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Grade 11 · Trigonometry · 12 min read

Sine, Cosine & Area Rules (Grade 11)

Choosing between the sine rule, the cosine rule and the area rule for non-right-angled triangles, and applying them to two-dimensional problems.

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.
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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

asin A = bsin B = csin C

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

A40°B75°C?12
In △ABC, Â = 40°, B̂ = 75° and AC = 12 units.
Worked ExampleCalculate BC in the triangle above
  1. 1
    BC is opposite Â, and AC (=12) is opposite B̂.
    Match each side to its opposite angle first.
  2. 2
    BCsin 40° = 12sin 75°.
    Sine rule.
  3. 3
    BC = 12 sin 40°sin 75°.
  4. 4
    = 7,99 units.
    Shorter than 12, and it faces the smaller angle. Sensible.

3The cosine rule

a² = b² + c² − 2bc·cos A

and for an angle: cos A = b² + c² − a²2bc

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

A60°BC5?8
In △ABC, AC = 8 units, AB = 5 units and  = 60°.
Worked ExampleCalculate BC in the triangle above
  1. 1
    BC² = 8² + 5² − 2(8)(5)cos60°.
    BC is opposite the known angle A.
  2. 2
    = 64 + 25 − 80(0,5).
    cos60° = 0,5.
  3. 3
    = 89 − 40 = 49.
  4. 4
    BC = 7 units.
    49 = 7 exactly.
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Do not work out 2bc − cos A separately. The whole term 2bc·cos A is one product — multiply all three, then subtract.

4The area rule

Area of △ABC = 12·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  = 60°
  1. 1
    Area = 12(8)(5)sin60°.
    Â lies between AC and AB.
  2. 2
    = 20 × 0,8660.
  3. 3
    = 17,32 square units.

Practice exercises

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

  1. 1
    In △ABC, Â = 40°, B̂ = 75° and AC = 12 units. Calculate the length of BC.
    A40°B75°C?12
    Show answer ▾
    BCsin40° = 12sin75° → BC = 7,99 units
  2. 2
    In △ABC, AC = 8 units, AB = 5 units and  = 60°. Calculate the length of BC.
    A60°BC5?8
    Show answer ▾
    BC² = 64 + 25 − 80cos60° = 49 → BC = 7 units
  3. 3
    Calculate the area of △ABC if AC = 8 units, AB = 5 units and  = 60°.
    A60°BC5?8
    Show answer ▾
    12(8)(5)sin60° = 17,32 square units
  4. 4
    Which rule would you use given three sides and no angles?
    Show answer ▾
    The cosine rule, rearranged as cos A = b² + c² − a²2bc
  5. 5
    Which rule would you use given two angles and one side?
    Show answer ▾
    The sine rule
  6. 6
    Write down the area rule for △PQR using sides p and q.
    Show answer ▾
    Area = 12·pq·sin R
  7. 7
    In △ABC, a = 7, b = 8, c = 5. Calculate  to one decimal.
    Show answer ▾
    cos A = 64 + 25 − 492(8)(5) = 4080 = 0,5 → Â = 60,0°
  8. 8
    Why can SOHCAHTOA not be used in △ABC above?
    Show answer ▾
    It is not right-angled — SOHCAHTOA is only defined for right-angled triangles.
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