A triangle has sides 8 cm and 10 cm, and the angle between them is 120 °. What is its area?

Correct — B. Two sides and the angle between them give area = ½ab sin C.

½ × 8 × 10 = 40
sin 120° = sin(180° − 60°) = sin 60° = √3⁄2
Area = 40 × √3⁄2 = 20√3 cm² → option (b)

Check: 20√3 ≈ 34.6 cm², less than the 40 cm² the same two sides would enclose at a right angle.

Why the others are wrong

• (a)
  40 cm² — 40 cm² is ½ × 8 × 10 with no sine, which is the area only when the angle is 90°. At 120°, sin 120° = √3⁄2, so the area is smaller.
• (c)
  40√3 cm² — 40√3 cm² is 8 × 10 × √3⁄2: the sine is right but the ½ is missing. At ≈69.3 cm² it is more than the 40 cm² any angle could give.
• (d)
  20 cm² — 20 cm² uses ½ in place of sin 120°. But ½ is sin 30° (and cos 120° = −½), while sin 120° = √3⁄2, which gives 20√3.

With two sides and the included angle known, area = ½ab sin C. The height onto side a is b sin C, so this is just ½ × base × height.

For an obtuse angle use sin(180° − C) = sin C: sin 120° = sin 60° = √3⁄2. The sine of an obtuse angle is positive.

The area is largest at C = 90°, where sin C = 1 and area = ½ab.

Key facts

• Area of a triangle = ½ab sin C, where C is the angle between sides a and b
• sin(180° − θ) = sin θ, so sin 120° = sin 60° = √3⁄2
• sin 30° = ½, sin 60° = √3⁄2, sin 90° = 1

Common traps

• Dropping the ½ and writing ab sin C, which doubles the area to 40√3
• Using ½ (sin 30°, or the size of cos 120°) in place of sin 120° = √3⁄2

No directly related past PYQ was found.