Triangle angle and side properties

Learning outcome

Find triangle angles, compare opposite sides and determine whether proposed lengths can form a triangle.

Concepts and assumptions

A Euclidean triangle joins three noncollinear vertices with straight segments. Lengths and interior angles are positive. ∠A means the interior angle at vertex A.

The three interior angles total 180°. To see why, draw a line through one vertex parallel to the opposite side. Alternate interior angles reproduce the other two angles along that straight line; all three together form 180°.

An exterior angle formed by extending one side is supplementary to the adjacent interior angle. It therefore equals the sum of the two nonadjacent interior angles. Use the non-reflex exterior angle.

Equal sides have equal opposite angles, and equal angles have equal opposite sides. Thus an isosceles triangle has two equal base angles; an equilateral triangle has three 60° angles. Compare opposite positions carefully: side AB is opposite ∠C, not ∠A or ∠B.

A larger angle faces a longer side, and conversely. Since the angle sum is 180°, a triangle can have at most one right or obtuse angle.

For side lengths a, b and c, the sum of any two must exceed the third. Equivalently, when a and b are known, |a - b| < c < a + b, where |a - b| denotes absolute difference. A nonstraight path is longer than the direct segment. Equality gives a straight, degenerate figure, not a triangle.

Worked examples

Example 1 — Missing angle and equal sides. In triangle ABC, ∠A = 46° and ∠B = 67°. Then ∠C = 180° - 46° - 67° = 67°. Since ∠B = ∠C, their opposite sides AC and AB are equal. The triangle is isosceles.

Example 2 — Exterior angle. In triangle PQR, extend QR beyond R to S, so Q-R-S are collinear in order. Given ∠PRS = 126°, ∠P = (2x + 6)° and ∠Q = (3x + 10)°, the exterior-angle rule gives 5x + 16 = 126. Thus x = 22, ∠P = 50°, ∠Q = 76° and ∠R = 54°. PR is longest because it faces 76°.

Example 3 — Possible lengths. Two sides are 8 cm and 13 cm; the third is an integer x cm. The bounds are 13 - 8 < x < 13 + 8, so 5 < x < 21. Thus x can be 6, 7, …, 20: 20 - 6 + 1 = 15 possibilities. Neither endpoint is allowed.

Common mistakes

Do not add an exterior angle to the three interior angles. Match equal sides with opposite angles. A pair of lengths summing exactly to the third does not form a triangle.

Practice questions

1. Triangle ABC has ∠A:∠B:∠C = 2:3:4. Find the angles and longest side.
2. In triangle ABC, AB = AC and ∠A = 38°. Find ∠B and ∠C.
3. In triangle PQR, Q-R-S are collinear in order, ∠PRS = 115° and ∠P = 47°. Find ∠Q and ∠R.
4. Two sides are 9 cm and 14 cm. Find all possible integer third-side lengths and their count.

Worked solutions

1. The 9 ratio parts total 180°, so each part is 20°. Angles are 40°, 60° and 80°; side AB faces 80° and is longest.
2. Equal sides give ∠B = ∠C. Each is (180° - 38°)/2 = 71°.
3. Exterior equality gives ∠Q = 115° - 47° = 68°. Adjacent ∠R = 180° - 115° = 65°.
4. The bounds are 5 < x < 23. Integer lengths are 6, 7, …, 22 cm, giving 22 - 6 + 1 = 17 possibilities.