Triangle properties — angle sum = 180°, exterior angle theorem: Exercises

This page is a self-test ladder for the parent topic. Every problem has a hidden full solution — try it first, then reveal. We climb five levels: L1 Recognition (spot the rule), L2 Application (plug in), L3 Analysis (combine two ideas), L4 Synthesis (build a multi-step argument), L5 Mastery (prove / general-case).

Before we start, two facts from the parent note that every solution leans on. Both are just words for now — the figures below make them pictures.

Look at the figure. The triangle has three inside angles $∠ A, ∠ B, ∠ C$ (black). At corner $C$ the bottom side $B C$ is pushed on to point $D$ (the dashed extension). The red angle $∠ A C D$ opens up outside the triangle — that is the exterior angle at $C$ . Notice it sits next to $∠ C$ on a straight line, so red $+ ∠ C = 180°$ . The two "remote" corners for that red angle are $A$ and $B$ (they don't touch the red angle). The theorem says $red = ∠ A + ∠ B$ .

Keep this single picture in your head — every exercise below is a variation on it.

L1 — Recognition

Goal: just identify which rule applies and read the numbers.

Recall Solution 1.1

WHAT rule? The three interior angles of one triangle sum to $180°$ . WHY subtract? To isolate the unknown we move the two known angles to the other side. $30° + 90° + x = 180°$ $x = 180° - 120° = 60°$ Check: $30° + 90° + 60° = 180°$ ✓

Recall Solution 1.2

WHAT looks like this? Exterior angle and its adjacent interior angle sit on a straight line — like the red and the $∠ C$ in figure s01. Angles on a straight line sum to $180°$ (this is the supplementary relationship). $interior = 180° - 115° = 65°$ Check: $115° + 65° = 180°$ ✓

L2 — Application

Goal: use the exterior-angle theorem directly, and handle the "given exterior, find interior" direction.

Recall Solution 2.1

WHY the theorem? The exterior angle at $C$ equals the two remote interior angles — those are $A$ and $B$ (they don't touch $C$ 's exterior angle, exactly like figure s01). $θ_{ext at C} = ∠ A + ∠ B = 52° + 71° = 123°$ Cross-check the slow way: interior $∠ C = 180° - 52° - 71° = 57°$ ; exterior $= 180° - 57° = 123°$ ✓

Recall Solution 2.2

Step 1 — remote sum. Exterior at $C$ $= ∠ A + ∠ B$ : $140° = 65° + ∠ B \Rightarrow ∠ B = 75°$ Step 2 — the third angle. $∠ C = 180° - ∠ A - ∠ B = 180° - 65° - 75° = 40°$ . Check: $65° + 75° + 40° = 180°$ ✓, and exterior $= 180° - 40° = 140°$ ✓

L3 — Analysis

Goal: combine angle-sum with isosceles or with a second triangle.

Recall Solution 3.1

Step 1 — apex interior angle. Exterior and interior at the apex are a linear pair: $apex = 180° - 110° = 70°$ Step 2 — base angles are equal. In an isosceles triangle the two angles opposite the equal sides are equal; call each $x$ . WHY equal? Reflect the triangle across its axis of symmetry — the two base angles swap onto each other, so they must be identical. $70° + x + x = 180° \Rightarrow 2 x = 110° \Rightarrow x = 55°$ Angles: $70°, 55°, 55°$ . Check: $70 + 55 + 55 = 180$ ✓

Recall Solution 3.2

Step 1 — angle $∠ A D B$ . In triangle $A B D$ : $∠ A D B = 180° - 40° - 30° = 110°$ . Step 2 — $∠ A D C$ is the straight-line partner (since $B, D, C$ are collinear): $∠ A D C = 180° - 110° = 70°$ Step 3 — $∠ D A C$ from triangle $A C D$ : $∠ D A C = 180° - ∠ C - ∠ A D C = 180° - 55° - 70° = 55°$ . Exterior-angle check: treat $∠ A D B$ as the exterior angle of triangle $A C D$ at $D$ (side $C D$ extended to $B$ ). Its remote interiors are $∠ C$ and $∠ D A C$ : $∠ C + ∠ D A C = 55° + 55° = 110° = ∠ A D B ✓$

L4 — Synthesis

Goal: build a several-step chain, sometimes with an unknown carried symbolically.

Recall Solution 4.1

Step 1 — let the parts be $2 k, 3 k, 4 k$ . They must total $180°$ : $2 k + 3 k + 4 k = 9 k = 180° \Rightarrow k = 20°$ Step 2 — the angles: $40°, 60°, 80°$ . Step 3 — exterior at the $80°$ corner (linear pair): $180° - 80° = 100°$ Remote-sum check: the two other angles $40° + 60° = 100°$ ✓ (matches the exterior-angle theorem).

Recall Solution 4.2

Step 1 — recover two interior angles via linear pairs: $∠ A = 180° - 130° = 50°, ∠ B = 180° - 120° = 60°$ Step 2 — third interior angle: $∠ C = 180° - 50° - 60° = 70°$ . Step 3 — third exterior angle: $180° - 70° = 110°$ . Nice check: the three exterior angles of any triangle sum to $360°$ : $130° + 120° + 110° = 360°$ ✓ WHY $360°$ ? Each exterior $= 180° -$ its interior, so the three sum to $540° - 180° = 360°$ — you've turned one full circle walking round the triangle.

L5 — Mastery

Goal: prove a general statement, and handle a degenerate/limiting case.

Recall Solution 5.1

Setup (figure s03). At $C$ , the straight line $B - C - D$ means $∠ A C B$ (interior) and $∠ A C D$ (red exterior) form a linear pair. Step 1 — straight line at $C$ : $∠ A C B + ∠ A C D = 180° (⋆)$ Step 2 — angle sum of the triangle: $∠ A + ∠ B + ∠ A C B = 180° (⋆ ⋆)$ Step 3 — both right sides equal $180°$ , so set the left sides equal: $∠ A C B + ∠ A C D = ∠ A + ∠ B + ∠ A C B$ Step 4 — cancel $∠ A C B$ : $∠ A C D = ∠ A + ∠ B ■$ Limiting case $∠ A \to 0°$ . If $A$ slides until $∠ A$ shrinks to nothing, the triangle collapses toward a straight segment. Then $∠ A C D \to 0° + ∠ B = ∠ B$ : the exterior angle just equals the one remaining corner $B$ . This is the smallest the exterior angle can be (since remote angles are $\geq 0$ ); it also confirms the exterior angle is always larger than either remote angle alone whenever both are positive — a useful ordering fact.

Recall Solution 5.2

Argument by contradiction. Suppose two angles are each $\geq 90°$ , say $α \geq 90°$ and $β \geq 90°$ . Then $α + β \geq 180°.$ But $α + β + γ = 180°$ and $γ > 0°$ (a real corner has positive opening), so $α + β = 180° - γ < 180°.$ These contradict each other ( $\geq 180°$ and $< 180°$ cannot both hold). Hence at most one angle is $\geq 90°$ . $■$ Reading it: the $180°$ "budget" is spent by all three corners; two large angles would overspend it.

Recall One-line self-quiz

Exterior angle at a corner equals ::: the sum of the two remote interior angles. The three exterior angles of any triangle sum to ::: $360°$ . Why can't a triangle have two obtuse angles? ::: their sum would already meet or exceed $180°$ , leaving nothing (or negative) for the third positive angle.

Related tools used above: Straight Lines and Angles, Supplementary and Complementary Angles, Parallel Lines and Transversals, Properties of Isosceles Triangles, Polygon Angle Sums, Triangle Congruence.