Angles

Study Sheet

Angles

Naming, classifying, and solving for angle measures

Naming Angles and Their Parts

Concept
What is an angle?

An angle is formed by two rays that share a common endpoint. That shared endpoint is the vertex, and the two rays are the sides of the angle. We measure the “opening” between the sides in degrees (∘^\circ).

There are three ways to name an angle:

  • Three points: ∠AOB\angle AOB or ∠BOA\angle BOA — the vertex letter always goes in the middle.
  • The vertex alone: ∠O\angle O — only when there is no confusion about which angle you mean.
  • A number: ∠1\angle 1 — when a small number is written inside the angle.

An angle is a pair of rays, but its measure is how far you would turn the first ray about the vertex to lay it on the second. That is why the size of an angle has nothing to do with how long its sides are drawn: extending the rays changes nothing. Pairs of angles get names for their relationships — complementary, supplementary, vertical, adjacent, linear pair — and most angle problems are one of those relationships written as an equation.

Example
Naming the angle below
1OAB
Vertex O; sides are the rays OA and OB.

The rays OA→\overrightarrow{OA} and OB→\overrightarrow{OB} in the figure meet at the vertex OO. We can call this angle ∠AOB\angle AOB, ∠BOA\angle BOA, ∠O\angle O, or ∠1\angle 1. It could not be called ∠OAB\angle OAB: that name puts AA in the middle, so it would mean an angle with its vertex at AA.

Side note
Where it goes wrong

Assuming two angles that look equal in a diagram are equal. Diagrams are not to scale; use only the given marks and the theorems.

Tip

Protractor tip. To measure an angle, place the protractor's center hole on the vertex and line up one side with the 0∘0^\circ mark. Read where the other side crosses the scale. Use the scale that starts at 0∘0^\circ on the side you lined up — most protractors have two rows of numbers!

Try it
Quick check

Three rays leave point YY: YX→\overrightarrow{YX}, YW→\overrightarrow{YW} and YZ→\overrightarrow{YZ}. Why is the name ∠Y\angle Y not good enough here, and what should you write instead?

Show the answer

Three rays from YY make three angles with vertex YY (∠XYW\angle XYW, ∠WYZ\angle WYZ and ∠XYZ\angle XYZ), so "∠Y\angle Y" does not say which one you mean. Use three letters with the vertex in the middle, such as ∠XYW\angle XYW.

Classifying Angles

Concept
Five kinds of angles
acute 40°right 90°obtuse 130°straight 180°reflex 250°
Each class is fixed by comparing with a quarter turn and a half turn.
  • Acute: greater than 0∘0^\circ and less than 90∘90^\circ.
  • Right: exactly 90∘90^\circ (marked with a small square).
  • Obtuse: greater than 90∘90^\circ and less than 180∘180^\circ.
  • Straight: exactly 180∘180^\circ (a straight line).
  • Reflex: greater than 180∘180^\circ and less than 360∘360^\circ.

The class boundaries are the two turns everyone can see without a protractor: a quarter turn (90∘90^\circ, the corner of a page) and a half turn (180∘180^\circ, a straight line). Acute and obtuse are the open intervals on either side of 90∘90^\circ; reflex angles are what you get going "the long way round", which is why two rays actually make two angles — one of θ\theta and one of 360∘−θ360^\circ - \theta.

Side note
Where it goes wrong

Calling 180∘180^\circ obtuse or 90∘90^\circ acute. The inequalities are strict: obtuse means strictly between 90∘90^\circ and 180∘180^\circ, so 90∘90^\circ is right and 180∘180^\circ is straight — each is its own class.

Tip

Remember. A right angle is your reference. If an angle looks “smaller than the corner of a page,” it is acute; if it looks “more open than a corner,” it is obtuse.

Try it
Quick check

Classify each angle: (a) 89.9∘89.9^\circ (b) 90∘90^\circ (c) 181∘181^\circ (d) the smaller angle between the hands of a clock at 4:00.

Show the answer

(a) acute (just under 90∘90^\circ). (b) right. (c) reflex (just over 180∘180^\circ). (d) At 4:00 the hands are 44 hour-marks apart, 4×30∘=120∘4 \times 30^\circ = 120^\circ: obtuse.

Adjacent Angles & the Angle Addition Postulate

Concept
Adjacent angles

Two angles are adjacent when they share a common vertex and a common side, but do not overlap (no interior points in common). Think of two angles sitting side by side.

Concept
Angle Addition Postulate

If point BB is in the interior of ∠AOC\angle AOC, then the two small angles add up to the big angle:

m∠AOB+m∠BOC=m∠AOC.m\angle AOB + m\angle BOC = m\angle AOC .

Like segment addition, this is an axiom about measurement: lay a protractor with OA→\overrightarrow{OA} at 0∘0^\circ; if BB is inside the angle, OB→\overrightarrow{OB} reads m∠AOBm\angle AOB and OC→\overrightarrow{OC} reads m∠AOCm\angle AOC, and the gap between the readings is m∠BOCm\angle BOC. The condition "BB in the interior" is essential — if OB→\overrightarrow{OB} lies outside, the two angles overlap or overshoot, and the measures subtract instead of add.

Example
Adding adjacent angles
35°40°OABC
Ray OB splits the 75∘^\circ angle AOC into 35∘^\circ and 40∘^\circ.

Ray OB→\overrightarrow{OB} lies inside ∠AOC\angle AOC. If m∠AOB=35∘m\angle AOB = 35^\circ and m∠BOC=40∘m\angle BOC = 40^\circ, then

m∠AOC=35∘+40∘=75∘.m\angle AOC = 35^\circ + 40^\circ = 75^\circ .
Example
Angle addition with algebra

BB is in the interior of ∠AOC\angle AOC, m∠AOC=120∘m\angle AOC = 120^\circ, m∠AOB=(2x+10)∘m\angle AOB = (2x + 10)^\circ and m∠BOC=(3x−5)∘m\angle BOC = (3x - 5)^\circ. Find both parts.

The parts add to the whole:

(2x+10)+(3x−5)=1205x+5=120x=23.\begin{aligned} (2x + 10) + (3x - 5) &= 120 \\ 5x + 5 &= 120 \\ x &= 23 . \end{aligned}

So m∠AOB=2(23)+10=56∘m\angle AOB = 2(23) + 10 = 56^\circ and m∠BOC=3(23)−5=64∘m\angle BOC = 3(23) - 5 = 64^\circ. Check: 56+64=12056 + 64 = 120.

Side note
Where it goes wrong

Adding two angles that overlap. If ∠AOB=35∘\angle AOB = 35^\circ and ∠AOC=75∘\angle AOC = 75^\circ share the side OA→\overrightarrow{OA}, they are not adjacent (one sits inside the other), and 35+75=11035 + 75 = 110 measures nothing. Adjacent angles share a side and lie on opposite sides of it.

Tip

Tip. The Angle Addition Postulate works both ways: you can add two pieces to get the whole, or subtract a known piece from the whole to find the missing piece.

Try it
Quick check

RR is in the interior of ∠PQS\angle PQS, m∠PQS=97∘m\angle PQS = 97^\circ and m∠PQR=58∘m\angle PQR = 58^\circ. Find m∠RQSm\angle RQS.

Show the answer

By angle addition, m∠PQR+m∠RQS=m∠PQSm\angle PQR + m\angle RQS = m\angle PQS, so m∠RQS=97∘−58∘=39∘m\angle RQS = 97^\circ - 58^\circ = \mathbf{39^\circ}.

Angle Bisectors

Concept
What is a bisector?

A ray that bisects an angle divides it into two congruent (equal) angles. If OB→\overrightarrow{OB} bisects ∠AOC\angle AOC, then

m∠AOB=m∠BOC=12 m∠AOC.m\angle AOB = m\angle BOC = \tfrac{1}{2}\, m\angle AOC .

Every angle has exactly one bisector: on a protractor the bisector is the ray at the halfway reading. With compass and straightedge it is found by symmetry — mark equal distances along both sides, then find a point equally far from those two marks; the ray to it splits the angle into two congruent triangles (SSS), so the two halves are equal.

Example
Using a bisector
40°40°OABC
The bisector OB cuts the 80∘^\circ angle into two 40∘^\circ angles.

OB→\overrightarrow{OB} bisects ∠AOC\angle AOC, and m∠AOC=80∘m\angle AOC = 80^\circ. Then each half is

m∠AOB=m∠BOC=12(80∘)=40∘.m\angle AOB = m\angle BOC = \tfrac{1}{2}(80^\circ) = 40^\circ .
Side note
Where it goes wrong

Doubling instead of halving. If OB→\overrightarrow{OB} bisects ∠AOC\angle AOC and m∠AOB=40∘m\angle AOB = 40^\circ, then m∠AOC=80∘m\angle AOC = 80^\circ — not 20∘20^\circ. The bisected angle is the whole; the two equal angles are its halves.

Tip

Remember. “Bisect” means “cut into two equal parts.” The two halves are always congruent, so you can set them equal to each other when solving with algebra.

Try it
Quick check

OB→\overrightarrow{OB} bisects ∠AOC\angle AOC. m∠AOB=(5x−8)∘m\angle AOB = (5x - 8)^\circ and m∠AOC=84∘m\angle AOC = 84^\circ. Find xx.

Show the answer

Each half is 84∘÷2=42∘84^\circ \div 2 = 42^\circ, so 5x−8=425x - 8 = 42, 5x=505x = 50, x=10x = \mathbf{10}.

Complementary & Supplementary Angles

Concept
Two special sums
  • Complementary angles have measures that add to 90∘\mathbf{90^\circ}.
  • Supplementary angles have measures that add to 180∘\mathbf{180^\circ}.

The angles do not have to be next to each other — only their measures matter.

The two sums come from the two turns that show up everywhere: a right angle and a straight line. That is why they appear in so many problems — a right triangle's two acute angles are complementary (they share the 90∘90^\circ left over from 180∘180^\circ), and any two angles that sit side by side on a line are supplementary. Every angle less than 180∘180^\circ has a supplement; only angles less than 90∘90^\circ have a complement.

Example
Finding a complement and a supplement
30°60°30°150°complementarysupplementary
A 30∘^\circ angle with its complement (fills a right angle) and its supplement (fills a straight line).

The complement of 30∘30^\circ is 90∘−30∘=60∘90^\circ - 30^\circ = 60^\circ.

The supplement of 30∘30^\circ is 180∘−30∘=150∘180^\circ - 30^\circ = 150^\circ.

Placed side by side, as in the figure, 30∘30^\circ and 60∘60^\circ fill a right angle, and 30∘30^\circ and 150∘150^\circ fill a straight line.

Example
An angle compared with its complement

An angle measures 24∘24^\circ more than its complement. Find it.

Call the angle xx; its complement is 90−x90 - x. "2424 more than the complement" is

x=(90−x)+24  ⇒  2x=114  ⇒  x=57.x = (90 - x) + 24 \;\Rightarrow\; 2x = 114 \;\Rightarrow\; x = 57 .

The angle is 57∘57^\circ and its complement is 33∘33^\circ; indeed 57−33=2457 - 33 = 24.

Side note
Where it goes wrong

Writing the complement of xx as x−90x - 90. The complement is what is left over to reach 90∘90^\circ, so it is 90−x90 - x; with x=57x = 57 that is 3333, while x−90=−33x - 90 = -33 is not an angle at all.

Tip

Memory trick. C comes before S in the alphabet, and 9090 comes before 180180: Complementary =90∘= 90^\circ, Supplementary =180∘= 180^\circ. (Or: “Corner” is 90∘90^\circ, “Straight” is 180∘180^\circ.)

Try it
Quick check

The supplement of an angle is four times its complement. Find the angle.

Show the answer

With angle xx: 180−x=4(90−x)180 - x = 4(90 - x), so 180−x=360−4x180 - x = 360 - 4x, 3x=1803x = 180 and x=60∘x = \mathbf{60^\circ}. Check: supplement 120∘120^\circ, complement 30∘30^\circ, and 120=4×30120 = 4 \times 30.

Linear Pairs & Vertical Angles

Concept
Linear pairs

When two angles are adjacent and their outer sides form a straight line, they make a linear pair. The angles of a linear pair are always supplementary (they add to 180∘180^\circ).

Example
A linear pair
12110°70°
Together the two angles make the straight angle along the line.

∠1\angle 1 and ∠2\angle 2 below form a linear pair, so m∠1+m∠2=180∘m\angle 1 + m\angle 2 = 180^\circ. If m∠1=110∘m\angle 1 = 110^\circ, then m∠2=180∘−110∘=70∘m\angle 2 = 180^\circ - 110^\circ = 70^\circ.

Concept
Vertical angles

When two straight lines cross, they form two pairs of vertical angles — the angles “across” from each other. Vertical angles are always congruent (equal).

Why they are equal: ∠1\angle 1 and ∠2\angle 2 form a linear pair, so ∠1=180∘−∠2\angle 1 = 180^\circ - \angle 2; ∠3\angle 3 and ∠2\angle 2 also form a linear pair, so ∠3=180∘−∠2\angle 3 = 180^\circ - \angle 2. Two things equal to the same quantity are equal. The proof never uses a measurement, which is why it holds for every pair of crossing lines.

Example
Vertical angles are equal
1234
Angles 1 and 3 are vertical; so are 2 and 4. Here angle 1 is 65∘^\circ.

Two lines cross at OO. Angles ∠1\angle 1 and ∠3\angle 3 are vertical (so m∠1=m∠3m\angle 1 = m\angle 3), and ∠2\angle 2 and ∠4\angle 4 are vertical (so m∠2=m∠4m\angle 2 = m\angle 4). If m∠1=65∘m\angle 1 = 65^\circ, then m∠3=65∘m\angle 3 = 65^\circ, while ∠2\angle 2 and ∠4\angle 4 each measure 180∘−65∘=115∘180^\circ - 65^\circ = 115^\circ.

Side note
Where it goes wrong

Calling any two angles that "point away from each other" vertical. Vertical angles must be formed by the same two lines crossing. Two angles at a vertex of a star, or at different crossings, can look opposite without being equal.

Tip

Key facts. Linear pair ⇒\Rightarrow angles are supplementary (180∘180^\circ). Vertical angles ⇒\Rightarrow angles are congruent (equal). These two rules solve most crossing-line puzzles!

Try it
Quick check

Two lines cross. One angle measures (4x+6)∘(4x + 6)^\circ and the angle vertical to it measures (6x−20)∘(6x - 20)^\circ. Find all four angles.

Show the answer

Vertical angles are equal: 4x+6=6x−204x + 6 = 6x - 20, so 2x=262x = 26 and x=13x = 13. That angle is 4(13)+6=58∘4(13) + 6 = 58^\circ, and so is its vertical partner. The other two form linear pairs with it: 180∘−58∘=122∘180^\circ - 58^\circ = 122^\circ each. So the angles are 58∘,122∘,58∘,122∘\mathbf{58^\circ, 122^\circ, 58^\circ, 122^\circ}.

Finding Unknown Angles with Algebra

Concept
Set up an equation

When an angle relationship gives you a total (90∘90^\circ, 180∘180^\circ, or a bisected angle), write an equation, then solve for the variable. Finally, substitute back to find each actual angle measure.

The algebra is never the hard part; choosing the equation is. Each relationship gives exactly one equation — equal (vertical angles, bisected halves) or a fixed sum (90∘90^\circ complementary, 180∘180^\circ linear pair or supplementary, 360∘360^\circ around a point). Name the relationship out loud first, write its equation, and only then solve. The final substitution matters because the question asks for angles, and a negative or oversized angle there tells you the equation was wrong.

Example
Complementary angles with algebra
x2x
x and 2x fill a right angle, drawn to scale at 30∘^\circ and 60∘^\circ.

Two complementary angles measure xx and 2x2x. Find each angle.

x+2x=903x=90x=30\begin{aligned} x + 2x &= 90 \\ 3x &= 90 \\ x &= 30 \end{aligned}

So the angles are x=30∘x = 30^\circ and 2x=60∘2x = 60^\circ. Check: 30∘+60∘=90∘30^\circ + 60^\circ = 90^\circ. ✓

Example
Vertical angles with algebra
(3x+10)°(x+50)°
The two labelled angles are vertical, so their expressions are equal (both 70∘^\circ).

Two vertical angles measure (3x+10)∘(3x+10)^\circ and (x+50)∘(x+50)^\circ. Because vertical angles are congruent, set them equal:

3x+10=x+502x=40x=20\begin{aligned} 3x + 10 &= x + 50 \\ 2x &= 40 \\ x &= 20 \end{aligned}

Each angle measures 3(20)+10=70∘3(20)+10 = 70^\circ (and 20+50=70∘20 + 50 = 70^\circ). ✓

Side note
Where it goes wrong

Adding vertical angles to 180180. Writing (3x+10)+(x+50)=180(3x + 10) + (x + 50) = 180 gives x=30x = 30 and angles of 100∘100^\circ and 80∘80^\circ — which cannot both be vertical angles, since vertical angles are equal. Adjacent angles on a line sum to 180∘180^\circ; opposite ones are equal.

Tip

Always check! After solving, plug your answer back in. Complementary angles should total 90∘90^\circ, supplementary and linear pairs should total 180∘180^\circ, and vertical angles should come out equal.

Try it
Quick check

∠1\angle 1 and ∠2\angle 2 form a linear pair, with m∠1=(3x+12)∘m\angle 1 = (3x + 12)^\circ and m∠2=(5x+8)∘m\angle 2 = (5x + 8)^\circ. Find both angles.

Show the answer

A linear pair is supplementary: (3x+12)+(5x+8)=180(3x + 12) + (5x + 8) = 180, so 8x+20=1808x + 20 = 180 and x=20x = 20. Then m∠1=72∘m\angle 1 = 72^\circ and m∠2=108∘m\angle 2 = 108^\circ; check 72+108=18072 + 108 = 180.

Going Deeper: Advanced Angle Ideas

Concept
Angle chasing: work one step at a time

Angle chasing means finding an unknown angle by filling in other angles one at a time, using the rules you already know, until the one you want is forced. At every step, ask which single rule turns something you know into something new:

  • angles on a straight line add to 180∘180^\circ (linear pair);
  • angles around a point add to 360∘360^\circ;
  • vertical angles are equal;
  • a bisector splits an angle into two equal halves.

Mark each angle on the figure as soon as you find it — one known angle usually unlocks the next.

Concept
Two counting rules that act as constraints

These two facts are the “budget” every angle problem must respect:

along a line:   ∑angles=180∘,around a point:   ∑angles=360∘.\text{along a line: } \; \sum \text{angles} = 180^\circ, \qquad \text{around a point: } \; \sum \text{angles} = 360^\circ .

If three rays leave a single point OO and go all the way around, the three angles between neighboring rays must total 360∘360^\circ. If several angles sit in a row on one side of a straight line, they must total 180∘180^\circ. Turning a picture into one of these equations is the whole game.

Example
Angle chasing around a point
x120°2x
Four angles fill the full turn around O.

Four rays leave point OO. Going around, the angles between neighbors are 90∘90^\circ, xx, 120∘120^\circ, and 2x2x. Find xx.

Everything around a point must add to 360∘360^\circ:

90+x+120+2x=3603x+210=360x=50.\begin{aligned} 90 + x + 120 + 2x &= 360 \\ 3x + 210 &= 360 \\ x &= 50 . \end{aligned}

So the unknown angles are 50∘50^\circ and 100∘100^\circ. Check: 90+50+120+100=36090 + 50 + 120 + 100 = 360. ✓

Concept
The Exterior Angle Theorem

Extend one side of a triangle to make an exterior angle. That exterior angle equals the sum of the two remote (non-adjacent) interior angles:

m∠ext=m∠A+m∠B.m\angle \text{ext} = m\angle A + m\angle B .

This follows from two facts working together: the three interior angles add to 180∘180^\circ, and the exterior angle forms a linear pair (180∘180^\circ) with the interior angle beside it. Subtracting the shared interior angle from both leaves exactly the two remote angles.

Example
Using the exterior angle theorem
50°65°4ABC
Side AC is extended past C; angle 4 sits outside the triangle.

In the figure, the two remote interior angles are 50∘50^\circ (at AA) and 65∘65^\circ (at BB), and ∠4\angle 4 is the exterior angle at CC. Instead of finding the third interior angle first, add the remote angles directly:

m∠4=50∘+65∘=115∘.m\angle 4 = 50^\circ + 65^\circ = 115^\circ .

Check the long way: the interior angle at CC is 180−115=65∘180 - 115 = 65^\circ, and 50+65+65=180∘50 + 65 + 65 = 180^\circ. ✓

Concept
Bisectors and the angle between them
70°110°
Bisect 70∘^\circ and 110∘^\circ: the dashed bisectors meet at 35∘^\circ + 55∘^\circ = 90∘^\circ.

Angle bisectors set up equal pieces you can chain together. A useful pattern: if two rays bisect neighboring angles that sit on a straight line, the angle between the two bisectors is always 90∘90^\circ. Reason: the four half-angles come in two equal pairs, and all four together fill the 180∘180^\circ line, so the middle two halves total exactly half of 180∘180^\circ. Whenever a problem mentions a bisector, replace the whole angle by “two equal halves” and the algebra usually falls out.

Example
A bisector condition with algebra

OB→\overrightarrow{OB} bisects ∠AOC\angle AOC. The two halves are given as m∠AOB=(4x−5)∘m\angle AOB = (4x - 5)^\circ and m∠BOC=(2x+15)∘m\angle BOC = (2x + 15)^\circ. Because a bisector makes the halves equal, set them equal:

4x−5=2x+152x=20x=10.\begin{aligned} 4x - 5 &= 2x + 15 \\ 2x &= 20 \\ x &= 10 . \end{aligned}

Each half is 4(10)−5=35∘4(10) - 5 = 35^\circ, so the whole angle is m∠AOC=70∘m\angle AOC = 70^\circ. ✓

Concept
Clock hands as angles

A clock face is a circle of 360∘360^\circ split into 1212 hours, so each hour mark is 360/12=30∘360 / 12 = 30^\circ apart. The two hands move at different, steady speeds:

minute hand: 360∘60 min=6∘ per minute,hour hand: 30∘60 min=0.5∘ per minute.\text{minute hand: } \tfrac{360^\circ}{60\text{ min}} = 6^\circ \text{ per minute}, \qquad \text{hour hand: } \tfrac{30^\circ}{60\text{ min}} = 0.5^\circ \text{ per minute}.

The minute hand gains on the hour hand at 6−0.5=5.5∘6 - 0.5 = 5.5^\circ per minute. To find the angle between the hands at a given time, compute each hand's position from the top (1212) and subtract. If the gap is more than 180∘180^\circ, use 360∘360^\circ minus that gap for the smaller angle.

Example
The angle at 3:40, and when the hands overlap
12369130°
At 3:40 the minute hand is at 240∘^\circ, the hour hand at 110∘^\circ (both from 12).

Part 1 — angle at 3:40. Measure each hand clockwise from the 1212:

minute hand=40×6=240∘,hour hand=3×30+40×0.5=90+20=110∘.\begin{aligned} \text{minute hand} &= 40 \times 6 = 240^\circ, \\ \text{hour hand} &= 3 \times 30 + 40 \times 0.5 = 90 + 20 = 110^\circ. \end{aligned}

The gap is 240−110=130∘240 - 110 = 130^\circ, which is already less than 180∘180^\circ, so the angle between the hands is 130∘\mathbf{130^\circ}.

Part 2 — when do the hands overlap after 12:00? At 12:0012{:}00 both hands sit at 0∘0^\circ. The minute hand closes the gap at 5.5∘5.5^\circ per minute, and it must make up a full 360∘360^\circ to lap the hour hand and line up again:

t=3605.5=36011/2=72011≈65.45 minutes.t = \frac{360}{5.5} = \frac{360}{11/2} = \frac{720}{11} \approx 65.45 \text{ minutes}.

So the hands first overlap again at about 1:055111{:}05\tfrac{5}{11}. In general, consecutive overlaps are exactly 72011=3605.5\dfrac{720}{11} = \dfrac{360}{5.5} minutes apart, which is why the hands line up only 1111 times every 1212 hours — not 1212.

Side note
Where it goes wrong

Leaving the hour hand on the 33. At 3:40 the hour hand has moved 40×0.5∘=20∘40 \times 0.5^\circ = 20^\circ past the 33; using 90∘90^\circ instead of 110∘110^\circ gives 240−90=150∘240 - 90 = 150^\circ instead of the true 130∘130^\circ.

Example
An angle condition that leads to a quadratic

An angle measures xx degrees. Its supplement is numerically equal to the square of its complement divided by 1010. Find xx.

The complement is (90−x)(90 - x) and the supplement is (180−x)(180 - x), so

180−x=(90−x)210.180 - x = \frac{(90 - x)^2}{10}.

Multiply through by 1010 and expand:

1800−10x=(90−x)21800−10x=8100−180x+x20=x2−170x+6300.\begin{aligned} 1800 - 10x &= (90 - x)^2 \\ 1800 - 10x &= 8100 - 180x + x^2 \\ 0 &= x^2 - 170x + 6300 . \end{aligned}

Factor (or use the quadratic formula): x2−170x+6300=(x−50)(x−120)x^2 - 170x + 6300 = (x - 50)(x - 120), so

x=50orx=120.x = 50 \qquad \text{or} \qquad x = 120 .

Reject the extra root. A complement only exists for an angle less than 90∘90^\circ, and x=120x = 120 has no complement (and a negative 90−x90 - x). So we discard x=120x = 120 and keep x=50∘\boxed{x = 50^\circ}. Check: complement =40= 40, supplement =130= 130, and 402/10=1600/10=13040^2 / 10 = 1600/10 = 130. ✓

Reminder — The quadratic formula:  x=−b±b2−4ac2a(a≠0)x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\qquad (a\neq 0)
Tip

Advanced checklist. (1) Chase angles one rule at a time and mark each result on the figure. (2) Treat “on a line =180∘= 180^\circ” and “around a point =360∘= 360^\circ” as equations. (3) An exterior angle equals the two remote interior angles. (4) Replace any bisected angle by two equal halves. (5) For clocks, use 6∘6^\circ/min and 0.5∘0.5^\circ/min, with overlaps every 72011\tfrac{720}{11} minutes. (6) When algebra gives two roots, keep only the one that makes every angle valid (complements need angles below 90∘90^\circ; all measures must be positive).

Try it
Quick check

Five rays leave a point and split the full turn into angles of xx, 2x2x, 3x3x, 4x4x and 5x5x degrees. Find the largest angle.

Show the answer

The angles fill the turn: x+2x+3x+4x+5x=360x + 2x + 3x + 4x + 5x = 360, so 15x=36015x = 360 and x=24x = 24. The largest is 5x=120∘5x = \mathbf{120^\circ}.

Why It Works

Theorem
The reason behind the rule
123
Angle 2 is shared: 1 + 2 = 180∘^\circ and 3 + 2 = 180∘^\circ.

Vertical angles are equal: at a crossing, angle 11 and angle 22 form a linear pair, so ∠1+∠2=180∘\angle1 + \angle2 = 180^\circ; angle 33 and angle 22 also form a linear pair, so ∠3+∠2=180∘\angle3 + \angle2 = 180^\circ. Both ∠1\angle1 and ∠3\angle3 equal 180∘−∠2180^\circ - \angle2, hence each other. Two facts about straight lines, one subtraction — that is the whole proof, and it is the template for dozens of angle-chasing arguments.

Tip
Reading a proof

Do not read a proof the way you read a rule. Cover the last line, ask what the previous line forces, and only then uncover it. If a step feels like magic, that is the step to work by hand with a concrete number.

Formulas, Proofs & Tips

Tip
Triangle angle sum and the exterior angle
A+B+C=180∘,exterior=sum of the two remote interior anglesA+B+C=180^\circ, \qquad \text{exterior} = \text{sum of the two remote interior angles}

What it means. Every triangle's angles total a straight angle; an exterior angle equals the two angles it is not next to.

Example. Angles 50∘50^\circ and 60∘60^\circ give a third of 70∘70^\circ; the exterior angle there is 110∘110^\circ.

Why it works. Draw a line through one vertex parallel to the opposite side. The two alternate interior angles equal the other two triangle angles, and together with the third they form a straight line — 180∘180^\circ. The exterior result follows since exterior =180∘−adjacent=180^\circ-\text{adjacent}, and the other two also total 180∘−adjacent180^\circ-\text{adjacent}.

Tip. The exterior angle shortcut saves a step: no need to find the third angle first.

Tip
Angle pair relationships
vertical angles are equal;linear pair sums to 180∘;complements sum to 90∘\text{vertical angles are equal};\quad \text{linear pair sums to }180^\circ;\quad \text{complements sum to }90^\circ

What it means. Angles formed by crossing lines come in predictable pairs.

Example. If two angles are supplementary and one is 110∘110^\circ, the other is 70∘70^\circ.

Why it works. Two angles on a straight line total 180∘180^\circ. If ∠1\angle 1 and ∠2\angle 2 form a line, and ∠2\angle 2 and ∠3\angle 3 also form a line, then both equal 180∘−∠2180^\circ-\angle 2 — so the vertical angles ∠1\angle 1 and ∠3\angle 3 are equal.

Tip. With parallel lines cut by a transversal, corresponding and alternate angles are equal, while same-side interior angles are supplementary.