Sine and Cosine Rules

Apply the sine rule and cosine rule to non-right-angled triangles for GCSE Maths. Find missing sides, angles, and areas.

SOH CAH TOA only works for right-angled triangles. For non-right-angled triangles, use the sine rule and cosine rule.

The Sine Rule

asin⁡A=bsin⁡B=csin⁡C\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}

Use when you have a side-angle pair and one other measurement.

Finding a Side

asin⁡A=bsin⁡B\frac{a}{\sin A} = \frac{b}{\sin B} → a=bsin⁡Asin⁡Ba = \frac{b \sin A}{\sin B}

Finding an Angle

sin⁡Aa=sin⁡Bb\frac{\sin A}{a} = \frac{\sin B}{b} → sin⁡A=asin⁡Bb\sin A = \frac{a \sin B}{b}

The Cosine Rule

Finding a Side

a2=b2+c2−2bccos⁡Aa^2 = b^2 + c^2 - 2bc\cos A

Use when you know two sides and the included angle (SAS).

Finding an Angle

cos⁡A=b2+c2−a22bc\cos A = \frac{b^2 + c^2 - a^2}{2bc}

Use when you know three sides (SSS).

Area of a Triangle

Area=12absin⁡C\text{Area} = \frac{1}{2}ab\sin C

Use when you know two sides and the included angle.

Worked Example: Sine Rule (Side

Problem

A=40°A = 40°, B=60°B = 60°, b=10b = 10. Find aa. a=10sin⁡40°sin⁡60°≈7.42a = \frac{10 \sin 40°}{\sin 60°} \approx 7.42

Solution

Worked Example: Cosine Rule (Side

Problem

b=8b = 8, c=6c = 6, A=50°A = 50°. Find aa. a2=64+36−96cos⁡50°=100−61.7=38.3a^2 = 64 + 36 - 96\cos 50° = 100 - 61.7 = 38.3. a≈6.19a \approx 6.19.

Solution

Worked Example: Area

Problem

a=7a = 7, b=9b = 9, C=65°C = 65°. Area = 12(7)(9)sin⁡65°≈28.6\frac{1}{2}(7)(9)\sin 65° \approx 28.6.

Solution

When to Use Which Rule

Known Use
Two angles + one side Sine rule
Two sides + opposite angle Sine rule
Two sides + included angle Cosine rule
Three sides Cosine rule

Practice Problems

    1. A=35°A = 35°, a=8a = 8, B=70°B = 70°. Find bb.
    1. Sides 5, 7, 9. Find the largest angle.
    1. Sides 6 and 10, included angle 45°. Find the area.

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Key Takeaways

  • ✓

    Sine rule for angle-side pairs.

  • ✓

    Cosine rule for SAS or SSS.

  • ✓

    Area = 12absin⁡C\frac{1}{2}ab\sin C.

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