Types of angles — acute, right, obtuse, straight, reflex, complete: Exercises

This page is a practice ladder. Each rung is harder than the last:

L1 Recognition — can you name the type?
L2 Application — can you compute with the rules?
L3 Analysis — can you reason backward or combine ideas?
L4 Synthesis — can you build a full argument from several facts?
L5 Mastery — can you handle the tricky, "everything at once" problem?

Every solution is hidden inside a collapsible callout so you can try first, then check. Work with pencil and paper — reveal only when you have committed to an answer.

Before we start, one shared picture that every problem leans on. It is a horizontal number line running from $0°$ to $360°$ , with the boundary marks $0°$ , $90°$ , $180°$ , $360°$ drawn as black ticks. Coloured bands sit above the line — blue for the acute zone, green for obtuse, orange for reflex — and red arrows point down to the exact-value angles (right $90°$ , straight $180°$ , complete $360°$ ). It shows the whole "opening door" progression so you always know which zone a measure lands in. Every "classify" step below is literally "drop the number onto this line and read the band it falls in" — keep it in view as you work.

Figure — Types of angles — acute, right, obtuse, straight, reflex, complete

Definition The classification rule (our only tool for naming)

Given an angle measure $θ$ (in degrees), we place it on the number line above and read off its zone:

$θ = 0°$ → zero angle (degenerate: the two rays lie exactly on top of each other, no opening at all)
$0° < θ < 90°$ → acute
$θ = 90°$ → right
$90° < θ < 180°$ → obtuse
$θ = 180°$ → straight
$180° < θ < 360°$ → reflex
$θ = 360°$ → complete

Notice the two degenerate endpoints, $θ = 0°$ and $θ = 360°$ , are special: at $0°$ there is no opening (the rays coincide, having never separated), and at $360°$ the moving ray has swept all the way back onto the fixed one. Neither is acute and neither is reflex — they bracket the whole scale.

The word degree just means "one $\frac{1}{360}$ slice of a full turn." A full spin around a point is cut into $360$ equal slices; each slice is $1°$ . (Why $360$ ? It divides evenly into halves, thirds, quarters, fifths, sixths… very handy.) If you want the physical picture of measuring $θ$ , see Measuring angles with a protractor.

Level 1 — Recognition

Here the only skill is: read the number, find the zone. Ignore how the picture looks — trust the measure.

L1.1

State the type of an angle measuring $θ = 3°$ .

Recall Solution

WHAT: Locate $3°$ on the zone line (figure s01). WHY: $3°$ is above $0°$ and below $90°$ , so it sits in the first band — the blue acute zone. $0° < 3° < 90° \Rightarrow acute$ Answer: acute. (A very sharp, almost-closed corner — the door barely cracked open.)

L1.2

State the type of an angle measuring $θ = 90°$ .

Recall Solution

WHAT: Check the exact boundary on the zone line (figure s01). WHY: $90°$ is not "less than $90°$ " and not "more than $90°$ " — it lands on the red boundary tick between the blue and green bands, which has its own name. $θ = 90° \Rightarrow right$ Answer: right angle. This is a perfect square corner, one-quarter of a full turn.

L1.3

State the type of an angle measuring $θ = 249°$ .

Recall Solution

WHAT: Drop $249°$ onto the zone line (figure s01) and see which band catches it. WHY: $249°$ lands in the orange band, strictly between "straight" ( $180°$ ) and "complete" ( $360°$ ) — the long-way-around zone. $180° < 249° < 360° \Rightarrow reflex$ Answer: reflex.

Level 2 — Application

Now we compute — usually with the reflex bridge $θ_{r} = 360° - θ$ .

L2.1

Find the reflex angle that corresponds to $θ = 75°$ .

Recall Solution

WHAT: Apply the bridge. WHY: The short opening is $75°$ ; the long way round is whatever completes the full $360°$ turn. $θ_{r} = 360° - 75° = 285°$ Check WHAT it looks like: dropping $285°$ onto the zone line (figure s01) lands it in the orange band, $180° < 285° < 360°$ ✓ — it is a reflex angle, as a reflex must be. Answer: $285°$ .

L2.2

An angle and its reflex differ by $150°$ . Find the smaller (non-reflex) angle $θ$ .

Recall Solution

WHAT: Turn the words into two equations. WHY: Two facts are given — they sum to a full turn, and they differ by $150°$ . Two facts, two unknowns, solvable. $θ + θ_{r} = 360°, θ_{r} - θ = 150°$ WHY write the difference as $θ_{r} - θ$ (this order)? The word reflex means "the long way around," so by definition the reflex $θ_{r}$ is the larger of the pair and the non-reflex $θ$ is the smaller. A difference should read "bigger minus smaller" to come out positive, so it must be $θ_{r} - θ = 150°$ , never $θ - θ_{r}$ (which would be $- 150°$ ). Add the equations to cancel $θ$ : $2 θ_{r} = 510° \Rightarrow θ_{r} = 255°$ Then subtract back: $θ = 360° - 255° = 105°$ Check: $255° - 105° = 150°$ ✓ and $105° + 255° = 360°$ ✓. Dropping $105°$ onto the zone line (figure s01) lands it in the green band. Answer: the smaller angle is $θ = 105°$ (which is itself obtuse, since $90° < 105° < 180°$ ).

L2.3

The hour and minute hands of a clock at 2:00 are $60°$ apart the short way. What is the reflex angle between them?

Recall Solution

WHAT: Same bridge, applied to a real gap. WHY: Between any two hands there are always two angles: the short one ( $60°$ ) and the long one that wraps the rest of the clock face. $θ_{r} = 360° - 60° = 300°$ Answer: $300°$ (reflex). This is why "how far apart" needs a direction — see the reflex point below.

Level 3 — Analysis

Now we reason: work backward from a type, or from a relationship, to a number.

L3.1

An angle is acute, and its supplement (the angle that completes a straight line with it) is obtuse. Prove this is always true for any acute angle.

Recall Solution

WHAT: Let the acute angle be $θ$ with $0° < θ < 90°$ . Its supplement is $s = 180° - θ$ . WHY start with the inequality: the type "acute" is exactly the statement $0° < θ < 90°$ . To find the type of $s$ we track what happens to $180° - θ$ as $θ$ ranges over that band. Subtract each part of $0° < θ < 90°$ from $180°$ . Subtracting flips the inequality direction: $180° - 0° > 180° - θ > 180° - 90°$ $180° > s > 90°$ Reading it forward: $90° < s < 180°$ , which is precisely the obtuse band on the zone line (figure s01). Answer: For every acute $θ$ , the supplement lands strictly between $90°$ and $180°$ → always obtuse. (No acute angle can have an acute or right supplement.)

L3.2

The reflex angle of some angle $θ$ is exactly three times the non-reflex angle. Find $θ$ and classify it.

Recall Solution

WHAT: Write the condition, then use the bridge. WHY: "three times" and "sum to $360°$ " are two facts about the same pair — enough to pin down $θ$ . $θ_{r} = 3 θ and θ + θ_{r} = 360°$ Substitute the first into the second: $θ + 3 θ = 360° \Rightarrow 4 θ = 360° \Rightarrow θ = 90°$ Check: $θ_{r} = 3 (90°) = 270°$ , and $90° + 270° = 360°$ ✓. Classify: $θ = 90°$ is a right angle; its reflex $270°$ is reflex. Answer: $θ = 90°$ (right); the reflex is $270°$ .

L3.3

Two angles are complementary (they sum to $90°$ ), and one is four times the other. Find both and name each type.

Recall Solution

WHAT: Let the smaller be $x$ ; the larger is $4 x$ . Complementary means: $x + 4 x = 90° \Rightarrow 5 x = 90° \Rightarrow x = 18°$ WHY complementary → $90°$ : two angles that "corner up" to a right angle sum to $90°$ by definition (see Complementary and supplementary angles). Then $4 x = 72°$ . Check: $18° + 72° = 90°$ ✓. Classify: $18°$ is acute; $72°$ is acute. (Both parts of a right angle must be acute — each is below $90°$ .) Answer: $18°$ (acute) and $72°$ (acute).

Level 4 — Synthesis

Here you combine the classification rule, the reflex bridge, and geometric relationships in one problem. The figure below shows three rays $O A$ , $O B$ , $O C$ leaving a single point $O$ : blue ray $O A$ points right, orange ray $O B$ sits $72°$ above it, and green ray $O C$ points left ( $180°$ from $O A$ ). Coloured arcs mark the three gaps — $72°$ (orange), $108°$ (green), and the $180°$ wrap (blue) — which together sweep the full circle.

L4.1

Three rays $O A$ , $O B$ , $O C$ leave a single point $O$ and fill the entire plane around it (they meet only at $O$ ). The three angles between consecutive rays are in the ratio $2 : 3 : 5$ . Find all three angles and classify each.

Recall Solution

WHAT: The three angles around a point sweep one complete turn, so they sum to $360°$ . WHY $360°$ : rays radiating from one point and returning to the start cover the whole circle exactly once — that is a complete angle by definition. Let the shares be $2 k$ , $3 k$ , $5 k$ : $2 k + 3 k + 5 k = 360° \Rightarrow 10 k = 360° \Rightarrow k = 36°$ So the angles are: $2 k = 72°, 3 k = 108°, 5 k = 180°$ Classify each by dropping onto the zone line (figure s01):

$72°$ : $0° < 72° < 90°$ → acute (blue band)
$108°$ : $90° < 108° < 180°$ → obtuse (green band)
$180°$ : exactly $180°$ → straight (rays point in opposite directions) Check: $72° + 108° + 180° = 360°$ ✓. Answer: $72°$ (acute), $108°$ (obtuse), $180°$ (straight).

L4.2

Continuing L4.1: consider the reflex angle on the outside of the $72°$ opening (the long way round from ray to ray). Find it, and find the non-reflex angle that a bisector would create if it split that original $72°$ angle in half.

Recall Solution

Part A — reflex of $72°$ : WHY the bridge here: the $72°$ opening is the short way between rays $O A$ and $O B$ ; the reflex is the long way around the outside. Short plus long wraps the point exactly once, so the long way is whatever completes $360°$ . $θ_{r} = 360° - 72° = 288°$ Check: dropping $288°$ onto the zone line (figure s01) lands it in the orange band, $180° < 288° < 360°$ ✓ reflex. Part B — bisector of $72°$ : a bisector is a ray that cuts an angle into two equal halves (see Angle bisectors). WHY divide by $2$ : "cut into two equal halves" means the two new angles are the same size and together rebuild the original $72°$ . If each half is $h$ , then $h + h = 72°$ , i.e. $2 h = 72°$ , so $h = \frac{72°}{2}$ — dividing by $2$ is exactly what "split evenly in two" demands. $h = \frac{72°}{2} = 36°$ Each half is $36°$ , still acute ( $0° < 36° < 90°$ , blue band on figure s01). Answer: reflex $= 288°$ ; each bisected half $= 36°$ (acute).

Level 5 — Mastery

One problem, everything at once: classification, the reflex bridge, backward reasoning, and a degenerate/limiting check.

L5.1

An angle $θ$ satisfies all of these at once: (i) $θ$ is obtuse; (ii) its reflex $θ_{r}$ is exactly $80°$ more than twice $θ$ ; Find $θ$ , verify it is obtuse, and state its reflex. Then investigate: for what single value of $θ$ would condition (ii) instead force $θ$ to be a right angle, and is that value consistent?

Recall Solution

Main solve. WHAT: Translate (ii) into symbols and use the bridge $θ_{r} = 360° - θ$ . $θ_{r} = 2 θ + 80°$ Replace $θ_{r}$ using the bridge, so everything is written in the single unknown $θ$ : $360° - θ = 2 θ + 80°$ WHY substitute: we have two expressions for the same quantity $θ_{r}$ ; setting them equal gives one equation in one unknown. Now solve step by step. First add $θ$ to both sides to gather the $θ$ -terms on the right (this clears the $- θ$ on the left): $360° = 2 θ + θ + 80° = 3 θ + 80°$ Next subtract $80°$ from both sides to isolate the $θ$ -term: $360° - 80° = 3 θ \Rightarrow 280° = 3 θ$ Finally divide both sides by $3$ to free $θ$ : $θ = \frac{280°}{3} \approx 93.33°$ Verify (i): dropping $93.33°$ onto the zone line (figure s01) lands it just inside the green band, $90° < 93.33° < 180°$ ✓ obtuse. Reflex: $θ_{r} = 360° - \frac{280°}{3} = \frac{1080° - 280°}{3} = \frac{800°}{3} \approx 266.67°$ . Cross-check (ii): $2 θ + 80° = \frac{560°}{3} + \frac{240°}{3} = \frac{800°}{3}$ ✓ matches $θ_{r}$ .

The "right-angle" investigation. Suppose instead we demand $θ = 90°$ and ask whether condition (ii) can hold. Then the bridge gives $θ_{r} = 360° - 90° = 270°$ , but condition (ii) would require $θ_{r} = 2 (90°) + 80° = 260°.$ These disagree ( $270° \neq = 260°$ ), so no: a right angle is not consistent with condition (ii). The only measure that satisfies (ii) is $θ = \frac{280°}{3}$ , and that value is obtuse, not right — the two demands cannot be met together. Answer: $θ = \frac{280°}{3} \approx 93.33°$ (obtuse); reflex $= \frac{800°}{3} \approx 266.67°$ . Forcing a right angle is impossible under (ii).

L5.2 (Degenerate / limiting check)

As an angle $θ$ shrinks toward $0°$ , what happens to its reflex $θ_{r}$ ? And as $θ$ grows toward $360°$ ? Describe the limiting pictures.

Recall Solution

WHAT: Track $θ_{r} = 360° - θ$ at the two extremes. WHY the bridge governs both ends: the reflex partner is defined as $360° - θ$ , so its behaviour at the edges is read straight off that formula.

As $θ \to 0°$ : $θ_{r} \to 360° - 0° = 360°$ . The "short way" vanishes to nothing (rays nearly overlap — the degenerate zero angle), and the "long way" swells to almost a complete turn. This is the degenerate edge where a reflex angle blends into a complete angle.
As $θ \to 360°$ : $θ_{r} \to 360° - 360° = 0°$ . Now the roles swap — the once-tiny gap has grown to a full turn while the reflex partner collapses to nothing. WHY this matters: it shows the reflex bridge is symmetric: whichever partner is small, the other is nearly complete. There is no "missing" scenario — every $θ$ from $0°$ to $360°$ has a well-behaved partner. Note the boundaries $0°$ and $360°$ are themselves not reflex (they are the degenerate "no angle" and "complete" cases); the reflex zone is the strict interior $180° < θ < 360°$ . Answer: $θ \to 0° \Rightarrow θ_{r} \to 360°$ ; and $θ \to 360° \Rightarrow θ_{r} \to 0°$ .

[!recall]- One-line self-test

Quick-fire — cover the answers.

Type of

3°

acute

Type of

90°

right angle

Type of

249°

reflex

Reflex of

75°

285°

Smaller angle when reflex

-

angle

= 150°

105°

Reflex between clock hands at 2:00

300°

Reflex

= 3 \times

angle, so angle

90°

(right)

Complementary pair in ratio

1 : 4

18°

and

72°

Three rays, ratio

2 : 3 : 5

around a point

72°, 108°, 180°

Reflex of

72°

288°

Bisected half of

72°

36°

L5 obtuse angle

θ

\frac{280°}{3} \approx 93.33°

Connections used: parent topic, Measuring angles with a protractor, Complementary and supplementary angles, Angle bisectors, Adjacent angles and linear pairs.