Session 06 · Angle Pairs

Lesson 5 · Complementary and supplementary angles

Page 105

A cyclist cornering hard leans 32° from the vertical, and the page asks the natural follow-up: what is the lean from the horizontal? Since vertical and horizontal meet at 90°, the answer is 90 − 32 = 58°, and that little subtraction gets a permanent name: two angles are complementary iff their sum is 90°, each the complement of the other.

Watch how the definition is built: it is about the two measures only. Nothing requires the angles to touch, share a side, or even live in the same figure, and page 106 says so explicitly. Keeping "what the definition says" separate from "what the picture happens to show" is this lesson's discipline, and it pays off two pages from now when the same idea has to survive being moved between figures.

New here complementary

Khan Academy Complementary and supplementary angles

Lesson 5 · Complementary and supplementary angles

Page 106

Measure the bicycle's angle on the other side of the horizontal instead and the pair sums to 180°: supplementary angles, each the supplement of the other. Then the page delivers two firsts. Theorem 3 (complements of the same angle are equal) comes with the proof vocabulary the book will use forever: the hypothesis is labeled "Given," the conclusion "Prove," and the two-column proof runs four lines of pure page 79 algebra.

Read the proof until you can see why it never mentions where the angles sit; that is the 50° triangle figure's point, where ∠DCB = ∠A despite the two angles living in different corners. Theorem 4 (supplements of the same angle are equal) is stated and left for you: page 109's question 46 collects it, with 180 replacing 90 as the only change.

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Lesson 5 · Exercises, Set I

Page 107

The dictionary questions land the mnemonic: complement completes a whole of 90, supplement strengthens it to 180 (and a compliment is the one you want from a friend). The astrology "sign" (360/12 = 30°) gets both treatments, and the bisect-then-trisect question builds a fan of 15° angles whose every combination is a multiple of 15, a fact worth noticing before protractor work.

The runway approach lights are the page's real-world teeth: a 3° glide path, so the angle from the vertical is its complement, 87°, and each question swaps between the pair. The skier's force diagram at the bottom (lift, drag, weight as rays with coordinates) sets up page 108; copy it carefully now and that page goes twice as fast.

Worked example · page 107

Questions 4 to 7: the 3° glide path

A plane approaches the runway at the safe angle of 3° above the horizontal (angle BLP). What angle does its path make with the vertical VL, what should ∠PLA be, and how do these angles relate to ∠BLP?

Question 4: vertical and horizontal are 90° apart, so ∠VLP = 90 − 3 = 87°.
Question 5: 87 + 3 = 90, so ∠VLP and ∠BLP are complementary, by the page 105 definition.
Question 6: A lies along the runway on the other side of L, so ∠PLA stretches from the path to the opposite ray: 180 − 3 = 177°.
Question 7: 177 + 3 = 180, so ∠PLA and ∠BLP are supplementary.
Self-check the pattern: one 3° angle generated an 87 and a 177, complement and supplement, subtractions from 90 and from 180. The table on page 108 (questions 36 to 39) will show the supplement always beats the complement by exactly 90, and here it does: 177 − 87 = 90.

Lesson 5 · Exercises, Sets I and II

Page 108

The skier wraps up with betweenness doing physics: OD-OW-OF gives ∠DOW + ∠WOF = ∠DOF, the coordinates hand you 117° and 63° (supplementary), and lift versus weight gives the complementary pair. Ollie's rope triangle is vocabulary discipline: "supplementary" is defined for two angles, so three angles cannot be it, two angles of a triangle cannot manage 180 (the third would be 0), but two summing to 90 happens in every right triangle.

The clock questions are sneaky: at the intended bedtime of 8:30 the little hand is not on the 8 at all but halfway to 9, so the description is never exactly true, and the hands make 75°, not 60. The hammer throw is the worked example, and the table in 36 to 39 proves the supplement-minus-complement gap of 90° with algebra.

Worked example · page 108

Questions 26 to 31: the ideal hammer throw

OH bisects the right angle YOX between vertical and horizontal. Supply the reason for each step proving ∠2 = 45°.

Question 26: OY-OH-OX, so ∠1 + ∠2 = ∠YOX by the Betweenness of Rays Theorem.
Questions 27 and 28: ∠YOX is a right angle, and a right angle is 90° by definition; substitute to get ∠1 + ∠2 = 90°. (In this lesson's language: the launch angles are complementary.)
Question 29: OH bisects ∠YOX, and a bisecting line divides an angle into two equal angles, so ∠1 = ∠2.
Question 30: substitute ∠2 for ∠1: ∠2 + ∠2 = 90°, so 2∠2 = 90°.
Question 31: divide both sides by 2: ∠2 = 45°. The ideal throw leaves at 45° above the horizontal, and the proof is the miter joint of page 103 with a hammer in it. Physics agrees about maximum range, but notice geometry got there with no physics at all.

Lesson 5 · Exercises, Sets II and III

Page 109

The planet takes the quiz: the Arctic Circle's 66.54° and the tropic of Capricorn's 23.46° sum to 90, and that complementarity is physical fact, which is why midwinter noon sun overhead at one means sun on the horizon at the other. The two 90° pole angles are supplementary, and the endless South Pole daylight has its mirror in the North Pole's endless night.

Questions 44 and 45 rehearse Theorem 3's exact content: two angles complementary to the same angle must be equal, yet nothing fixes their size; the theorem constrains the pair, not the numbers. Then question 46 asks for your first fully self-written theorem proof. The worked example scaffolds it; write yours before opening.

Worked example · page 109

Question 46: proving Theorem 4

Supplements of the same angle are equal. Given: ∠1 and ∠2 are supplements of ∠3. Prove: ∠1 = ∠2. Write the two-column proof, using Theorem 3's proof as the guide.

Statement 1: ∠1 and ∠2 are supplements of ∠3. Reason: Given.
Statement 2: ∠1 + ∠3 = 180° and ∠2 + ∠3 = 180°. Reason: if two angles are supplementary, their sum is 180° (the definition).
Statement 3: ∠1 + ∠3 = ∠2 + ∠3. Reason: substitution, since both sides equal 180°.
Statement 4: ∠1 = ∠2. Reason: subtraction, taking ∠3 from both sides.
Compare line by line with Theorem 3 on page 106: one number changed. That near-identity is worth noticing as a habit; when two theorems differ by a constant, one proof is both proofs, and page 112 is about to lean on this theorem to prove vertical angles equal.

Lesson 6 · Linear pairs and vertical angles

Page 110

An aerial photograph of Castleton, North Dakota: section lines ruling the prairie into squares, and one railroad slicing through at a sharp angle. Where railroad meets road, four angles appear, and the crossing point P splits each line into two opposite rays, rays pointing opposite directions along one line.

Two new pair-names ride on that: angles sharing one side whose other sides are opposite rays are a linear pair; angles whose sides are opposite rays to each other's are vertical angles. Read the book's own footnote grumble: "vertical" here has nothing to do with vertical lines (the dictionary's "opposite angles" would have been better), so do not let the word paint a picture the definition never drew. Lesson 5 paired angles by measure and said nothing about sides; this lesson pairs them by sides and says nothing about measure, yet.

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Lesson 6 · Linear pairs and vertical angles

Page 111

The definitions go on the books (linear pair: common side, other sides opposite rays; vertical angles: each side opposite a side of the other), and then the page does what this chapter always does with a picture that "seems" true: it proves it. Theorem 5, the angles in a linear pair are supplementary, and the proof's one clever move is step 3: let the Protractor Postulate assign coordinates 0, n, 180. Then ∠1 = n, ∠2 = 180 − n, and the sum is 180 before you know what n is.

That "let" is new equipment: you are allowed to install coordinates wherever a postulate permits, and the proof works for every n at once, which no protractor measurement of any single figure could do. Theorem 6 waits on the next page with a proof that never touches a number at all.

New here linear pair · vertical angles

Worked example · page 112

Questions 1 and 2: the folding chair

The chair's leg braces cross like scissors. As the chair opens and closes, why are some brace angles always equal, and some always supplementary?

Model the braces: two straight braces crossing at a pivot are two intersecting lines, making four angles at every opening width.
Question 1: the angles across the pivot from each other are vertical angles, and Theorem 6 says vertical angles are equal, at every position of the chair. Which numbers they equal changes as the chair folds; that they are equal never does.
Question 2: adjacent angles at the pivot share a side, and their other sides run opposite ways along one brace: a linear pair. Theorem 5 makes each such pair supplementary, again at every position.
The punchline to carry: the theorems are about the crossing, not the crossing angle. Open the chair 10 degrees or 80 and both facts hold, which is why the book keeps saying proofs buy every case at once. The hedge shears on page 115 run the same idea with a subtraction twist.

Lesson 6 · Exercises, Set I

Page 113

The dangerous intersection runs both new theorems in three questions (55° makes its linear partner 125°), and question 10's answer is the acute angle: the sharper the turn, the more the crossing traffic sits behind your shoulder. The two stars in question 16 make a converse-trap: all three angles equal, yet ∠1 and ∠2 are not vertical angles, so equal never implies vertical, and questions 17 to 20 chase that converse through every combination, page 47's lesson hunting in new territory.

The Stonehenge-flavored sun directions are the worked example. The ship-position figure at the bottom (three concurrent lines, six angles) continues overleaf; the word question 21 wants is "concurrent," from all the way back on page 9.

Worked example · page 113

Questions 11 to 15: the solstice sun

Lines AD and BC mark midsummer and midwinter sunrise and sunset; the compass directions bisect the angles they form. With a circular protractor at O, ON = 0 and OA = 50. Find every coordinate, then ∠AOB, ∠COD, ∠AOC, and ∠BOD.

OE (due east, 90) bisects ∠AOB, so B sits as far past east as A sits before it: ∠AOE = 40, so OB = 130.
Opposite rays differ by 180 on a circular scale: A and D are the same line (line AD), so OD = 230; B and C likewise give OC = 310. The compass rays fill in at 0, 90, 180, 270.
Question 12: ∠AOB = 130 − 50 = 80°. Question 13: ∠COD = 310 − 230 = 80°.
Question 14: they had to match: sunrise pair and sunset pair are vertical angles (each side of one is the opposite ray of a side of the other), and Theorem 6 says equal. The bisecting compass directions were a hint, not a necessity.
Question 15: ∠AOC and ∠BOD are the other vertical pair: each is 180 − 80 = 100° by Theorem 5. Check the family: 80 + 100 = 180 twice over, and the four angles sum to 360 around O. Stonehenge's builders aligned stones on exactly these lines four thousand years before the theorem had a number.

Worked example · page 114

Questions 26 to 33: the navigator's triangle

Three slightly-off position lines form a small triangle with nine numbered angles: interior angles 3, 6, 9, and pairs 1-2, 4-5, 7-8 around each corner. Find the totals asked for.

Questions 26 and 27: angles 3, 6, 9 are the little triangle's interior angles, so ∠3 + ∠6 + ∠9 = 180° by the Triangle Angle Sum Theorem from page 66.
Questions 28 and 29: at each corner, the angle numbered 1 (or 4, or 7) is the vertical angle of the triangle's interior angle there, equal by Theorem 6. Three vertical copies of the interior angles: ∠1 + ∠4 + ∠7 = 180° too.
Question 30: each remaining pair (∠2 with ∠3, ∠5 with ∠6, ∠8 with ∠9) is a linear pair along one of the position lines, so each sums to 180° by Theorem 5.
Question 31: add the three linear-pair equations: ∠2 + ∠3 + ∠5 + ∠6 + ∠8 + ∠9 = 540°.
Questions 32 and 33: subtract step 1's 180 from the 540: ∠2 + ∠5 + ∠8 = 360°. Every total came from exactly three theorems, and none required measuring a single angle of a figure drawn by a shaky hand at sea, which is rather the point of having theorems.

Lesson 6 · Exercises, Set II

Page 115

The hedge shears are the folding chair with a subtlety: each rigid half of the shears keeps its own handle-to-blade angle fixed, and the two moving angles (handle gap and blade gap) stay equal because both are the same fixed angle minus the shared middle. Question 43 wants exactly that subtraction argument in proof form. Einstein's bent light ray then puts page 96's minutes and seconds to work: subtract 1.75 seconds from a straight angle and the deflection reads 179° 59′ 58.25″, a bend of two second-of-arc that took a solar eclipse to observe.

The pinhole camera is the worked example: two theorems meet one given fact and out falls the angle of view. The SAT problem closes with pure linear-pair algebra; find z first, then w rides free.

Worked example · page 115

Questions 46 and 47: the pinhole camera

∠1 and ∠2 are vertical angles, and they are also complementary. Complete the proof and find the camera's angle of view.

Line 2: ∠1 = ∠2. Reason: vertical angles are equal, Theorem 6.
Line 4: ∠1 + ∠2 = 90°. Reason: two complementary angles sum to 90°, the definition from page 105.
Line 5: substitute ∠1 for ∠2: ∠1 + ∠1 = 90°, so 2∠1 = 90°. Reason: substitution.
Line 6: ∠1 = 45°. Reason: division property. Question 46's answer: the angle of view is 45°.
Notice the proof's shape is page 108's hammer throw with the bisector swapped for Theorem 6: whenever two equal angles must share a fixed total, each is half of it. You now own that argument in three costumes.

Worked example · page 116

Set III: reading the Chinese proof

Boxes A and B are translated for you ("Theorem 3. Vertical angles are equal" and "Given"). Reconstruct C through J.

Box C sits with the given figure setup: "Two lines AB and CD intersect at point O." Box D finishes the given: "∠AOD and ∠BOC are vertical angles; ∠AOC and ∠BOD are vertical angles."
Box E introduces the goal, our "Prove:" (the equations after it are exactly Theorem 6's conclusion), and box F is "Proof."
Box G's equations say ∠AOD + ∠AOC = 2∠R and ∠BOC + ∠AOC = 2∠R. Question 5: 2∠R is two right angles, 180°. Box H, the reason: angles whose outer sides form a straight line are supplementary, our Theorem 5, the linear pair theorem.
Box I justifies setting the two sums equal: quantities equal to the same quantity are equal, our substitution step.
Box J justifies dropping ∠AOC from both sides: equals subtracted from equals leave equals, the subtraction property, and the proof ends where page 112's did. Same theorems, same order, same logic, different alphabet; Euclid's system is the shared language.