The area of a triangle is half a b sin C

Two sides of a triangle and the angle between them are enough to find its area:

\text{Area}=\tfrac12 ab\sin C

Here $a$ and $b$ are the two sides, and $C$ is the angle between them.

Before you start

You need three things:

The area of a triangle is half its base times its perpendicular height. That is $\text{Area}=\frac12\times\text{base}\times\text{height}$ , where the height is measured at right angles to the base.
The sine ratio. In a right-angled triangle, $\sin\theta=\dfrac{\text{opposite}}{\text{hypotenuse}}$ .
An angle and the angle that takes it up to $180^\circ$ have the same sine: $\sin(180^\circ-\theta)=\sin\theta$ . The third fact is needed for one case, and this page comes back to it.

Predict first

The figure shows a triangle with the angle at $C$ changing. A line drops from $B$ at right angles onto the line through $A$ and $C$ .

Before you move the slider, predict: while $C$ is acute, the foot of that line lands on the side $AC$ . When $C$ becomes obtuse, where does the foot land — on the side $AC$ , on the line beyond $A$ , or on the line beyond $C$ ? Write your prediction down, then move the slider past $90^\circ$ and test it.

The proof

Take a triangle with corners $A$ , $B$ and $C$ . Use the usual labels: side $a$ is opposite $A$ , side $b$ is opposite $B$ , and side $c$ is opposite $C$ . So $a$ is the length $BC$ , $b$ is the length $CA$ , and the angle $C$ is the angle between those two sides.

Drop a perpendicular from $B$ onto the line $AC$ . Call the point where it lands $D$ , and call the length $BD$ the height $h$ . For now, suppose $D$ lands on the side $AC$ itself. The next section deals with the other case.

Start from the area you know. The area of a triangle is half its base times its height. Take the side $AC$ as the base. It has length $b$ , so

\text{Area}=\frac12\times b\times h.

Write the height using the sine ratio. Look at the right-angled triangle $BCD$ .

Its hypotenuse is $BC$ , which has length $a$ .
The side opposite the angle at $C$ is $BD$ , which has length $h$ .

The sine ratio gives

\sin C=\frac{h}{a}\qquad\text{so}\qquad h=a\sin C.

Put the height back into the area. Substituting $h=a\sin C$ into $\text{Area}=\frac12 bh$ gives

\text{Area}=\frac12 b\,(a\sin C)=\frac12 ab\sin C.

That is the rule.

One step of this proof is done by the figure. The sine ratio was applied to the angle at $C$ inside the right-angled triangle $BCD$ . That angle is the triangle's own angle $C$ only while $D$ sits on the side $AC$ . Move the slider and watch $D$ leave the side: the figure is showing you the step that needs care.

When C is obtuse

Move the slider past $90^\circ$ . The foot $D$ is now on the line beyond $C$ , outside the side $AC$ .

The step that needs care is the sine ratio. In the right-angled triangle $BCD$ , the angle at $C$ is now $180^\circ-C$ , not $C$ : the two angles lie on a straight line through $C$ and add to $180^\circ$ . So the ratio gives

\sin(180^\circ-C)=\frac{h}{a}\qquad\text{so}\qquad h=a\sin(180^\circ-C).

An obtuse angle and the angle that takes it up to $180^\circ$ have the same sine, so $\sin(180^\circ-C)=\sin C$ . The height is $h=a\sin C$ after all, and the area is still $\frac12 ab\sin C$ .

This case is worth a section of its own. A proof that handled only acute angles would have shown the rule for some triangles, not for every triangle.

Check your understanding

A triangle has sides $6\text{ cm}$ and $7\text{ cm}$ with an angle of $50^\circ$ between them. Work out its area, to 3 significant figures.
In the formula $\frac12 ab\sin C$ , which angle is $C$ ? Explain how to find it on a diagram.
Explain why an obtuse $C$ still gives a positive area.

Answers

$\frac12(6)(7)\sin 50^\circ\approx16.1$ , so the area is $16.1\text{ cm}^2$ to 3 significant figures.
$C$ is the angle between the two sides $a$ and $b$ . On a diagram, find the two lengths you are using and take the angle where they meet. If the angle you have is opposite one of those sides, it is the wrong one.
$\sin C$ is positive for every angle between $0^\circ$ and $180^\circ$ , and every angle of a triangle lies in that range. So the product $\frac12 ab\sin C$ is positive whatever the shape.

One more thing

Here is the same rule seen from a different direction. Keep the two sides $a$ and $b$ fixed and change the angle between them. The area then follows $\frac12 ab\sin C$ , and the sine is largest when $C=90^\circ$ . So a triangle with two fixed sides has its largest area when those sides meet at a right angle. You can watch that happen in the figure: slide toward $90^\circ$ and the height is at its longest.