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

This page is the "throw anything at me" companion to the parent note. We will not learn a new rule here. We will take the two rules you already have and drag them through every kind of question a triangle can ask, so that no exam question surprises you.

The two rules, restated in plain words first:

Every symbol above is a corner angle measured in degrees. Nothing else enters this page except addition and subtraction. That is the whole toolkit — the skill is knowing which subtraction each scenario wants.

The scenario matrix

Think of every triangle-angle question as landing in one of these cells. The worked examples below are tagged with the cell they cover, and together they fill the whole grid.

Cell	What makes it this case	Covered by
A. Two angles given	plainest angle-sum: find the third	Example 1
B. Ratio / algebra angles	angles given as $x, 2 x, \dots$ — no numbers yet	Example 2
C. Isosceles symmetry	two angles secretly equal	Example 3
D. Degenerate / boundary	an angle hits $0°$ or $180°$ — is it still a triangle?	Example 4
E. Right-angle special	one angle fixed at $90°$ , other two must fit	Example 5
F. Exterior forward	remote angles known → find exterior	Example 6
G. Exterior backward	exterior known → recover an interior	Example 7
H. Real-world word problem	angles hidden inside a story	Example 8
I. Exam twist (chained triangles)	one angle feeds the next triangle	Example 9

Cell A — two angles given

Worked example Example 1 (Cell A)

A triangle has angles $38°$ and $91°$ . Find the third angle.

Forecast: Guess the missing angle before reading on. Is the triangle "pointy" or "flat"?

Write the rule: $α + β + γ = 180°$ . Why this step? Every angle-sum problem starts by naming the total we are allowed to use — $180°$ , no more, no less.
Put in what we know: $38° + 91° + γ = 180°$ . Why this step? Two of the three slots are filled, so only $γ$ is unknown.
Add the knowns: $129° + γ = 180°$ . Why this step? Combine everything certain onto one side so the unknown stands alone.
Subtract: $γ = 180° - 129° = 51°$ . Why this step? Subtracting the known part from the total leaves exactly the missing part.

Verify: $38° + 91° + 51° = 180°$ ✓. Also $91°$ is the biggest angle → the triangle is obtuse (one angle over $90°$ ), which matches a "wide" shape.

Cell B — angles given as algebra

See Supplementary and Complementary Angles for the same "unknown-as- $x$ " trick applied to pairs of angles.

Cell C — isosceles symmetry

Here two angles are secretly equal because two sides are equal — see Properties of Isosceles Triangles.

Worked example Example 3 (Cell C)

An isosceles triangle has a vertex (apex) angle of $34°$ . Find the two base angles.

Forecast: The two base angles share whatever is left after $34°$ . Guess: are they bigger or smaller than the apex?

Name the base angles $x$ and $x$ . Why this step? In an isosceles triangle the two angles opposite the two equal sides are equal — look at the mirror line (dashed) in the figure; the left half is a reflection of the right half.
Angle sum: $34° + x + x = 180°$ . Why this step? Even a symmetric triangle still obeys the $180°$ total.
Combine: $34° + 2 x = 180°$ . Why this step? Two equal $x$ 's add to $2 x$ .
Isolate: $2 x = 146°$ . Why this step? Remove the known apex so only the equal pair remains.
Solve: $x = 73°$ . Why this step? Split the leftover equally between the two identical base angles.

Verify: $34° + 73° + 73° = 180°$ ✓. The base angles ( $73°$ ) are bigger than the apex ( $34°$ ) — makes sense, a narrow tip forces wide base corners.

Cell D — degenerate / boundary

Worked example Example 4 (Cell D)

Two students argue. Aki says "a triangle can have angles $0°, 90°, 90°$ ." Bina says "a triangle can have $1°, 1°, 178°$ ." Who is right?

Forecast: Both sets add to $180°$ . Does passing the sum test guarantee a real triangle?

Check Aki: $0° + 90° + 90° = 180°$ — sum is fine, but a $0°$ angle means two sides lie on top of each other; the triangle has been squashed flat. Look at the left picture: the corner has collapsed onto the line. Why this step? The sum rule is necessary but not sufficient — a real triangle also needs every angle strictly between $0°$ and $180°$ .
A $0°$ angle gives no enclosed area, so it is not a triangle (it is a "degenerate" one). Aki is wrong. Why this step? "Triangle" means three genuine corners enclosing space.
Check Bina: $1° + 1° + 178° = 180°$ , and every angle is strictly between $0°$ and $180°$ . Why this step? Same boundary test — Bina's angles are extreme but all legal.
Bina's triangle is a real (very thin, very "flat") triangle. Bina is right. Why this step? Thin is allowed; collapsed is not.

Verify: Both triples sum to $180°$ ✓, but only Bina's has all angles in the open range $(0°, 180°)$ . Boundary rule: $0° < each angle < 180°$ .

Cell E — right-angle special

Cell F — exterior angle, forward

Worked example Example 6 (Cell F)

In triangle $A B C$ , $∠ A = 47°$ and $∠ B = 58°$ . Side $B C$ is extended past $C$ to point $D$ . Find the exterior angle $∠ A C D$ .

Forecast: The exterior angle equals a sum, so it should be bigger than either $47°$ or $58°$ . Guess a value above $58°$ .

Identify the remote interior angles for the exterior at $C$ : they are $∠ A$ and $∠ B$ (the two corners not at $C$ ). Why this step? "Remote" means far from where the side was extended — in the figure, $A$ and $B$ are the two corners the red exterior angle does not touch.
Apply the theorem: $∠ A C D = ∠ A + ∠ B = 47° + 58°$ . Why this step? The theorem says the exterior angle equals the sum of exactly those two.
Add: $∠ A C D = 105°$ .

Verify (two ways): Interior at $C = 180° - 47° - 58° = 75°$ . Exterior and interior at $C$ form a straight line: $75° + 105° = 180°$ ✓. And $105°$ is indeed bigger than both $47°$ and $58°$ ✓.

Cell G — exterior angle, backward

Worked example Example 7 (Cell G)

The exterior angle at vertex $P$ of triangle $P QR$ is $132°$ . One remote interior angle, $∠ Q$ , is $54°$ . Find $∠ R$ and the interior angle at $P$ .

Forecast: Since $132°$ is the sum of $∠ Q$ and $∠ R$ , and $∠ Q$ already takes $54°$ , guess $∠ R$ around $80°$ .

Exterior theorem at $P$ : $132° = ∠ Q + ∠ R$ . Why this step? The exterior angle at $P$ equals the two remote corners, which are $Q$ and $R$ .
Substitute the known remote: $132° = 54° + ∠ R$ .
Solve: $∠ R = 132° - 54° = 78°$ . Why this step? Subtract the known remote angle from the exterior total to isolate the other remote angle.
Interior at $P$ is the linear pair of the exterior: $∠ P = 180° - 132° = 48°$ . Why this step? The exterior angle and the interior angle at the same corner sit on a straight line, so they add to $180°$ .

Verify: Interior angles $∠ P + ∠ Q + ∠ R = 48° + 54° + 78° = 180°$ ✓.

Cell H — real-world word problem

Worked example Example 8 (Cell H)

A drone flies a triangular route between towers $X$ , $Y$ , $Z$ . At tower $X$ the drone turns so that the interior angle of the path is $65°$ . At tower $Y$ the interior angle is $52°$ . The pilot needs the turning angle of the flight path itself at $Z$ — that is, the exterior angle at $Z$ (how sharply the drone must swing to keep flying straight on toward $X$ ). Find it.

Forecast: It's an exterior angle, so it equals a sum of remotes — expect it to be larger than either $65°$ or $52°$ .

Strip the story: interior angles at $X$ and $Y$ are $65°$ and $52°$ ; we want the exterior angle at $Z$ . Why this step? Word problems hide a matrix cell — here it's Cell F (exterior forward).
The remote interiors for the exterior at $Z$ are $X$ and $Y$ . Why this step? They are the two corners not at $Z$ .
Apply the theorem: exterior at $Z = 65° + 52° = 117°$ .

Verify: Interior at $Z = 180° - 65° - 52° = 63°$ ; then $63° + 117° = 180°$ (straight line) ✓, and total interior $65° + 52° + 63° = 180°$ ✓. Units: all in degrees, and a turn of $117°$ is a sharp-but-sensible swing.

Cell I — exam twist (chained triangles)

Worked example Example 9 (Cell I)

Two triangles share a straight line. In triangle $A B C$ , $∠ A = 40°$ and $∠ B = 75°$ . The exterior angle of $A B C$ at $C$ becomes an interior angle of a second triangle $C D E$ , where another angle $∠ D = 30°$ . Find $∠ E$ .

Forecast: We must find one angle to pass into the next triangle. Two triangles, so expect two rounds of the same rule.

Triangle 1 — exterior angle at $C$ : $∠ A + ∠ B = 40° + 75° = 115°$ . Why this step? The exterior angle at $C$ equals the two remote interiors — this is the bridge value the problem hands to triangle 2.
In triangle $C D E$ this $115°$ is the interior angle at $C$ . Now use angle sum: $115° + 30° + ∠ E = 180°$ . Why this step? Triangle 2 is a fresh triangle, so its own three interior angles must total $180°$ .
Combine: $145° + ∠ E = 180°$ .
Solve: $∠ E = 35°$ . Why this step? Subtract the two known angles of triangle 2 from $180°$ .

Verify: Triangle 1 interior at $C = 180° - 115° = 65°$ , check $40° + 75° + 65° = 180°$ ✓. Triangle 2: $115° + 30° + 35° = 180°$ ✓. The exterior angle of triangle 1 correctly became an interior angle of triangle 2.

This "one angle feeds the next" pattern generalises to Polygon Angle Sums and shows up whenever Parallel Lines and Transversals chain shapes together.

Decision flow — which cell am I in?

Recall Quick self-test

The exterior angle theorem gives $θ_{ext} = α + β$ . What is the interior angle at that same corner? ::: $180° - θ_{ext}$ , because the interior and exterior angles form a straight line (a linear pair). A triangle's angles are $x, x, 4 x$ . What is $x$ ? ::: $6 x = 180°$ so $x = 30°$ . Angles $0°, 90°, 90°$ sum to $180°$ — is it a triangle? ::: No; a $0°$ angle makes it degenerate (collapsed flat), and every angle must be strictly between $0°$ and $180°$ . A right triangle has one acute angle of $28°$ . The other acute angle is? ::: $62°$ , since the two acute angles are complementary (add to $90°$ ).