Area of a Triangle (½ab sin C)

\( A = \tfrac{1}{2}ab\sin C \)
Trigonometry GCSE

\( Triangle with a=10 cm, b=14 cm, angle C=45°. Find its area. \)

Hint (H)
\( Use A = 1/2 ab sin C \)

Explanation

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Statement

The area of a triangle can be calculated if you know two sides and the included angle. The formula is:

\[ A = \tfrac{1}{2} ab \sin C \]

Here, \(a\) and \(b\) are two sides of the triangle, and \(C\) is the angle between them.

Why it’s true (short reason)

  • In any triangle, area is half base times height.
  • If you take side \(a\) as the base, the perpendicular height can be expressed as \(b \sin C\).
  • Substituting gives \(A = \tfrac{1}{2} \times a \times b \sin C\).

Recipe (how to use it)

  1. Identify two sides and the included angle.
  2. Check your calculator is in degrees (for GCSE) or radians (if specified).
  3. Substitute values into the formula: \(A = \tfrac{1}{2} ab \sin C\).
  4. Work out the sine first, then multiply.
  5. Give the answer with correct square units.

Spotting it

This formula is useful when:

  • You are given two sides and the angle between them.
  • No perpendicular height is provided directly.
  • The triangle is not right-angled, so simple base–height methods do not apply.

Common pairings

  • The sine rule and cosine rule, when solving non-right-angled triangles.
  • Pythagoras’ theorem, for cross-checking side lengths in special cases.
  • Coordinate geometry, where trigonometry is used to find heights.

Mini examples

  1. Given: Triangle with sides \(a=7\) cm, \(b=10\) cm, angle \(C=30^\circ\). Find: Area. Answer: \(A = \tfrac{1}{2} \times 7 \times 10 \times \sin 30^\circ = 17.5\ \text{cm}^2\).
  2. Given: Triangle with sides \(a=12\) m, \(b=15\) m, angle \(C=90^\circ\). Find: Area. Answer: \(A = \tfrac{1}{2} \times 12 \times 15 \times \sin 90^\circ = 90\ \text{m}^2\).

Pitfalls

  • Forgetting that the angle must be the included angle between the two sides.
  • Leaving the calculator in radians when the question is in degrees (or vice versa).
  • Mixing up sine with cosine — the correct formula is always with sine.
  • Forgetting units — always give area in cm², m², etc.

Exam strategy

  • Underline the given sides and angle to confirm they form an included pair.
  • Check if the triangle is right-angled — the formula still works but reduces to \(\tfrac{1}{2} ab\).
  • Use at least 3–4 significant figures until the final answer to avoid rounding errors.
  • Always label your substitution clearly to gain method marks.

Summary

The formula \(A = \tfrac{1}{2} ab \sin C\) is a powerful method for finding triangle area when no perpendicular height is available. It works by expressing height in terms of sine. In exams, look for two sides and their included angle, substitute carefully, and check your calculator mode. Remember that the formula is always based on sine, never cosine or tangent.

Worked examples

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  1. \( Find the area of a triangle with sides a=7 cm, b=10 cm, and angle C=30°. \)
    1. \( A = 1/2 ab sin C \)
    2. \( A = 1/2 × 7 × 10 × sin 30° \)
    3. \( A = 35 × 0.5 \)
    4. \( A = 17.5 \)
    Answer: \( 17.5\ \text{cm}^2 \)
  2. \( Find the area of a triangle with sides a=8 m, b=12 m, angle C=90°. \)
    1. \( A = 1/2 × 8 × 12 × sin 90° \)
    2. \( A = 48 × 1 \)
    3. \( A = 48 \)
    Answer: \( 48\ \text{m}^2 \)
  3. \( Triangle with a=9 cm, b=14 cm, angle C=45°. Find its area. \)
    1. \( A = 1/2 × 9 × 14 × sin 45° \)
    2. \( A = 63 × 0.707 \)
    3. A ≈ 44.5
    Answer: \( 44.5\ \text{cm}^2 \)
  4. \( Find area of a triangle with a=20 m, b=25 m, angle C=60°. \)
    1. \( A = 1/2 × 20 × 25 × sin 60° \)
    2. \( A = 250 × 0.866 \)
    3. A ≈ 216.5
    Answer: \( 216.5\ \text{m}^2 \)
  5. \( Triangle has a=5 cm, b=13 cm, angle C=120°. Find its area. \)
    1. \( A = 1/2 × 5 × 13 × sin 120° \)
    2. \( A = 32.5 × 0.866 \)
    3. A ≈ 28.1
    Answer: \( 28.1\ \text{cm}^2 \)
  6. \( Triangle with sides a=11 m, b=15 m, angle C=75°. Find its area. \)
    1. \( A = 1/2 × 11 × 15 × sin 75° \)
    2. \( A = 82.5 × 0.966 \)
    3. A ≈ 79.7
    Answer: \( 79.7\ \text{m}^2 \)
  7. \( Triangle with a=6 cm, b=10 cm, angle C=90°. Find its area. \)
    1. \( A = 1/2 × 6 × 10 × sin 90° \)
    2. \( A = 30 × 1 \)
    3. \( A = 30 \)
    Answer: \( 30\ \text{cm}^2 \)
  8. \( Triangle with a=18 m, b=22 m, angle C=45°. Find area. \)
    1. \( A = 1/2 × 18 × 22 × sin 45° \)
    2. \( A = 198 × 0.707 \)
    3. A ≈ 139.9
    Answer: \( 139.9\ \text{m}^2 \)
  9. \( Triangle with a=9 cm, b=12 cm, angle C=60°. Find its area. \)
    1. \( A = 1/2 × 9 × 12 × sin 60° \)
    2. \( A = 54 × 0.866 \)
    3. A ≈ 46.8
    Answer: \( 46.8\ \text{cm}^2 \)
  10. \( Triangle with a=16 m, b=30 m, angle C=40°. Find its area. \)
    1. \( A = 1/2 × 16 × 30 × sin 40° \)
    2. \( A = 240 × 0.643 \)
    3. A ≈ 154.3
    Answer: \( 154.3\ \text{m}^2 \)