Math Core

Almost every geometry problem on MATHCOUNTS and the AMC 8 eventually asks for an angle, and nearly all of them come down to a handful of facts used over and over. The skill is called angle chasing: write every angle you know on the figure, then use the facts below to fill in the rest until the one you want appears.

The basic angle facts

These are the tools. Know them cold.

Angles that make a straight line add to $180^\circ$ .
Angles all the way around a point add to $360^\circ$ .
Vertical angles (opposite angles where two lines cross) are equal.
When a line crosses two parallel lines, corresponding angles and alternate interior angles are equal, and the two interior angles on the same side add to $180^\circ$ .
The angles of a triangle add to $180^\circ$ .
In an isosceles triangle, the angles opposite the equal sides (the base angles) are equal. An equilateral triangle has three $60^\circ$ angles.

One consequence saves a lot of time. An exterior angle of a triangle (formed by extending one side) equals the sum of the two interior angles that are not next to it. Why? The exterior angle and its neighbor make $180^\circ$ , and so do all three interior angles, so the exterior angle equals the other two interior angles together.

Angles of polygons

Pick one vertex of an $n$ -sided polygon and draw every diagonal from it. That cuts the polygon into $n - 2$ triangles, and their angles make up exactly the polygon's angles. So the interior angles add to $(n-2) \cdot 180^\circ$ .

Exterior angles are even simpler. Walk around any convex polygon: at each corner you turn by the exterior angle, and after one lap you have turned exactly once around. So the exterior angles always add to $360^\circ$ , no matter how many sides.

Polygon angle facts

For a polygon with $n$ sides:

The interior angles add to $(n - 2) \cdot 180^\circ$ .
The exterior angles (one at each vertex) add to $360^\circ$ .
In a regular polygon, each exterior angle is $\dfrac{360^\circ}{n}$ and each interior angle is $180^\circ - \dfrac{360^\circ}{n}$ .

For regular polygons, work with the exterior angle first. It is one short division, and the interior angle is just $180^\circ$ minus it. Going backward is just as easy: if you know an interior angle, subtract from $180^\circ$ and divide $360^\circ$ by the result to get $n$ .

You will also see diagonal counts. Each of the $n$ vertices connects to $n - 3$ others by a diagonal (not itself and not its two neighbors), and each diagonal gets counted from both ends, so a polygon has $\dfrac{n(n-3)}{2}$ diagonals.

Worked example: Back to the number of sides

Each interior angle of a regular polygon measures $156^\circ$ . How many sides does it have?

Each exterior angle is $180^\circ - 156^\circ = 24^\circ$ . The exterior angles add to $360^\circ$ , so $n = 360 \div 24 = 15$ sides.

Worked example: Chasing through an isosceles triangle

In triangle $ABC$ , point $D$ lies on side $BC$ with $AD = BD$ . If $\angle B = 32^\circ$ and $\angle C = 48^\circ$ , find $\angle DAC$ .

Since $AD = BD$ , triangle $ABD$ is isosceles, so $\angle BAD = \angle B = 32^\circ$ . The whole angle at $A$ is $180^\circ - 32^\circ - 48^\circ = 100^\circ$ . So $\angle DAC = 100^\circ - 32^\circ = 68^\circ$ .

Check with the exterior angle fact: $\angle ADC$ is exterior to triangle $ABD$ , so it equals $32^\circ + 32^\circ = 64^\circ$ . Then in triangle $ADC$ , $\angle DAC = 180^\circ - 64^\circ - 48^\circ = 68^\circ$ . Same answer.

Worked example: Two regular polygons sharing a side

A regular hexagon and a square share a side, and they lie on opposite sides of it. At the top end of the shared side, a segment connects the next vertex of the hexagon to the next vertex of the square. What angle does this segment make with the top side of the square?

A regular hexagon (left) and a square (right) share a side. The segment joins a hexagon vertex to a square vertex.

At the shared top vertex, three angles fit around the point: the hexagon's $120^\circ$ , the square's $90^\circ$ , and the angle of the new triangle. So the triangle's angle there is $360^\circ - 120^\circ - 90^\circ = 150^\circ$ .

The triangle's two sides at that vertex are a side of the hexagon and a side of the square, which have the same length. So the triangle is isosceles, and each of its other angles is $\dfrac{180^\circ - 150^\circ}{2} = 15^\circ$ . The segment makes a $15^\circ$ angle with the top of the square.

Worked example: A bend between parallel lines

Two horizontal lines are parallel. Point $A$ is on the top line and $B$ is on the bottom line, and segments $AP$ and $PB$ meet at a point $P$ between the lines, to the right of $A$ and $B$ . The angle between the top line and $AP$ is $35^\circ$ , and the angle between the bottom line and $BP$ is $50^\circ$ . Find $\angle APB$ .

Two parallel lines with a bend at P between them (not to scale).

Draw a third line through $P$ , parallel to the other two. It splits $\angle APB$ into two pieces. The top piece and the $35^\circ$ angle are alternate interior angles, so the top piece is $35^\circ$ . Likewise the bottom piece is $50^\circ$ . So $\angle APB = 35^\circ + 50^\circ = 85^\circ$ .

Tip

When a figure has parallel lines and a "bent" path between them, draw an extra parallel line through each bend. It turns the problem into several easy alternate-interior-angle steps.

Common mistake

"Regular" matters. The formula $180^\circ - \dfrac{360^\circ}{n}$ gives each angle only when all the angles are equal. For a polygon that isn't regular, you only know the sum of the angles, $(n-2)\cdot 180^\circ$ .