Math Core

You know the area of a triangle is $\tfrac{1}{2}bh$ . The trouble is that the height is rarely given. It is usually a length nobody measured. Trigonometry fixes this: you can find the area from two sides and an angle, or from the three sides alone.

Area from two sides and the included angle

Take any triangle and use side $b$ as the base. The height $h$ is the perpendicular distance from vertex $B$ down to the line containing side $b$ .

The height from B is h = a sin C, so the area is one-half b times a sin C.

The height is a leg of a right triangle whose hypotenuse is side $a$ and whose angle at the base is $C$ . So $\sin C = \dfrac{h}{a}$ , which gives $h = a \sin C$ . Substitute into $\tfrac{1}{2}bh$ :

\text{Area} = \frac{1}{2} \cdot b \cdot a \sin C = \frac{1}{2} ab \sin C.

If angle $C$ is obtuse, the height falls outside the triangle, but it still equals $a \sin(180^\circ - C) = a \sin C$ , so the formula holds.

SAS area formula

The area of a triangle is half the product of two sides times the sine of the angle between them:

\text{Area} = \frac{1}{2}ab \sin C = \frac{1}{2}bc \sin A = \frac{1}{2}ac \sin B.

Worked example: Two sides and the included angle

A triangle has sides $a = 9$ and $b = 14$ with $C = 40^\circ$ between them. Find its area to the nearest tenth.

\text{Area} = \frac{1}{2}(9)(14)\sin 40^\circ = 63 \sin 40^\circ \approx 63(0.6428) \approx 40.5 \text{ square units}.

Common mistake

The angle must be the one between the two sides you use. If you know $a$ , $b$ and angle $A$ , then $\tfrac{1}{2}ab\sin A$ is wrong. Find the included angle $C$ first (or find a different pair of sides that surrounds a known angle).

When you know two angles

If you know two angles and a side, first use the law of sines to get a second side. Then you have two sides and the angle between them.

Worked example: ASA area

In triangle $ABC$ , $A = 50^\circ$ , $B = 70^\circ$ and $c = 12$ . Find the area to the nearest tenth.

The third angle is $C = 60^\circ$ . By the law of sines, $b = \dfrac{12 \sin 70^\circ}{\sin 60^\circ} \approx 13.02$ . Sides $b$ and $c$ surround angle $A$ :

\text{Area} = \frac{1}{2}bc \sin A \approx \frac{1}{2}(13.02)(12)\sin 50^\circ \approx 59.8 \text{ square units}.

Area from three sides: Heron's formula

If you know all three sides, you could find an angle with the law of cosines and then use the SAS formula. A formula credited to Heron of Alexandria does all of that in one step. (It can be derived by combining the law of cosines with $\sin^2 C = 1 - \cos^2 C$ and factoring.)

Heron's formula

For a triangle with sides $a$ , $b$ and $c$ , let $s$ be the semiperimeter, half the perimeter:

s = \frac{a + b + c}{2}.

Then

\text{Area} = \sqrt{s(s - a)(s - b)(s - c)}.

Worked example: Using Heron's formula

Find the area of a triangle with sides $10$ , $17$ and $21$ .

A triangle with sides 10, 17 and 21.

The semiperimeter is $s = \dfrac{10 + 17 + 21}{2} = 24$ . The three differences are $24 - 10 = 14$ , $24 - 17 = 7$ and $24 - 21 = 3$ .

\text{Area} = \sqrt{24 \cdot 14 \cdot 7 \cdot 3} = \sqrt{7056} = 84 \text{ square units}.

Tip

Check your semiperimeter: the three differences $s - a$ , $s - b$ and $s - c$ always add up to $s$ . Above, $14 + 7 + 3 = 24$ . If one difference is zero or negative, the three lengths can't form a triangle.

Working backward

The SAS formula can also run in reverse: if you know the area and two sides, you can find the angle between them. Watch for the same ambiguity as in SSA problems, since an acute angle and its supplement have the same sine.

Worked example: Finding the included angle

A triangle has sides $a = 8$ and $b = 10$ and area $30$ . Find all possible values of angle $C$ to the nearest tenth of a degree.

30 = \frac{1}{2}(8)(10)\sin C = 40 \sin C, \qquad \sin C = \frac{30}{40} = 0.75.

So $C \approx \sin^{-1}(0.75) \approx 48.6^\circ$ or $C \approx 180^\circ - 48.6^\circ = 131.4^\circ$ . Both are possible: a skinny acute triangle and a wide obtuse one have the same area.