Area of a Triangle Using Sine: the 1/2 ab sin C Formula

For any triangle, if you know two side lengths and the size of the angle trapped between them, you can find the area without ever measuring a height. The formula is:

\( \)Area\( = \frac{1}{2}ab\sin(C) \)

Here \(a\) and \(b\) are two sides of the triangle, and \(C\) is the angle between them, the "included angle." This is often written as the \(\frac{1}{2}ab\sin(C)\) formula, and it works for any triangle, not just right triangles.

The familiar area formula for a triangle is \(\)Area\( = \frac{1}{2} \times \)base\( \times \)height\(\). The trick behind \(\frac{1}{2}ab\sin(C)\) is that \(\sin(C)\) lets you write the height in terms of a side and an angle, without drawing a separate right triangle every time.

Picture a triangle with vertex \(C\) at the top, and sides \(a\) and \(b\) running down to the base. If you drop a perpendicular height \(h\) from \(C\) straight down to the base, that height, side \(a\), and the angle \(C\) form a right triangle where \(\sin(C) = \frac{h}{a}\), so \(h = a\sin(C)\).

Substituting \(h = a\sin(C)\) into \(\)Area\( = \frac{1}{2} \times b \times h\) gives \(\)Area\( = \frac{1}{2}ab\sin(C)\). Notice the base used is side \(b\), and the height comes from side \(a\) and angle \(C\), which is exactly why \(C\) must be the angle between \(a\) and \(b\), not one of the other two angles.

Find the area of a triangle with \(a = 8\), \(b = 10\), and included angle \(C = 30^\circ\).

\( \)Area\( = \frac{1}{2}(8)(10)\sin(30^\circ) = \frac{1}{2}(80)(0.5) = 20 \)

The area is \(20\) square units. If the angle is not one of the common ones like \(30^\circ\), \(45^\circ\), or \(60^\circ\), you can review how to find the exact value of trigonometric ratios before evaluating \(\sin(C)\).

Find the area when \(a = 7\), \(b = 9\), and \(C = 110^\circ\).

\( \)Area\( = \frac{1}{2}(7)(9)\sin(110^\circ) \approx \frac{1}{2}(63)(0.9397) \approx 29.6 \)

The formula still works for an obtuse angle because \(\sin(C)\) is positive for every angle between \(0^\circ\) and \(180^\circ\), which covers every possible triangle angle.

Sometimes the area is given and you need to find a missing piece. Suppose a triangle has \(a = 12\), \(C = 40^\circ\), and an area of \(50\) square units. Solve for \(b\):

\( 50 = \frac{1}{2}(12)(b)\sin(40^\circ) \)

\( 50 = 6b\sin(40^\circ) \)

\( b = \frac{50}{6\sin(40^\circ)} \approx 12.96 \)

The same rearranging idea works if the missing piece is the angle instead of a side; you isolate \(\sin(C)\) and use an inverse sine.

The \(\frac{1}{2}ab\sin(C)\) formula only finds area. If a problem instead asks you to find a missing side or angle in a triangle that is not right-angled, you will usually need the law of cosines or the sine law, especially in mixed problems covered under applications of the sine law and cosine law. A useful check: this area formula needs two sides and the included angle; if a problem gives you two angles and one side, or all three sides, you're looking at a different tool.

The most frequent error is using an angle that is not actually between the two chosen sides. If \(a\) and \(b\) meet at vertex \(C\), only that angle belongs in the formula, not one of the other two triangle angles. Also double check your calculator's angle mode (degrees versus radians) before evaluating \(\sin(C)\), since a mismatched mode gives a completely wrong area.

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