🏰 Geometry Kingdom · Geometry

Triangles

Classify triangles, use the 180-degree angle sum, and apply the triangle inequality and exterior angle theorem.

In short

The interior angles of any triangle add to 180°.
Classify by sides (equilateral, isosceles, scalene) or by angles (acute, right, obtuse) — the two are independent.
Equal sides face equal angles, so an isosceles triangle has equal base angles.
An exterior angle equals the sum of the two remote interior angles, and three lengths form a triangle only if the two shorter ones together exceed the longest.

The angle sum is always 180°

Every triangle, no matter how it is stretched or squashed, has three interior angles that add to exactly 180°.

You can see why with a paper triangle: tear off the three corners and lay them side by side. They fit together along a straight line — and a straight line is 180°.

This one fact solves an enormous number of problems. If two angles of a triangle are 52° and 61°, the third must be 180 - 52 - $61 = 67°$ .

It also has immediate consequences. A triangle can have at most one angle of 90° or more, because two such angles would already use up 180° with nothing left for the third. So every triangle has at least two acute angles.

Classifying triangles

Triangles are sorted in two independent ways.

By sides:

equilateral — all three sides equal (and therefore all three angles 60°)
isosceles — exactly two sides equal
scalene — no two sides equal

By angles:

acute — all three angles less than 90°
right — one angle exactly 90°
obtuse — one angle greater than 90°

The two systems are independent: a triangle can be right AND isosceles (angles $90°, 45°, 45°)$ , or obtuse and scalene, and so on.

When classifying by angles, look only at the largest angle. The other two are always acute, so they tell you nothing.

Isosceles triangles: equal sides, equal angles

In an isosceles triangle the two angles facing the equal sides — the base angles — are equal. This works in both directions: equal sides force equal angles, and equal angles force equal sides.

Suppose AB $=$ AC and the angle at A is 40°. The angles at B and C are equal, call each of them b:

$40 + b + b = 180 2b = 140 b = 70$

So each base angle is 70°. The commonest slip here is stopping at 140 and forgetting to halve.

An equilateral triangle is the extreme case: all three sides are equal, so all three angles are equal, and $180 / 3 = 60°$ each.

Exterior angles and the triangle inequality

Extend one side of a triangle past a vertex and you create an exterior angle. It is supplementary to the interior angle beside it, and it equals the sum of the two remote interior angles.

Why? If the interior angles are $A, B$ and C, the exterior angle at C is 180 - C. But $A + B + C = 180$ , so 180 - $C = A + B$ . That is the whole proof.

A different question is whether three given lengths can even form a triangle. The triangle inequality says: the two shorter sides added together must be strictly greater than the longest side. Sides $3, 4$ and 9 are impossible — a 3-stick and a 4-stick reach only 7, and cannot span a gap of 9.

Worked examples

Example 1

In triangle ABC, angle $A = 43°$ and angle $B = 88°$ . Find angle C.

The three interior angles add to 180°.
So $43 + 88 + C = 180$ .
Add the known angles: $43 + 88 = 131$ .
$C = 180$ - $131 = 49°$ .

Example 2

The angles of a triangle are $(2x)°, (3x + 10)°$ and $(x + 20)°$ . Find x, then the largest angle.

Set the sum equal to $180: 2x + (3x + 10) + (x + 20) = 180$ .
Collect like terms: $6x + 30 = 180$ .
Subtract 30 and divide by $6: 6x = 150$ , so $x = 25$ .
The angles are $50°, 85°$ and 45°, so the largest is 85°.

Practice problems, with solutions

Three problems of increasing difficulty, each with the full working. In the game these are generated fresh every time; these three are fixed so this page always shows the same ones.

Problem 1

Difficulty 1 of 5

In triangle ABC, angle $A = 110°$ and angle $B = 50°$ . How many degrees is angle C?

Answer: 20 degrees

Angle sum of a triangle: $A + B + C = 180°$ .
$110 + 50 + C = 180$
$160 + C = 180$
$C = 180$ - $160 = 20°$

Problem 2

Difficulty 3 of 5

A triangle has angles $83°, 48°$ and 49°. Classify it by its angles.

acute
obtuse
right

Answer: A. acute

The angles are $83°, 48°$ and 49°; the largest is 83°.
$83° < 90°$ , so every angle is acute and the triangle is acute.

Problem 3

Difficulty 4 of 5

In triangle ABC, angle $A = 76°$ and angle $B = 5°$ . Side BC is extended past C. How many degrees is the exterior angle at C?

Answer: 81 degrees

Interior angle $C = 180$ - 76 - $5 = 99°$ .
The exterior angle at C is 180 - $99 = 81°$ .
That equals angle $A +$ angle $B = 76 + 5 = 81°$ — the exterior angle theorem.
Exterior angle $= 81°$ .

Common mistakes

Using 360° instead of 180° for the angle sum.
Subtracting only one of the two known angles from 180.
Forgetting to halve when sharing the remaining degrees between two equal base angles.
Naming a triangle after a small angle instead of its largest one.

What you should be able to do

Classify triangles by sides and by angles.
Find a missing angle using the 180-degree angle sum.
Apply the exterior angle theorem.
Decide whether three lengths can form a triangle.

Where this fits in the curriculum

Common Core

4.G.A.2
Grade 4 — Classify two-dimensional figures by the presence of parallel or perpendicular lines and by angle size.
8.G.A.5
Grade 8 — Establish facts about the angle sum and the exterior angle of triangles.
7.G.A.2
Grade 7 — Draw triangles from three given measures and notice when they determine one triangle, many, or none.
The triangle inequality is not a Common Core standard by name; 7.G.A.2 is where "can these three lengths make a triangle?" is actually asked.

Ontario

G5.E1.1
Grade 5 — Identify geometric properties of triangles, and construct triangles from given side or angle measurements.
G8.E2.2
Grade 8 — Solve problems involving the angle properties of polygons, where the 180° angle sum of a triangle is applied.
Ontario has no expectation that states "the angles of a triangle sum to 180°"; it is used inside the angle-property work of G6.E2.3 and G8.E2.2, so this pairing is a judgement call.