---
title: "Triangles and Angles"
book: "Primary & Middle School Mathematics"
subject: math
language: en
chapter: 51
exercises: 11
source: https://one-course.com/books/math/1/en/chapter/51-triangles-and-angles
---

# Chapter 51 — Triangles and Angles

Can you draw a triangle with sides $10$, $3$ and $2$? With angles $80^\circ$, $70^\circ$ and $50^\circ$? This chapter answers both questions with two of the most useful facts of plane geometry: the triangle inequality and the $180^\circ$ angle [sum](https://one-course.com/books/math/1/en/chapter/2-addition-first-steps#def-g1-addition-def) — plus the angle vocabulary that comes with crossing and [parallel lines](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-perp).

## 51.1 Angles around crossing lines

**Definition 51.1 (Angle pairs).**

- Two angles are *complementary* when their measures add up to $90^\circ$ , *supplementary* when they add up to $180^\circ$ .
- When two lines cross, the angles facing each other across the crossing point are *vertically opposite* .
- When a line crosses two other lines, angles located in the same position at each crossing are *corresponding angles* ; angles between the two lines on opposite sides of the crossing line are *alternate angles* .

**Theorem 51.2 (Angle equalities).**

1. [Vertically opposite angles](#def-g7-triangles-pairs) are equal.
2. If two lines are [parallel](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-perp) , corresponding angles are equal, and [alternate angles](#def-g7-triangles-pairs) are equal. Conversely, equal corresponding (or alternate) angles force the lines to be [parallel](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-perp) .

**Proof of 1.** The two angles $\widehat{1}$ and $\widehat{2}$ on one side of a line make a straight angle: $\widehat 1 + \widehat 2 = 180^\circ$. The [vertically opposite](#def-g7-triangles-pairs) angle $\widehat 3$ of $\widehat 1$ also satisfies $\widehat 2 + \widehat 3 = 180^\circ$ (same reason). So $\widehat 1 = 180^\circ - \widehat 2 = \widehat 3$. Point 2 is admitted at this level. ∎

![Left: vertically opposite angles 1 = 3 (each is supplementary to 2). Right: parallel lines cut by a third line — the corresponding angles a and b are equal.](https://one-course.com/images/onecourse/chapters/math-1/g7-triangles/fig-3b89e20a134d.svg)

*Left: [vertically opposite angles](#def-g7-triangles-pairs) $\widehat 1 = \widehat 3$ (each is supplementary to $\widehat 2$). Right: [parallel lines](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-perp) cut by a third line — the corresponding angles $\widehat a$ and $\widehat b$ are equal.*

## 51.2 The sum of the angles of a triangle

**Theorem 51.3 (Angle sum).**

In every triangle, the three angles add up to $180^\circ$.

**Proof.** Let $ABC$ be a triangle. Draw the line through $A$ [parallel](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-perp) to $(BC)$. At the [vertex](https://one-course.com/books/math/1/en/chapter/35-cubes-and-boxes#def-g5-solids-def) $A$, three angles line up along this [parallel](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-perp) and make a straight angle, $180^\circ$: the middle one is $\widehat{A}$, and the two outer ones are [alternate angles](#def-g7-triangles-pairs) with $\widehat B$ and $\widehat C$ ([parallel lines](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-perp)!), so they equal $\widehat B$ and $\widehat C$. Hence $\widehat B + \widehat A + \widehat C = 180^\circ$. ∎

![The proof in one picture: along the parallel to (BC) through A, the angles B, A, C (alternate angles in matching colors) fill a straight angle.](https://one-course.com/images/onecourse/chapters/math-1/g7-triangles/fig-f3ecc75ad45a.svg)

*The proof in one picture: along the [parallel](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-perp) to $(BC)$ through $A$, the angles $\widehat B$, $\widehat A$, $\widehat C$ ([alternate angles](#def-g7-triangles-pairs) in matching colors) fill a straight angle.*

**Example 51.4.**

Two angles of a triangle measure $67^\circ$ and $58^\circ$. The third measures

$$
180 - (67 + 58) = 180 - 125 = 55^\circ .
$$

Special cases worth memorizing: each angle of an [equilateral triangle](https://one-course.com/books/math/1/en/chapter/41-triangles-and-quadrilaterals#def-g6-shapes-triangles) is $180 \div 3 = 60^\circ$; in a [right triangle](https://one-course.com/books/math/1/en/chapter/41-triangles-and-quadrilaterals#def-g6-shapes-triangles) the two acute angles are complementary (they add up to $180 - 90 = 90^\circ$); the base angles of an [isosceles triangle](https://one-course.com/books/math/1/en/chapter/41-triangles-and-quadrilaterals#def-g6-shapes-triangles) are equal, so each is $\frac{180 - \text{apex}}{2}$.

**Example 51.5 (No triangle with two right angles).**

A triangle with two angles of $90^\circ$ would already use up $180^\circ$, leaving $0^\circ$ for the third — impossible. The angle [sum](https://one-course.com/books/math/1/en/chapter/2-addition-first-steps#def-g1-addition-def) forbids many triangles at once.

## 51.3 The triangle inequality

**Theorem 51.6 (Triangle inequality).**

In every triangle, each side is shorter than the [sum](https://one-course.com/books/math/1/en/chapter/2-addition-first-steps#def-g1-addition-def) of the two others: for a triangle with sides $a$, $b$, $c$,

$$
a < b + c .
$$

Conversely, three lengths with the *largest* smaller than the [sum](https://one-course.com/books/math/1/en/chapter/2-addition-first-steps#def-g1-addition-def) of the two others can always be made into a triangle. If the largest *equals* the [sum](https://one-course.com/books/math/1/en/chapter/2-addition-first-steps#def-g1-addition-def) of the others, the “triangle” is flat: the three [vertices](https://one-course.com/books/math/1/en/chapter/35-cubes-and-boxes#def-g5-solids-def) are aligned.

**Proof.** *Admitted at this level.* ∎

**Example 51.7.**

Can sides be $10$, $3$ and $2$? Test the largest: is $10 < 3 + 2 = 5$? No — no such triangle exists. The compass shows why: arcs of radii $3$ and $2$ drawn from the two ends of a [segment](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-objects) of length $10$ never meet.

Sides $7$, $5$ and $4$? Largest: $7 < 5 + 4 = 9$: yes, the triangle exists (checking the largest side is enough).

![Trying to build a triangle with sides 10, 3, 2: the two compass arcs stay hopelessly far apart, because 3 + 2 < 10.](https://one-course.com/images/onecourse/chapters/math-1/g7-triangles/fig-408c26ed4520.svg)

*Trying to build a triangle with sides $10$, $3$, $2$: the two compass arcs stay hopelessly far apart, because $3 + 2 < 10$.*

**Method 51.8 (Existence and construction of a triangle).**

Given three lengths:

1. compare the largest with the [sum](https://one-course.com/books/math/1/en/chapter/2-addition-first-steps#def-g1-addition-def) of the two others; if it is not strictly smaller, stop: no triangle;
2. otherwise construct as in [Method 41.2](https://one-course.com/books/math/1/en/chapter/41-triangles-and-quadrilaterals#met-g6-shapes-construct) (base [segment](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-objects) , two compass arcs);
3. after any construction from angles, check mentally that the given angles sum to less than $180^\circ$ — with the third angle making exactly $180^\circ$ .

## 51.4 Exercises

**Exercise 51.1 ★.**

Give the complement (to $90^\circ$) and the supplement (to $180^\circ$) of: $30^\circ$; $45^\circ$; $72^\circ$.

**Solution of Exercise 51.1.**

Complements: $60^\circ$; $45^\circ$; $18^\circ$. Supplements: $150^\circ$; $135^\circ$; $108^\circ$.

**Exercise 51.2 ★.**

Two lines cross; one of the four angles measures $35^\circ$. Give the measures of the three others, with a reason for each.

**Solution of Exercise 51.2.**

The [vertically opposite](#def-g7-triangles-pairs) angle also measures $35^\circ$ ([Theorem 51.2](#thm-g7-triangles-equalities)); the two remaining angles are supplementary to it: $180 - 35 = 145^\circ$ each.

**Exercise 51.3 ★.**

Compute the third angle of a triangle whose first two angles measure: (a) $40^\circ$ and $60^\circ$; (b) $90^\circ$ and $28^\circ$; (c) $75^\circ$ and $75^\circ$.

**Solution of Exercise 51.3.**

(a) $180 - 100 = 80^\circ$. (b) $180 - 118 = 62^\circ$. (c) $180 - 150 = 30^\circ$.

**Exercise 51.4 ★.**

An [isosceles triangle](https://one-course.com/books/math/1/en/chapter/41-triangles-and-quadrilaterals#def-g6-shapes-triangles) has its apex angle equal to $40^\circ$. Compute its two base angles. Another [isosceles triangle](https://one-course.com/books/math/1/en/chapter/41-triangles-and-quadrilaterals#def-g6-shapes-triangles) has a base angle of $70^\circ$: compute its apex angle.

**Solution of Exercise 51.4.**

Apex $40^\circ$: the base angles share $180 - 40 = 140^\circ$ equally: $70^\circ$ each.

Base angle $70^\circ$: the two base angles are equal, so the apex is $180 - 2 \times 70 = 40^\circ$.

**Exercise 51.5 ★.**

Which of these triples can be the sides of a triangle? Justify by the triangle inequality (largest side!):

$$
(6,\ 8,\ 12); \qquad
(5,\ 5,\ 11); \qquad
(4,\ 9,\ 13); \qquad
(7,\ 7,\ 7).
$$

**Solution of Exercise 51.5.**

$(6, 8, 12)$: $12 < 6 + 8 = 14$: triangle exists.

$(5, 5, 11)$: $11 > 5 + 5 = 10$: impossible.

$(4, 9, 13)$: $13 = 4 + 9$: flat “triangle” — the points are aligned, no genuine triangle.

$(7, 7, 7)$: $7 < 14$: [equilateral triangle](https://one-course.com/books/math/1/en/chapter/41-triangles-and-quadrilaterals#def-g6-shapes-triangles).

**Exercise 51.6 ★.**

A triangle has angles $x$, $2x$ and $3x$. Find $x$ and the three angles. What kind of triangle is it?

**Solution of Exercise 51.6.**

$x + 2x + 3x = 6x = 180^\circ$, so $x = 30^\circ$: the angles are $30^\circ$, $60^\circ$ and $90^\circ$ — a [right triangle](https://one-course.com/books/math/1/en/chapter/41-triangles-and-quadrilaterals#def-g6-shapes-triangles).

**Exercise 51.7 ★.**

Construct a triangle $ABC$ with $BC = 6$ cm, $\widehat{ABC} = 50^\circ$ and $\widehat{ACB} = 60^\circ$ (protractor at $B$ and at $C$). Before constructing, predict the measure of $\widehat{BAC}$, then check on your figure.

**Solution of Exercise 51.7.**

Prediction: $\widehat{BAC} = 180 - (50 + 60) = 70^\circ$; the protractor on the finished figure confirms it.

**Exercise 51.8 ★★.**

Two [parallel lines](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-perp) are cut by a third line, making an angle of $54^\circ$ at the first crossing (between the crossing line and one [parallel](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-perp)). Draw the situation and give the measures of all eight angles formed.

**Solution of Exercise 51.8.**

At the first crossing, the four angles are $54^\circ$, $126^\circ$, $54^\circ$, $126^\circ$ ([vertically opposite](#def-g7-triangles-pairs) pairs and supplements). The [parallel](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-perp) line reproduces the same four measures at the second crossing (corresponding angles): eight angles in all, four of $54^\circ$ and four of $126^\circ$.

**Exercise 51.9 ★★.**

Two sides of a triangle measure $8$ cm and $3$ cm. Between which two values must the third side lie? Give a whole-number length that works and one that does not.

**Solution of Exercise 51.9.**

Let $c$ be the third side. The triangle inequality demands $c < 8 + 3 = 11$ and also $8 < c + 3$, i.e. $c > 5$. So $5 < c < 11$: the length $7$ cm works; the length $12$ cm (or $4$ cm) does not.

**Exercise 51.10 ★★.**

In a triangle $ABC$, $\widehat A = 2\widehat B$ and $\widehat C = 90^\circ$. Compute $\widehat A$ and $\widehat B$.

**Solution of Exercise 51.10.**

$\widehat A + \widehat B = 180 - 90 = 90^\circ$, and $\widehat A = 2\widehat B$, so $3\widehat B = 90^\circ$: $\widehat B = 30^\circ$ and $\widehat A = 60^\circ$.

**Exercise 51.11 ★★★.**

The three angles of any [quadrilateral](https://one-course.com/books/math/1/en/chapter/26-lines-and-polygons#def-g4-geometry-polygon) $ABCD$ can be studied by cutting it along a diagonal into two triangles. Use this to prove that the four angles of every [quadrilateral](https://one-course.com/books/math/1/en/chapter/26-lines-and-polygons#def-g4-geometry-polygon) add up to $360^\circ$, and deduce the angle [sum](https://one-course.com/books/math/1/en/chapter/2-addition-first-steps#def-g1-addition-def) of a [pentagon](https://one-course.com/books/math/1/en/chapter/26-lines-and-polygons#def-g4-geometry-polygon) (five sides).

**Solution of Exercise 51.11.**

The diagonal $[AC]$ splits $ABCD$ into triangles $ABC$ and $ACD$. The four angles of the [quadrilateral](https://one-course.com/books/math/1/en/chapter/26-lines-and-polygons#def-g4-geometry-polygon) are exactly the six angles of the two triangles, regrouped (the angles at $A$ and at $C$ are each split in two). Total: $180 + 180 = 360^\circ$. A [pentagon](https://one-course.com/books/math/1/en/chapter/26-lines-and-polygons#def-g4-geometry-polygon) splits from one [vertex](https://one-course.com/books/math/1/en/chapter/35-cubes-and-boxes#def-g5-solids-def) into three triangles: $3 \times 180 = 540^\circ$.

## 51.5 Problem: Walking around the block

**Problem 51.1.**

Weekend problem — the angles of any polygon: $(n-2) \times 180^\circ$ inside, always $360^\circ$ of turning outside, and why only three regular shapes tile a floor

[Exercise 51.11](#exo-g7-triangles-11) cut a [quadrilateral](https://one-course.com/books/math/1/en/chapter/26-lines-and-polygons#def-g4-geometry-polygon) into two triangles and found $360^\circ$. This problem pushes the idea to [polygons](https://one-course.com/books/math/1/en/chapter/26-lines-and-polygons#def-g4-geometry-polygon) with any number of sides, then discovers a second, completely different proof — by *walking* around the shape and adding up the turns — and ends on the floor of your bathroom: among all regular [polygons](https://one-course.com/books/math/1/en/chapter/26-lines-and-polygons#def-g4-geometry-polygon), exactly three can tile a plane without gaps or overlaps, and the bees have chosen theirs.

**Part I — The angles inside.**

1. From one [vertex](https://one-course.com/books/math/1/en/chapter/35-cubes-and-boxes#def-g5-solids-def) of a [hexagon](https://one-course.com/books/math/1/en/chapter/26-lines-and-polygons#def-g4-geometry-polygon) (six sides), draw all the diagonals leaving that [vertex](https://one-course.com/books/math/1/en/chapter/35-cubes-and-boxes#def-g5-solids-def) . Into how many triangles is the [hexagon](https://one-course.com/books/math/1/en/chapter/26-lines-and-polygons#def-g4-geometry-polygon) cut, and what is the [sum](https://one-course.com/books/math/1/en/chapter/2-addition-first-steps#def-g1-addition-def) of *all* its interior angles ( [Theorem 51.3](#thm-g7-triangles-sum) )?
2. Generalize: from one [vertex](https://one-course.com/books/math/1/en/chapter/35-cubes-and-boxes#def-g5-solids-def) of a [polygon](https://one-course.com/books/math/1/en/chapter/26-lines-and-polygons#def-g4-geometry-polygon) with $n$ sides, how many triangles does the fan of diagonals produce? Deduce the formula for the [sum](https://one-course.com/books/math/1/en/chapter/2-addition-first-steps#def-g1-addition-def) of the interior angles, and check it for $n = 3$ , $4$ , $5$ , $6$ .
3. Compute the [sum](https://one-course.com/books/math/1/en/chapter/2-addition-first-steps#def-g1-addition-def) of the interior angles of a decagon ( $10$ sides) and of a dodecagon ( $12$ sides).
4. In a *regular* [polygon](https://one-course.com/books/math/1/en/chapter/26-lines-and-polygons#def-g4-geometry-polygon) all interior angles are equal. Compute the interior angle of the [equilateral triangle](https://one-course.com/books/math/1/en/chapter/41-triangles-and-quadrilaterals#def-g6-shapes-triangles) , the square, the regular [pentagon](https://one-course.com/books/math/1/en/chapter/26-lines-and-polygons#def-g4-geometry-polygon) , [hexagon](https://one-course.com/books/math/1/en/chapter/26-lines-and-polygons#def-g4-geometry-polygon) and [octagon](https://one-course.com/books/math/1/en/chapter/26-lines-and-polygons#def-g4-geometry-polygon) .
5. Explain, without any formula, why the angle [sum](https://one-course.com/books/math/1/en/chapter/2-addition-first-steps#def-g1-addition-def) grows by exactly $180^\circ$ each time the [polygon](https://one-course.com/books/math/1/en/chapter/26-lines-and-polygons#def-g4-geometry-polygon) gains one [vertex](https://one-course.com/books/math/1/en/chapter/35-cubes-and-boxes#def-g5-solids-def) .

**Part II — The walk around the block.** Walk along the boundary of a polygonal block, always forward. At each corner you *turn* by some angle — the *exterior angle* of that corner; turning continues until you are back at your starting point, facing your starting direction.

6. At a corner whose interior angle is $\widehat A$ , by how much do you turn? (Interior and exterior angle are supplementary.) Compute the three turns for a triangle with angles $40^\circ$ , $60^\circ$ , $80^\circ$ .
7. Add up the three turns of question 6. What full-circle fact do you observe?
8. Explain why the turns of a complete walk around *any* [polygon](https://one-course.com/books/math/1/en/chapter/26-lines-and-polygons#def-g4-geometry-polygon) — three sides or thirty — always total exactly $360^\circ$ . (What has your nose done, all turns combined, when you arrive back?)
9. Deduce the interior-angle formula a second time: at each of the $n$ corners, interior $+$ turn $= 180^\circ$ ; [sum](https://one-course.com/books/math/1/en/chapter/2-addition-first-steps#def-g1-addition-def) over all corners and use question 8.
10. In a regular [polygon](https://one-course.com/books/math/1/en/chapter/26-lines-and-polygons#def-g4-geometry-polygon) , all the turns are equal, so each exterior angle is $\frac{360^\circ}{n}$ . Use this to answer instantly: which regular [polygon](https://one-course.com/books/math/1/en/chapter/26-lines-and-polygons#def-g4-geometry-polygon) has interior angles of $150^\circ$ ? And why does *no* regular [polygon](https://one-course.com/books/math/1/en/chapter/26-lines-and-polygons#def-g4-geometry-polygon) have interior angles of $155^\circ$ ?

**Part III — Tiling the floor.** Around every point of a tiled floor, the corners of the meeting tiles must total exactly $360^\circ$ — no gap, no overlap ([Problem 40.1](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#pb-g6-lines-1)).

11. For each of the [equilateral triangle](https://one-course.com/books/math/1/en/chapter/41-triangles-and-quadrilaterals#def-g6-shapes-triangles) , the square and the regular [hexagon](https://one-course.com/books/math/1/en/chapter/26-lines-and-polygons#def-g4-geometry-polygon) : how many copies meet at a corner point of the tiling? Verify the $360^\circ$ each time.
12. Show that regular [pentagons](https://one-course.com/books/math/1/en/chapter/26-lines-and-polygons#def-g4-geometry-polygon) cannot tile the floor: what do three corners total, and what would four total?
13. Show that no regular [polygon](https://one-course.com/books/math/1/en/chapter/26-lines-and-polygons#def-g4-geometry-polygon) with *more* than six sides can tile: at least three tiles must meet at each corner (each angle is less than $180^\circ$ ), and what happens to three corners of $135^\circ$ or more? Conclude the complete list of regular tilers.
14. Bathroom floors often mix regular [octagons](https://one-course.com/books/math/1/en/chapter/26-lines-and-polygons#def-g4-geometry-polygon) with small squares. Verify that this corner works: $135^\circ + 135^\circ + 90^\circ$ . Another classic mixes a square, a regular [hexagon](https://one-course.com/books/math/1/en/chapter/26-lines-and-polygons#def-g4-geometry-polygon) and a regular dodecagon at each corner: verify it too (the dodecagon’s interior angle follows from question 10’s method).
15. The bees’ choice: honeycomb cells are regular [hexagons](https://one-course.com/books/math/1/en/chapter/26-lines-and-polygons#def-g4-geometry-polygon) . Verify their corners ( $3$ angles), and explain in one sentence — with the help of Dido’s discovery ( [Problem 43.1](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#pb-g6-measure-1) ) — why, of the three possible regular tiles, the [hexagon](https://one-course.com/books/math/1/en/chapter/26-lines-and-polygons#def-g4-geometry-polygon) is the cheapest in wax for the honey it holds.

**Solution of Problem 51.1.**

**1.** The diagonals from one [vertex](https://one-course.com/books/math/1/en/chapter/35-cubes-and-boxes#def-g5-solids-def) of a [hexagon](https://one-course.com/books/math/1/en/chapter/26-lines-and-polygons#def-g4-geometry-polygon) cut it into $4$ triangles, so the interior angles total $4 \times 180 = 720^\circ$.

**2.** From one [vertex](https://one-course.com/books/math/1/en/chapter/35-cubes-and-boxes#def-g5-solids-def), diagonals go to all [vertices](https://one-course.com/books/math/1/en/chapter/35-cubes-and-boxes#def-g5-solids-def) except itself and its two neighbours: $n - 3$ diagonals, cutting the [polygon](https://one-course.com/books/math/1/en/chapter/26-lines-and-polygons#def-g4-geometry-polygon) into $n - 2$ triangles. Every interior angle of the [polygon](https://one-course.com/books/math/1/en/chapter/26-lines-and-polygons#def-g4-geometry-polygon) is distributed among the triangles, so the [sum](https://one-course.com/books/math/1/en/chapter/2-addition-first-steps#def-g1-addition-def) is

$$
(n - 2) \times 180^\circ .
$$

Check: $n = 3$: $180^\circ$; $n = 4$: $360^\circ$ ([Exercise 51.11](#exo-g7-triangles-11)); $n = 5$: $540^\circ$; $n = 6$: $720^\circ$.

**3.** Decagon: $8 \times 180 = 1\,440^\circ$. Dodecagon: $10 \times 180 = 1\,800^\circ$.

**4.** Dividing by $n$: triangle $60^\circ$; square $90^\circ$; [pentagon](https://one-course.com/books/math/1/en/chapter/26-lines-and-polygons#def-g4-geometry-polygon) $\frac{540}{5} = 108^\circ$; [hexagon](https://one-course.com/books/math/1/en/chapter/26-lines-and-polygons#def-g4-geometry-polygon) $\frac{720}{6} = 120^\circ$; [octagon](https://one-course.com/books/math/1/en/chapter/26-lines-and-polygons#def-g4-geometry-polygon) $\frac{1080}{8} =
135^\circ$.

**5.** Adding a [vertex](https://one-course.com/books/math/1/en/chapter/35-cubes-and-boxes#def-g5-solids-def) adds one more triangle to the fan: the new [polygon](https://one-course.com/books/math/1/en/chapter/26-lines-and-polygons#def-g4-geometry-polygon) needs $n - 1$ triangles where the old needed $n - 2$. One extra triangle, one extra $180^\circ$.

**6.** The walker turns by the supplement: $180^\circ - \widehat A$. For the $40$–$60$–$80$ triangle, the turns are $140^\circ$, $120^\circ$ and $100^\circ$.

**7.** $140 + 120 + 100 = 360^\circ$: one full turn.

**8.** Over the whole walk, your nose starts and ends pointing the same way, having swept around exactly once: all the turning done at the corners amounts to one complete revolution, $360^\circ$ — whatever the number of corners and the shape of the block. (The fact is beautifully independent of $n$.)

**9.** At each corner, interior $+$ turn $= 180^\circ$. Summing over the $n$ corners:

$$
\text{(sum of interiors)} + 360^\circ = 180^\circ \times n,
\qquad\text{so}\qquad
\text{sum of interiors} = (n - 2) \times 180^\circ
$$

— the formula of question 2, re-proved by walking.

**10.** Interior $150^\circ$ means each turn is $30^\circ$, and $n = \frac{360}{30} = 12$: the regular dodecagon. Interior $155^\circ$ would mean turns of $25^\circ$, and $\frac{360}{25} = 14.4$ is not a whole number of corners: no such regular [polygon](https://one-course.com/books/math/1/en/chapter/26-lines-and-polygons#def-g4-geometry-polygon) exists.

**11.** Triangle ($60^\circ$): $6$ meet, $6 \times 60 = 360$. Square ($90^\circ$): $4$ meet. [Hexagon](https://one-course.com/books/math/1/en/chapter/26-lines-and-polygons#def-g4-geometry-polygon) ($120^\circ$): $3$ meet, $3 \times 120 = 360$. All three close up perfectly.

**12.** Three [pentagon](https://one-course.com/books/math/1/en/chapter/26-lines-and-polygons#def-g4-geometry-polygon) corners: $3 \times 108 = 324^\circ$ — a gap of $36^\circ$ remains. Four corners: $4 \times 108 = 432^\circ > 360^\circ$ — overlap. Neither works: regular [pentagons](https://one-course.com/books/math/1/en/chapter/26-lines-and-polygons#def-g4-geometry-polygon) cannot tile the plane.

**13.** At a tiling corner at least $3$ tiles meet. A regular [polygon](https://one-course.com/books/math/1/en/chapter/26-lines-and-polygons#def-g4-geometry-polygon) with more than six sides has interior angles *greater* than $120^\circ$ (question 4 and the growth of the angle with $n$), so three corners exceed $3 \times 120 = 360^\circ$: overlap, impossible. With six sides exactly, $3 \times 120 = 360$ works; below six, questions 11 and 12 sort the cases. The complete list of regular tilers: triangle, square, [hexagon](https://one-course.com/books/math/1/en/chapter/26-lines-and-polygons#def-g4-geometry-polygon).

**14.** [Octagons](https://one-course.com/books/math/1/en/chapter/26-lines-and-polygons#def-g4-geometry-polygon) and squares: $135 + 135 + 90 = 360^\circ$: the corner closes — the classic street pattern. Square, [hexagon](https://one-course.com/books/math/1/en/chapter/26-lines-and-polygons#def-g4-geometry-polygon), dodecagon: the dodecagon’s turn is $\frac{360}{12} = 30^\circ$, so its interior angle is $150^\circ$, and $90 + 120 + 150 = 360^\circ$: it works too.

**15.** Honeycomb corners: $3 \times 120 = 360^\circ$, perfect. Among triangle, square and [hexagon](https://one-course.com/books/math/1/en/chapter/26-lines-and-polygons#def-g4-geometry-polygon) tiles of equal [area](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-area), the [hexagon](https://one-course.com/books/math/1/en/chapter/26-lines-and-polygons#def-g4-geometry-polygon) has the shortest boundary — it is the closest of the three to Dido’s circle ([Problem 43.1](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#pb-g6-measure-1)) — so hexagonal cells enclose the same honey with the least wax: the bees tile like geometers.