Math Core

On the AMC 12 especially, many geometry problems are designed to be solved with trigonometry: the Law of Cosines turns an angle into a length, the Law of Sines connects sides to the circumcircle, and special angles keep the numbers clean. Knowing when to reach for each one saves a lot of clever construction.

The Law of Cosines

Law of Cosines

In any triangle with sides $a, b, c$ opposite angles $A, B, C$ :

$c^2 = a^2 + b^2 - 2ab \cos C.$

It is the Pythagorean Theorem with a correction term: when $C = 90^\circ$ the correction is $0$ . If $C$ is obtuse, $\cos C$ is negative and $c$ is longer than it would be in a right triangle.

Use it two ways:

Two sides and the included angle give the third side.
Three sides give any angle: $\cos C = \dfrac{a^2 + b^2 - c^2}{2ab}$ .

A 5-7-8 triangle: the 60° angle at the bottom left is between the sides of length 5 and 8.

Worked example: The third side

Two sides of a triangle are $5$ and $8$ , with a $60^\circ$ angle between them. Then the third side is $\sqrt{25 + 64 - 2 \cdot 5 \cdot 8 \cdot \tfrac12} = \sqrt{49} = 7$ .

Worked example: An angle from three sides

A triangle has sides $7$ , $8$ and $13$ . Find its largest angle.

The largest angle $C$ is opposite the side $13$ : $\cos C = \dfrac{49 + 64 - 169}{2 \cdot 7 \cdot 8} = \dfrac{-56}{112} = -\dfrac12$ , so $C = 120^\circ$ .

The Law of Sines

Extended Law of Sines

In any triangle with circumradius $R$ :

$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R.$

Why it works. Draw the diameter from $B$ through the center, ending at $A'$ . The inscribed angle $\angle BA'C$ equals $\angle A$ (or its supplement), and triangle $BA'C$ has a right angle at $C$ . So $a = BA' \sin A = 2R \sin A$ .

Use the Law of Sines when you know two angles and a side, or when the problem mentions the circumcircle.

Worked example: Circumradius from one side and its angle

In $\triangle ABC$ , $\angle A = 30^\circ$ and $BC = 6$ . Then $2R = \dfrac{6}{\sin 30^\circ} = 12$ , so $R = 6$ , no matter what the other angles are.

Medians and cevians

A cevian is a segment from a vertex to the opposite side. To find its length, apply the Law of Cosines to the two small triangles it creates, using the fact that the two angles at its foot are supplementary (so their cosines are opposites). For a median this gives the formula

$m_a^2 = \frac{2b^2 + 2c^2 - a^2}{4}.$

Worked example: A median

In the $13$ - $14$ - $15$ triangle, find the median to the side of length $14$ .

$m^2 = \dfrac{2 \cdot 169 + 2 \cdot 225 - 196}{4} = \dfrac{592}{4} = 148$ , so $m = 2\sqrt{37}$ .

Tip

Memorize the exact values at $30^\circ$ , $45^\circ$ , $60^\circ$ , $120^\circ$ , $135^\circ$ and $150^\circ$ . AMC problems choose these angles on purpose, and $\cos 120^\circ = -\tfrac12$ turns $c^2 = a^2 + b^2 - 2ab\cos C$ into $c^2 = a^2 + b^2 + ab$ .

Common mistake

The Law of Sines can hide a second solution. If you solve $\sin B = k$ for an angle, both $B$ and $180^\circ - B$ are candidates. Check whether each one leaves room for the third angle.

Practice

Practice 1

A triangle has side lengths $3$ , $5$ and $7$ . What is the measure of its largest angle?

Practice 2

In $\triangle ABC$ , $\angle A = 45^\circ$ , $\angle B = 60^\circ$ and $BC = 6$ . What is $AC$ ?

Practice 3

A triangle has a $150^\circ$ angle, and the side opposite that angle has length $10$ . What is the radius of the circle that passes through all three vertices?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 4

In $\triangle ABC$ , $AB = 7$ , $AC = 9$ and $BC = 8$ . What is the length of the median from $A$ ?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 5

A parallelogram has sides $5$ and $8$ and one angle of $60^\circ$ . What is the length of its longer diagonal?

Practice 6

What is the circumradius of a triangle with side lengths $5$ , $6$ and $7$ ?

Practice 7

In $\triangle ABC$ , $AB = 4$ , $AC = 6$ and $\angle BAC = 120^\circ$ . The bisector of angle $A$ meets $\overline{BC}$ at $D$ . What is $AD$ ?

Practice 8

In $\triangle ABC$ , $\angle A = 60^\circ$ , $BC = 7$ and $AB + AC = 13$ . What is the area of the triangle?

Practice 9

In $\triangle ABC$ , $\tan A = \dfrac12$ , $\tan B = \dfrac13$ and $AB = 5$ . What is the area of the triangle?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Real contest practice

2019 AMC 12A, Problem 19: a triangle with integer sides and given angles; trigonometry fixes the ratio of the sides, then you find the smallest perimeter.
2020 AMC 12B, Problem 12: a chord meets a diameter at a $45^\circ$ angle; the special angle turns the chord lengths into right-triangle computations.