Area of a Triangle Using Sine (½ab sin C)

You have a triangle with two known sides and the angle between them. You need the area, but the triangle is not a right triangle. The rule is area = ½ab sin C. This is the area of triangle using sine, and it works for any triangle where you know two sides and the included angle. The rule comes from dropping an altitude from the vertex C to side c. In standard precalculus texts like Stewart/Redlin/Watson's Precalculus (9th ed., Section 6.3), the derivation shows that the altitude h equals b sin C. Multiply by half the base a to get ½ab sin C. The same logic gives equivalent forms ½bc sin A and ½ac sin B. The rule applies to oblique and right triangles alike. When C = 90°, sin 90° = 1, and the rule reduces to ½ab, the standard right-triangle area.

When solving an oblique triangle using the law of sines, you often find two sides and an included angle as a byproduct. This is the sine area formula used after working an AAS case. In AAS (two angles and a non-included side), you first find the third angle using the sum of angles (180°). Then apply the law of sines to find the missing sides. Once you have two sides and the included angle, plug into ½ab sin C for the area.

Example: A triangle has angles 50° and 70°, and side a = 10 cm opposite the 50° angle. Find angle C = 180° - 50° - 70° = 60°.The included angle between sides a and b is angle C = 60°. Area = ½ × 10 × 12.26 × sin(60°) = 61.3 × 0.8660 ≈ 53.09 cm².

When you know all three sides of a triangle but no angles, use Heron's rule. The area of oblique triangle from side lengths only is √[s(s-a)(s-b)(s-c)], where s is the semi-perimeter (a+b+c)/2. For an oblique triangle with sides 7, 8, 9, s = 12. Area = √[12(12-7)(12-8)(12-9)] = √[12 × 5 × 4 × 3] = √720 ≈ 26.83 square units. Heron's rule is your backup when you cannot get an angle. It avoids the law of sines and law of cosines entirely. Use it for SSS problems only.

The sine area rule demands the included angle between the two known sides. Using a non-included angle gives a wrong answer. In SSA (two sides and a non-included angle), you do not have the included angle, and the area is not directly calculable with ½ab sin C. You must first solve the triangle using the law of sines, check for the ambiguous case (0, 1, or 2 triangles), find the missing side and included angle, then apply the area rule. The ambiguous case shows that SSA can produce zero triangles (when the sine of the angle would need to exceed 1 in the law of sines), one triangle, or two triangles. The case table from Stewart/Redlin/Watson's Precalculus (9th ed., Section 6.5, Table 1) gives the rules: if A < 90°, a < b sin A means no triangle; a = b sin A gives one right triangle; b sin A < a < b gives two triangles; a ≥ b gives one triangle. If A ≥ 90°, a ≤ b gives no triangle; a > b gives one triangle. The area rule is never applied directly to SSA input.