When you know two sides of a triangle and the angle between them (but not a perpendicular height), the area is Area = ½ab sin C, where a and b are the two sides and C is the angle sandwiched between them. It works for right-angled, acute, and obtuse triangles alike, and it needs no height at all.
Most students meet triangle area as Area = ½ × base × height, the same right-angled setup behind Pythagoras' theorem. That formula is fine when a height is drawn, marked, or easy to find. The problem is that many O-Level questions give you two sides and an angle instead, with no height in sight, and dropping a perpendicular just to find one wastes time you don't have in an exam.
½ab sin C solves this directly. It comes from the same base-times-height idea (sin C gives you the height in disguise), but it lets you skip drawing it. Two sides and the angle between them go straight in, and the area comes straight out.
Step 1
Label the two known sides a and b
Step 2
Confirm C is the angle between a and b
Step 3
Area = ½ab sin C
Step 4
Calculate and round to 3 sf
The whole formula depends on one condition: C must be the angle between sides a and b, called the included angle. If the angle you're given sits opposite one of the sides instead of between them, ½ab sin C does not apply, and you need the sine rule or cosine rule instead. Before substituting anything, trace the triangle: does the angle sit exactly where the two known sides meet? If yes, you're clear to use ½ab sin C.
As covered in trigonometric ratios of an obtuse angle, sine stays positive for every angle between 0° and 180°, so ½ab sin C gives a correct, positive area even when C is obtuse (greater than 90°). You do not need to subtract C from 180° or make any other adjustment: just enter C exactly as given into your calculator's sin function.
A triangle has sides a = 8 cm and b = 10 cm, with an included angle of C = 55°. Find the area, correct to 3 significant figures.
Solution
Area = ½ × 8 × 10 × sin 55°Area = 40 × 0.8192 = 32.8 cm²Calculator check: make sure your calculator is in degree mode (look for a small "D" in the display) before evaluating sin 55°. A calculator stuck in radian mode gives a completely different, wrong number without any warning.
A triangle has area 15 cm². One side is 6 cm, and the included angle between this side and the unknown side b is 130°. Find b, correct to 3 significant figures.
Solution
15 = ½ × 6 × b × sin 130°15 = 3b × 0.7660 = 2.298bb = 15 ÷ 2.298 = 6.53 cm (3 sf)The step students get wrong
Using an angle that is not between the two named sides. If a question gives you three pieces of information (two sides and an angle) but the angle sits opposite one of the sides rather than between them, ½ab sin C gives the wrong area, even though every number you typed in was correct. Always check which two sides the angle is sandwiched between before you substitute. If the angle isn't included, switch to the sine rule to find the missing side or angle first.
Do I need to know the height of the triangle to use this formula?
No. That's the entire point of ½ab sin C: it replaces the height with sin C, so you never need to draw, calculate, or estimate a perpendicular height. You only need two sides and the angle between them.
What if I'm only given three sides and no angle?
Then ½ab sin C cannot be used directly, since there is no angle to substitute. Find the included angle first using the cosine rule (from sine rule vs cosine rule), then use that angle in ½ab sin C.
Why does the formula still work when C is obtuse?
Sine is positive for every angle from 0° up to 180°, which covers every possible triangle angle. So sin C never goes negative inside this formula, and the area comes out correctly for acute and obtuse triangles alike without any extra steps.
How is ½ab sin C different from SOH-CAH-TOA?
SOH-CAH-TOA (see SOH-CAH-TOA basics) works on right-angled triangles to find a missing side or angle. ½ab sin C works on any triangle, right-angled or not, and it finds area rather than a side or angle length.
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