Sine, Cosine & Area Rules — Grade 11 & 12 Maths (CAPS)

The sine rule, the cosine rule and the area of a triangle: which one to use, and how to solve any non-right-angled triangle, with worked exam examples.

Also known as: law of sines · law of cosines · non-right triangles · oblique triangles

SOHCAHTOA only works in a right-angled triangle. For any other triangle you need these three rules.

1Which rule do I use?

Sine rule — when you have an angle opposite a known side (two angles and a side, or two sides and a non-included angle).
Cosine rule — when you have three sides, or two sides and the angle between them.
Area rule — when you have two sides and the angle between them and you want the area.

💡

Quick test: if a side and its opposite angle are both known, reach for the sine rule. If not, it is the cosine rule.

2The sine rule

a/(sin A) = b/(sin B) = c/(sin C)

Lower-case a is the side opposite angle A. Turn it upside down when you are solving for an angle.

In △ABC, Â = 40°, B̂ = 75° and AC = 12 units.

Worked ExampleCalculate BC in the triangle above

1
BC is opposite Â, and AC (=12) is opposite B̂.
Match each side to its opposite angle first.
2
BC/(sin 40°) = 12/(sin 75°).
Sine rule.
3
BC = (12 sin 40°)/(sin 75°).
4
= 7,99 units.
Shorter than 12, and it faces the smaller angle. Sensible.

⚠️

The ambiguous case. When you use the sine rule to find an ANGLE from two sides and a non-included angle, sin B gives you an acute angle B — but 180° − B has the same sine. Check whether 180° − B also fits with the given angle (the three angles must still add up to less than 180°). If it does, two different triangles are possible; a sketch or the question will tell you which one is meant.

3The cosine rule

a² = b² + c² − 2bc·cos A

and for an angle: cos A = (b² + c² − a²)/2bc

It is Pythagoras with a correction term. If A = 90° then cos A = 0 and it collapses back to a² = b² + c².

In △ABC, AC = 8 units, AB = 5 units and Â = 60°.

Worked ExampleCalculate BC in the triangle above

1
BC² = 8² + 5² − 2(8)(5)cos 60°.
BC is opposite the known angle A.
2
= 64 + 25 − 80(0,5).
cos 60° = 0,5.
3
= 89 − 40 = 49.
4
BC = 7 units.
√49 = 7 exactly.

⚠️

Do not work out b² + c² − 2bc first and then multiply by cos A. The whole term 2bc·cos A is one product — multiply all three, then subtract it from b² + c².

4The area rule

Area of △ABC = 1/2·ab·sin C

The two sides must sandwich the angle. Any pair works, as long as the angle sits between them.

Worked ExampleCalculate the area of the triangle with AC = 8, AB = 5 and Â = 60°

1
Area = 1/2(8)(5)sin 60°.
Â lies between AC and AB.
2
= 20 × 0,8660.
3
= 17,32 square units.

5Proving the three rules

The ATP says prove and apply. A proof is examinable and worth up to five marks, and all three start the same way: drop a perpendicular from one vertex to the opposite side, creating two right-angled triangles you already know how to handle.

Worked ExampleProve the AREA rule: area of △ABC = ½ · ab · sin C

1
Construction: draw AD ⊥ BC, with D on BC. Let AD = h.
The height is what the ordinary area formula needs.
2
Area of △ABC = ½ × BC × h = ½ · a · h
½ × base × ⊥ height
3
In right-angled △ADC: sin C = h/b
SOH — b is the hypotenuse AC.
4
∴ h = b sin C
Make h the subject.
5
Area = ½ · a · b sin C
Substitute back.

Worked ExampleProve the SINE rule: a/(sin A) = b/(sin B)

1
Construction: draw CD ⊥ AB, with D on AB. Let CD = h.
2
In △ACD: sin A = h/b, so h = b sin A
3
In △BCD: sin B = h/a, so h = a sin B
The same h, from the other triangle.
4
∴ b sin A = a sin B
Both equal h.
5
Dividing both sides by sin A · sin B: b/(sin B) = a/(sin A)

Worked ExampleProve the COSINE rule: a² = b² + c² − 2bc cos A

1
Construction: draw CD ⊥ AB with D on AB. Let AD = x, so DB = c − x, and CD = h.
Naming the two pieces of AB is the key step.
2
In △ACD: b² = x² + h² (Pythagoras) and cos A = x/b, so x = b cos A
3
In △BCD: a² = (c − x)² + h²
Pythagoras in the second triangle.
4
a² = c² − 2cx + x² + h²
Expand.
5
But x² + h² = b², so a² = c² − 2cx + b²
Substitute from the first triangle.
6
a² = b² + c² − 2bc cos A
Replace x with b cos A.

💡

All three proofs begin with “construct the perpendicular” and end by eliminating h. If you remember only that, you can rebuild any of them under exam pressure.

Practice exercises

Work each one out, then click to reveal the answer.

📄 Worksheet ✅ Memo 📝 Exam paper 🗝️ Marking guide

1
In △ABC, Â = 40°, B̂ = 75° and AC = 12 units. Calculate the length of BC.
Show answer ▾Hide answer ▴
BC/(sin 40°) = 12/(sin 75°) → BC = 7,99 units
2
In △ABC, AC = 8 units, AB = 5 units and Â = 60°. Calculate the length of BC.
Show answer ▾Hide answer ▴
BC² = 64 + 25 − 80cos60° = 49 → BC = 7 units
3
Calculate the area of △ABC if AC = 8 units, AB = 5 units and Â = 60°.
Show answer ▾Hide answer ▴
1/2(8)(5)sin 60° = 17,32 square units
4
Which rule would you use given three sides and no angles?
Show answer ▾Hide answer ▴
The cosine rule, rearranged as cos A = (b² + c² − a²)/2bc
5
Which rule would you use given two angles and one side?
Show answer ▾Hide answer ▴
The sine rule
6
Write down the area rule for △PQR using sides p and q.
Show answer ▾Hide answer ▴
Area = 1/2·pq·sin R
7
In △ABC, a = 7, b = 8, c = 5. Calculate Â to one decimal.
Show answer ▾Hide answer ▴
cos A = (64 + 25 − 49)/(2(8)(5)) = 40/80 = 0,5 → Â = 60,0°
8
Why can SOHCAHTOA not be used in △ABC above?
Show answer ▾Hide answer ▴
It is not right-angled — SOHCAHTOA is only defined for right-angled triangles.
9
Prove that the area of △ABC = ½ab sin C.
Show answer ▾Hide answer ▴
Draw AD ⊥ BC, AD = h ✓; area = ½ × a × h ✓; in △ADC, sin C = h/b so h = b sin C ✓; ∴ area = ½ab sin C ✓
10
Prove the sine rule for △ABC.
Show answer ▾Hide answer ▴
Draw CD ⊥ AB, CD = h ✓; in △ACD, h = b sin A ✓; in △BCD, h = a sin B ✓; ∴ b sin A = a sin B ✓; dividing by sin A sin B gives a/(sin A) = b/(sin B) ✓
11
What single construction begins the proof of all three rules?
Show answer ▾Hide answer ▴
Drop a perpendicular from one vertex to the opposite side ✓, creating two right-angled triangles ✓