MPM2D The Cosine Law: Notes and Practice

A community garden has two straight fences that meet at a corner: one is 12 m long, the other 15 m, and the angle between them is $7 0^{\circ}$ . You want to close off the triangle with a third fence running straight from the end of one fence to the end of the other. How long must it be? Try the sine law: you know one angle ( $7 0^{\circ}$ ), but the side opposite it is exactly what you are looking for, and the sides you know (12 m and 15 m) sit across from angles you don't know. There is no complete opposite pair, so the sine law is stuck. You need a formula that links two sides, the angle between them, and the side across from that angle. That formula is the cosine law.

a^{2} = b^{2} + c^{2} - 2 b c cos A

The cosine law, written to find side a, which is opposite angle A. Angle A is the angle between sides b and c.

Where the cosine law comes from

Once again, make right triangles with an altitude. In acute triangle $A BC$ , drop the altitude $h$ from $C$ to point $D$ on side $A B$ . Let $x$ be the length $A D$ . Because the whole side $A B$ is $c$ , the other piece is $D B = c - x$ . Now use the two right triangles. In the left one, $A C D$ , the Pythagorean theorem gives $b^{2} = x^{2} + h^{2}$ , and the cosine ratio gives $cos A = \frac{x}{b}$ , so $x = b cos A$ . In the right one, $BC D$ , the Pythagorean theorem gives $a^{2} = (c - x)^{2} + h^{2}$ .

Fig 1Left: the altitude h splits side AB into x and c − x, and the two right triangles lead to a² = b² + c² − 2bc cos A. Right: two triangles with the same sides b and c; at 90° the cosine law is the Pythagorean theorem, and at 55° side a is shorter.

a^{2} = (c - x)^{2} + h^{2} = c^{2} - 2 c x + x^{2} + h^{2} = c^{2} - 2 c x + b^{2} = b^{2} + c^{2} - 2 b c cos A since x^{2} + h^{2} = b^{2} since x = b cos A

Look at what happened. Expanding $(c - x)^{2}$ gave $c^{2} - 2 c x + x^{2}$ . The pair $x^{2} + h^{2}$ was replaced by $b^{2}$ from the left triangle, and $x$ was replaced by $b cos A$ . Both right triangles were needed, and both tools from right-triangle trigonometry appeared: the Pythagorean theorem and the cosine ratio. That is where the law gets its name. You don't need to reproduce this argument, but it shows exactly why the formula has the shape it does.

The link to the Pythagorean theorem

The cosine law is the Pythagorean theorem plus a correction term, $- 2 b c cos A$ . If $∠ A$ were exactly $9 0^{\circ}$ , then $cos 9 0^{\circ} = 0$ , the correction term vanishes, and the law becomes $a^{2} = b^{2} + c^{2}$ : the Pythagorean theorem, with $a$ as the hypotenuse. When $∠ A$ is acute, $cos A$ is positive, so something is subtracted and $a^{2}$ is less than $b^{2} + c^{2}$ . That matches the picture: closing the angle between two sides of fixed length brings their ends closer together. This also gives a quick test for an acute triangle: if the square of the longest side is less than the sum of the squares of the other two sides, the largest angle is acute, and so all three angles are acute.

To find	Formula	The angle is between
side $a$	$a^{2} = b^{2} + c^{2} - 2 b c cos A$	sides $b$ and $c$
side $b$	$b^{2} = a^{2} + c^{2} - 2 a c cos B$	sides $a$ and $c$
side $c$	$c^{2} = a^{2} + b^{2} - 2 ab cos C$	sides $a$ and $b$
$∠ A$	$cos A = \frac{b ^{2} + c ^{2} - a ^{2}}{2 b c}$	three sides known

The three versions of the cosine law, and the version rearranged to find an angle. The side on the left always sits across from the angle on the right.

When to use the cosine law

Use it for SAS: two sides and the angle between them, to find the third side. Use it for SSS: all three sides, to find any angle. In both cases there is no complete opposite pair yet, so the sine law can't start. After one use of the cosine law you have a complete pair, and you can finish with the sine law or the cosine law again.

VideoUsing the cosine law to find the third side when two sides and the angle between them are known.Watch on YouTube

Finding an angle from three sides

To find an angle, rearrange the law so the cosine is alone: start from $a^{2} = b^{2} + c^{2} - 2 b c cos A$ , add $2 b c cos A$ to both sides and subtract $a^{2}$ to get $2 b c cos A = b^{2} + c^{2} - a^{2}$ , then divide by $2 b c$ . The side in the subtracted spot, $a^{2}$ , is the side opposite the angle you want. Then use $cos^{- 1}$ . Unlike the inverse sine, $cos^{- 1}$ gives the correct angle in any triangle, because no two different angles between $0^{\circ}$ and $18 0^{\circ}$ share a cosine. A good habit: find the largest angle first (it is opposite the longest side). If it comes out under $9 0^{\circ}$ , the triangle is acute.

Loading interactive…

Choose Solve, then Acute, and pick SAS or SSS: type in the measures and watch the triangle drawn to scale and solved step by step, starting with the cosine law.

Order of operations

b^{2} + c^{2} - 2 b c cos A

, the product

2 b c cos A

is worked out first, then subtracted. Do not compute

b^{2} + c^{2} - 2 b c

and then multiply by

cos A

. On a calculator, type the whole right side in one line, or find

2 b c cos A

on its own first.

The Cosine Law

Key things to remember

Where the cosine law comes from

The link to the Pythagorean theorem

Finding an angle from three sides

Worked examples

Where marks get lost

Flashcards 7

Practice