, cosine and area rules are examinable in the IEB exam, in acute-angled triangles only. The rules themselves are on the formula sheet.

The sine, cosine and area rules solve triangles that are not right-angled. In Paper 2 they are often combined in a single question about a field, a roof or a structure, so choosing the right rule at each step matters as much as the calculation.

The sine rule and the ambiguous case

In any triangle $A B C$ :

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

Use it when you know a side and its opposite angle, plus one more side or angle.

Example. $a = 8$ , $B = 50°$ , $C = 70°$ . Then $A = 60°$ and

$b = \frac{8 s i n 50°}{s i n 60°} \approx 7, 08$

The ambiguous case. When you know two sides and a non-included angle, and you are finding an angle with the sine rule, there may be two triangles. If $sin B = 0, 86$ , then $B \approx 59, 36°$ or $B \approx 120, 64°$ . Keep the obtuse option only if the angles still add up to less than 180°.

The cosine rule

$a^{2} = b^{2} + c^{2} - 2 b c cos A$

Use it when you know:

two sides and the included angle (to find the third side), or
all three sides (to find an angle).

Rearranged for an angle:

$cos A = \frac{b ^{2} + c ^{2} - a ^{2}}{2 b c}$

Example. Sides 7, 9 and 12. The largest angle is opposite the longest side:

$cos C = \frac{49 + 81 - 144}{2 ( 7 ) ( 9 )} = - \frac{14}{126} \Rightarrow C \approx 96, 38°$

A negative cosine means an obtuse angle. The cosine rule never gives an ambiguous case.

The area rule

$Area = \frac{1}{2} ab sin C$

Use any two sides and the angle between them.

Example. Sides 40 m and 55 m with an included angle of 75°:

$Area = \frac{1}{2} (40) (55) sin 75° \approx 1062, 52 m^{2}$

For a quadrilateral, split it with a diagonal into two triangles, find the diagonal with the cosine rule, and add the two areas.

Key terms

Included angle: The angle between two given sides.
Ambiguous case: When two different triangles fit the given information, giving an acute and an obtuse option for an angle.
Non-right-angled triangle: A triangle where none of the angles is 90°, so SOH CAH TOA does not apply directly.

Common mistakes

Using the cosine rule with an angle that is not between the two known sides.
Missing the obtuse option in the ambiguous case.
Rounding intermediate answers too early, which changes the final answer.
Using the area rule with an angle that is not between the two sides.
Calculator in radian mode.

Practice questions

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

Question 1

4 marks total

routine

In $△ A B C$ , $a = 8$ cm, $\hat{B} = 50°$ and $\hat{C} = 70°$ . Calculate the length of $b$ .

Worked solution

Hidden - try the question first

Solution

$\hat{A} = 180° - 50° - 70° = 60°$
$\frac{b}{sin 50°} = \frac{8}{sin 60°}$
$b = \frac{8 sin 50°}{sin 60°}$
$b \approx 7, 08$ cm

Question 2

5 marks total

routine

A triangle has sides of 7 cm, 9 cm and 12 cm. Calculate the size of its largest angle.

Worked solution

Hidden - try the question first

Solution

The largest angle is opposite the 12 cm side
$cos C = \frac{7 ^{2} + 9 ^{2} - 1 2 ^{2}}{2 ( 7 ) ( 9 )}$
$= - \frac{14}{126}$
$C \approx 96, 38°$

Question 3

6 marks total

complex

In $△ P QR$ , $\hat{P} = 35°$ , $p = 6$ cm and $q = 9$ cm.

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

(b) Calculate the area of the triangle for each case. (3)

Worked solution

Hidden - try the question first

Solution

(a) $\frac{sin Q}{9} = \frac{sin 35°}{6} \Rightarrow sin Q \approx 0, 8604$
$\hat{Q} \approx 59, 36°$
or $\hat{Q} \approx 120, 64°$ (valid, since $35° + 120, 64° < 180°$ )
(b) $\hat{R} \approx 85, 64°$ or $\hat{R} \approx 24, 36°$
Area $= \frac{1}{2} (6) (9) sin 85, 64° \approx 26, 92$ cm²
or Area $= \frac{1}{2} (6) (9) sin 24, 36° \approx 11, 14$ cm²

Question 4

9 marks total

problem-solving

A plot of land $A B C D$ is a quadrilateral. $A B = 40$ m, $A D = 55$ m, $\hat{A} = 75°$ , $B C = 35$ m and $C D = 45$ m.

(a) Calculate the length of the diagonal $B D$ . (3)

(b) Calculate $\hat{C}$ . (3)

Worked solution

Hidden - try the question first

Solution

(a) $\frac{sin Q}{9} = \frac{sin 35°}{6} \Rightarrow sin Q \approx 0, 8604$
$= 4625 - 4400 cos 75° \approx 3486, 20$
$B D \approx 59, 04$ m
(b) $cos C = \frac{3 5 ^{2} + 4 5 ^{2} - B D ^{2}}{2 ( 35 ) ( 45 )}$
$\approx \frac{3 250 - 3 486 , 20}{3 150} \approx - 0, 0750$
$\hat{C} \approx 94, 30°$
(c) Area $△ A B D = \frac{1}{2} (40) (55) sin 75° \approx 1062, 52$ m²
Area $△ B C D = \frac{1}{2} (35) (45) sin 94, 30° \approx 785, 28$ m²
Total $\approx 1847, 80$ m²

FAQs

How do I know whether to use the sine rule or the cosine rule?

If you know a side and the angle opposite it, use the sine rule. If you know two sides and the angle between them, or all three sides, use the cosine rule.

When does the ambiguous case happen?

When you use the sine rule to find an angle from two sides and an angle that is not between them. Check whether the obtuse option also fits.

Do I need to prove the sine, cosine and area rules?

Yes. The proofs of the sine, cosine and area rules are examinable in the IEB exam, in acute-angled triangles only. The rules themselves are on the formula sheet.

Sine rule, cosine rule and area rule

The sine rule and the ambiguous case

The cosine rule

The area rule

Key terms

Common mistakes

Practice questions

FAQs

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