---
title: "The Right Triangle and the Cosine"
book: "Primary & Middle School Mathematics"
subject: math
language: en
chapter: 60
exercises: 11
source: https://one-course.com/books/math/1/en/chapter/60-the-right-triangle-and-the-cosine
---

# Chapter 60 — The Right Triangle and the Cosine

Two beautiful facts tie the [right triangle](https://one-course.com/books/math/1/en/chapter/41-triangles-and-quadrilaterals#def-g6-shapes-triangles) to the circle: a [right triangle](https://one-course.com/books/math/1/en/chapter/41-triangles-and-quadrilaterals#def-g6-shapes-triangles) fits exactly in a half-circle, its hypotenuse being a [diameter](https://one-course.com/books/math/1/en/chapter/26-lines-and-polygons#def-g4-geometry-circle). And one number, the *cosine* of an angle, encodes the shape of every [right triangle](https://one-course.com/books/math/1/en/chapter/41-triangles-and-quadrilaterals#def-g6-shapes-triangles) with that angle — the first trigonometric ratio, ahead of its siblings sine and tangent ([Chapter 69](https://one-course.com/books/math/1/en/chapter/69-trigonometry-in-the-right-triangle#ch-g9-trig)).

## 60.1 The right triangle and its circle

**Theorem 60.1 (Circle theorem).**

1. If a triangle $ABC$ is [right-angled](https://one-course.com/books/math/1/en/chapter/41-triangles-and-quadrilaterals#def-g6-shapes-triangles) at $A$ , then $A$ lies on the circle whose [diameter](https://one-course.com/books/math/1/en/chapter/26-lines-and-polygons#def-g4-geometry-circle) is the hypotenuse $[BC]$ .
2. Conversely, if $A$ lies on a circle of [diameter](https://one-course.com/books/math/1/en/chapter/26-lines-and-polygons#def-g4-geometry-circle) $[BC]$ (with $A \neq B, C$ ), then the triangle $ABC$ is [right-angled](https://one-course.com/books/math/1/en/chapter/41-triangles-and-quadrilaterals#def-g6-shapes-triangles) at $A$ .

Equivalently: in a [right triangle](https://one-course.com/books/math/1/en/chapter/41-triangles-and-quadrilaterals#def-g6-shapes-triangles), the midpoint of the hypotenuse is at equal distance from the three [vertices](https://one-course.com/books/math/1/en/chapter/35-cubes-and-boxes#def-g5-solids-def) — the median from the [right angle](https://one-course.com/books/math/1/en/chapter/18-shapes-and-right-angles#def-g3-shapes-rightangle) measures [half](https://one-course.com/books/math/1/en/chapter/17-sharing-and-division#def-g3-division-half) the hypotenuse.

**Proof of 1.** Let $O$ be the midpoint of $[BC]$ and $D$ the [symmetric](https://one-course.com/books/math/1/en/chapter/52-central-symmetry-and-parallelograms#def-g7-central-def) of $A$ about $O$. The diagonals of $ABDC$ cut at their common midpoint $O$, so $ABDC$ is a [parallelogram](https://one-course.com/books/math/1/en/chapter/52-central-symmetry-and-parallelograms#def-g7-central-parallelogram) ([Definition 52.7](https://one-course.com/books/math/1/en/chapter/52-central-symmetry-and-parallelograms#def-g7-central-parallelogram)) — with a [right angle](https://one-course.com/books/math/1/en/chapter/18-shapes-and-right-angles#def-g3-shapes-rightangle) at $A$: it is a rectangle. The diagonals of a rectangle are equal, so $OA = \frac{AD}{2} = \frac{BC}{2}$: the point $A$ is at distance $\frac{BC}{2}$ from $O$, i.e. on the circle of [diameter](https://one-course.com/books/math/1/en/chapter/26-lines-and-polygons#def-g4-geometry-circle) $[BC]$. (Point 2 is proved by running the argument backwards.) ∎

![Wherever A sits on the circle, the angle BAC is right — the conjecture of , now a theorem. The red median (OA) is a radius: half the hypotenuse.](https://one-course.com/images/onecourse/chapters/math-1/g8-cosine/fig-104da44b3199.svg)

*Wherever $A$ sits on the circle, the angle $\widehat{BAC}$ is right — the conjecture of [Exercise 41.10](https://one-course.com/books/math/1/en/chapter/41-triangles-and-quadrilaterals#exo-g6-shapes-10), now a theorem. The red median $[OA]$ is a [radius](https://one-course.com/books/math/1/en/chapter/18-shapes-and-right-angles#def-g3-shapes-shapes): [half](https://one-course.com/books/math/1/en/chapter/17-sharing-and-division#def-g3-division-half) the hypotenuse.*

**Example 60.2.**

A triangle has a hypotenuse of $10$ cm. Without knowing anything else, the median from the [right angle](https://one-course.com/books/math/1/en/chapter/18-shapes-and-right-angles#def-g3-shapes-rightangle) measures $5$ cm, and the [circumscribed circle](https://one-course.com/books/math/1/en/chapter/42-axial-symmetry#pb-g6-symmetry-1) of the triangle has [radius](https://one-course.com/books/math/1/en/chapter/18-shapes-and-right-angles#def-g3-shapes-shapes) $5$ cm, centered at the midpoint of the hypotenuse.

## 60.2 The cosine of an acute angle

**Definition 60.3 (Cosine).**

In a [right triangle](https://one-course.com/books/math/1/en/chapter/41-triangles-and-quadrilaterals#def-g6-shapes-triangles), for an acute angle $\theta$:

$$
\cos\theta = \frac{\text{side adjacent to } \theta}{\text{hypotenuse}}
$$

— the leg touching $\theta$, divided by the hypotenuse. This ratio depends only on the angle, not on the size of the triangle: all [right triangles](https://one-course.com/books/math/1/en/chapter/41-triangles-and-quadrilaterals#def-g6-shapes-triangles) with the same acute angle are enlargements of one another, and enlargements preserve ratios of lengths ([Chapter 59](https://one-course.com/books/math/1/en/chapter/59-midpoints-and-parallels#ch-g8-midpoints) began this story; [Chapter 68](https://one-course.com/books/math/1/en/chapter/68-thales-theorem#ch-g9-thales) finishes it).

![Two right triangles sharing the angle : the small one is a reduction of the large one, so adjacent hypotenuse is the same for both — that common value is .](https://one-course.com/images/onecourse/chapters/math-1/g8-cosine/fig-a17d051dd077.svg)

*Two [right triangles](https://one-course.com/books/math/1/en/chapter/41-triangles-and-quadrilaterals#def-g6-shapes-triangles) sharing the angle $\theta$: the small one is a reduction of the large one, so $\frac{\text{adjacent}}{\text{hypotenuse}}$ is the same for both — that common value is $\cos\theta$.*

**Proposition 60.4 (First values and bounds).**

For every acute angle $\theta$: $0 < \cos\theta < 1$, and the cosine *decreases* as the angle opens: a wider angle has a smaller cosine. Landmarks: $\cos 0^\circ = 1$, $\cos 60^\circ = \frac12$, $\cos 90^\circ = 0$ (the extreme values corresponding to flattened triangles).

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

**Method 60.5 (Using the cosine).**

In a [right triangle](https://one-course.com/books/math/1/en/chapter/41-triangles-and-quadrilaterals#def-g6-shapes-triangles), when the known and wanted quantities are an acute angle, its adjacent side, and the hypotenuse:

1. write the definition: $\cos\theta = \frac{\text{adjacent}}{\text{hypotenuse}}$ with the known values in place;
2. solve the small [equation](https://one-course.com/books/math/1/en/chapter/57-literal-calculation-and-equations#def-g8-equations-def) for the unknown (multiply or divide);
3. for an unknown *angle* , apply the calculator’s $\cos^{-1}$ to the computed ratio;
4. sanity checks: a length must come out shorter than the hypotenuse; an angle strictly between $0^\circ$ and $90^\circ$ .

**Example 60.6 (Finding a side).**

A $6$ m ramp makes an angle of $20^\circ$ with the horizontal ground. Horizontal distance covered (adjacent to $20^\circ$, hypotenuse $6$ m):

$$
\cos 20^\circ = \frac{d}{6}
\quad\Longrightarrow\quad
d = 6 \cos 20^\circ \approx 6 \times 0.940 \approx 5.6 \text{ m}.
$$

**Example 60.7 (Finding an angle).**

In a [right triangle](https://one-course.com/books/math/1/en/chapter/41-triangles-and-quadrilaterals#def-g6-shapes-triangles), the side adjacent to the angle $\theta$ measures $3.5$ and the hypotenuse $5$:

$$
\cos\theta = \frac{3.5}{5} = 0.7,
\qquad
\theta = \cos^{-1}(0.7) \approx 45.6^\circ .
$$

## 60.3 Exercises

**Exercise 60.1 ★.**

A [right triangle](https://one-course.com/books/math/1/en/chapter/41-triangles-and-quadrilaterals#def-g6-shapes-triangles) has a hypotenuse of $13$ cm. What is the length of the median from the right-angle [vertex](https://one-course.com/books/math/1/en/chapter/35-cubes-and-boxes#def-g5-solids-def)? What is the [radius](https://one-course.com/books/math/1/en/chapter/18-shapes-and-right-angles#def-g3-shapes-shapes) of the circle through the three [vertices](https://one-course.com/books/math/1/en/chapter/35-cubes-and-boxes#def-g5-solids-def)?

**Solution of Exercise 60.1.**

Median from the [right angle](https://one-course.com/books/math/1/en/chapter/18-shapes-and-right-angles#def-g3-shapes-rightangle): [half](https://one-course.com/books/math/1/en/chapter/17-sharing-and-division#def-g3-division-half) the hypotenuse, $6.5$ cm ([Theorem 60.1](#thm-g8-cosine-circle)). The [circumscribed circle](https://one-course.com/books/math/1/en/chapter/42-axial-symmetry#pb-g6-symmetry-1) is centered at the midpoint of the hypotenuse with [radius](https://one-course.com/books/math/1/en/chapter/18-shapes-and-right-angles#def-g3-shapes-shapes) $6.5$ cm.

**Exercise 60.2 ★.**

Draw a circle of [diameter](https://one-course.com/books/math/1/en/chapter/26-lines-and-polygons#def-g4-geometry-circle) $8$ cm with a [diameter](https://one-course.com/books/math/1/en/chapter/26-lines-and-polygons#def-g4-geometry-circle) $[BC]$, choose any point $A$ on the circle and draw $ABC$. Which angle is right? Cite the theorem. Measure $AB$ and $AC$ and check Pythagoras.

**Solution of Exercise 60.2.**

The angle at $A$ is right ($[BC]$ is a [diameter](https://one-course.com/books/math/1/en/chapter/26-lines-and-polygons#def-g4-geometry-circle): point 2 of [Theorem 60.1](#thm-g8-cosine-circle)). The measures satisfy $AB^2 + AC^2 \approx 64 = BC^2$, up to measuring precision.

**Exercise 60.3 ★.**

In a triangle $DEF$ [right-angled](https://one-course.com/books/math/1/en/chapter/41-triangles-and-quadrilaterals#def-g6-shapes-triangles) at $D$, name the side adjacent to the angle $\widehat E$, the side opposite it, and the hypotenuse. Write $\cos \widehat E$ as a ratio.

**Solution of Exercise 60.3.**

Hypotenuse: $[EF]$ (opposite the [right angle](https://one-course.com/books/math/1/en/chapter/18-shapes-and-right-angles#def-g3-shapes-rightangle) $D$). Adjacent to $\widehat E$: $[ED]$. Opposite: $[DF]$. So $\cos\widehat E = \dfrac{ED}{EF}$.

**Exercise 60.4 ★.**

Compute the missing quantity ($\cos 35^\circ \approx 0.819$, $\cos 50^\circ \approx 0.643$):

1. hypotenuse $10$ , angle $35^\circ$ : adjacent side?
2. adjacent side $6$ , angle $50^\circ$ : hypotenuse?

**Solution of Exercise 60.4.**

*1.* adjacent $= 10 \cos 35^\circ \approx 8.2$.

*2.* hypotenuse $= \dfrac{6}{\cos 50^\circ} \approx
\dfrac{6}{0.643} \approx 9.3$.

**Exercise 60.5 ★.**

In a [right triangle](https://one-course.com/books/math/1/en/chapter/41-triangles-and-quadrilaterals#def-g6-shapes-triangles), the adjacent side to $\theta$ measures $8$ and the hypotenuse $10$. Compute $\cos\theta$, then $\theta$ ($\cos^{-1}(0.8) \approx 36.9^\circ$).

**Solution of Exercise 60.5.**

$\cos\theta = \frac{8}{10} = 0.8$, so $\theta = \cos^{-1}(0.8) \approx 37^\circ$.

**Exercise 60.6 ★.**

Explain why $\cos\theta$ can never equal $1.2$ for an acute angle of a [right triangle](https://one-course.com/books/math/1/en/chapter/41-triangles-and-quadrilaterals#def-g6-shapes-triangles).

**Solution of Exercise 60.6.**

$\cos\theta$ is a leg divided by the hypotenuse, and the hypotenuse is the *longest* side of a [right triangle](https://one-course.com/books/math/1/en/chapter/41-triangles-and-quadrilaterals#def-g6-shapes-triangles): the [quotient](https://one-course.com/books/math/1/en/chapter/17-sharing-and-division#def-g3-division-remainder) is always smaller than $1$. A value of $1.2$ would mean a leg longer than the hypotenuse — impossible.

**Exercise 60.7 ★★.**

A $25$ m zip line descends from a platform to the ground, making an angle of $12^\circ$ with the horizontal ($\cos 12^\circ \approx
0.978$). What horizontal distance does it span?

**Solution of Exercise 60.7.**

Horizontal span $= 25 \cos 12^\circ \approx 25 \times 0.978 \approx
24.5$ m.

**Exercise 60.8 ★★.**

Order without a calculator: $\cos 20^\circ$, $\cos 70^\circ$, $\cos 45^\circ$ (use [Proposition 60.4](#prop-g8-cosine-bounds)). Then check with a calculator.

**Solution of Exercise 60.8.**

The cosine decreases as the angle grows ([Proposition 60.4](#prop-g8-cosine-bounds)):

$$
\cos 70^\circ < \cos 45^\circ < \cos 20^\circ .
$$

(Calculator: $0.342 < 0.707 < 0.940$.)

**Exercise 60.9 ★★.**

A ladder of length $L$ leans against a wall with an angle $\theta$ between ladder and *ground*. Its foot is $1.2$ m from the wall and $\theta = 68^\circ$ ($\cos 68^\circ \approx 0.375$). Compute $L$, then the height reached (Pythagoras).

**Solution of Exercise 60.9.**

The ground distance is adjacent to $\theta$: $\cos 68^\circ = \frac{1.2}{L}$, so $L = \frac{1.2}{0.375} = 3.2$ m. Height (Pythagoras): $\sqrt{3.2^2 - 1.2^2} = \sqrt{10.24 - 1.44} = \sqrt{8.8} \approx 3.0$ m.

**Exercise 60.10 ★★.**

$[BC]$ is a [diameter](https://one-course.com/books/math/1/en/chapter/26-lines-and-polygons#def-g4-geometry-circle) of a circle of center $O$ and [radius](https://one-course.com/books/math/1/en/chapter/18-shapes-and-right-angles#def-g3-shapes-shapes) $4.5$ cm, and $A$ is a point of the circle with $AB = 5.4$ cm.

1. Why is $ABC$ [right-angled](https://one-course.com/books/math/1/en/chapter/41-triangles-and-quadrilaterals#def-g6-shapes-triangles) at $A$ ?
2. Compute $AC$ , then $\cos \widehat B$ .

**Solution of Exercise 60.10.**

*1.* $A$ is on a circle of [diameter](https://one-course.com/books/math/1/en/chapter/26-lines-and-polygons#def-g4-geometry-circle) $[BC]$: [right angle](https://one-course.com/books/math/1/en/chapter/18-shapes-and-right-angles#def-g3-shapes-rightangle) at $A$ ([Theorem 60.1](#thm-g8-cosine-circle)).

*2.* $BC = 9$ cm (twice the [radius](https://one-course.com/books/math/1/en/chapter/18-shapes-and-right-angles#def-g3-shapes-shapes)). Pythagoras: $AC = \sqrt{9^2 - 5.4^2} = \sqrt{81 - 29.16} = \sqrt{51.84} = 7.2$ cm. And $\cos\widehat B = \dfrac{AB}{BC} = \dfrac{5.4}{9} = 0.6$.

**Exercise 60.11 ★★★.**

Using [half](https://one-course.com/books/math/1/en/chapter/17-sharing-and-division#def-g3-division-half) an [equilateral triangle](https://one-course.com/books/math/1/en/chapter/41-triangles-and-quadrilaterals#def-g6-shapes-triangles) of side $1$ (cut along a height), justify the landmark $\cos 60^\circ = \frac12$ of [Proposition 60.4](#prop-g8-cosine-bounds). (Where does the foot of the height fall on the base? The full trigonometric version is [Example 69.9](https://one-course.com/books/math/1/en/chapter/69-trigonometry-in-the-right-triangle#ex-g9-trig-special).)

**Solution of Exercise 60.11.**

In the [equilateral triangle](https://one-course.com/books/math/1/en/chapter/41-triangles-and-quadrilaterals#def-g6-shapes-triangles) of side $1$, the height from one [vertex](https://one-course.com/books/math/1/en/chapter/35-cubes-and-boxes#def-g5-solids-def) lands on the *midpoint* of the opposite side ([axis of symmetry](https://one-course.com/books/math/1/en/chapter/42-axial-symmetry#def-g6-symmetry-axis)). [Half](https://one-course.com/books/math/1/en/chapter/17-sharing-and-division#def-g3-division-half) the triangle is [right-angled](https://one-course.com/books/math/1/en/chapter/41-triangles-and-quadrilaterals#def-g6-shapes-triangles) with hypotenuse $1$ (a full side), an angle of $60^\circ$ at the base [vertex](https://one-course.com/books/math/1/en/chapter/35-cubes-and-boxes#def-g5-solids-def), and adjacent side $\frac12$ ([half](https://one-course.com/books/math/1/en/chapter/17-sharing-and-division#def-g3-division-half) the base). Hence

$$
\cos 60^\circ = \frac{1/2}{1} = \frac12 .
$$

## 60.4 Problem: Euclid’s relations, and Pythagoras all over again

**Problem 60.1.**

Weekend problem — the altitude of a right triangle: $AB^2 = BH \times BC$, $AH^2 = BH \times HC$, a second proof of Pythagoras, and a machine for constructing square roots

Drop, from the [right angle](https://one-course.com/books/math/1/en/chapter/18-shapes-and-right-angles#def-g3-shapes-rightangle) of a [right triangle](https://one-course.com/books/math/1/en/chapter/41-triangles-and-quadrilaterals#def-g6-shapes-triangles), the [perpendicular](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-perp) to the hypotenuse: this short [segment](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-objects) — the *altitude* — satisfies relations so useful that Euclid put them at the heart of his *Elements*. In this problem the cosine’s defining property, “the ratio depends only on the angle” ([Definition 60.3](#def-g8-cosine-def)), proves all of them; on the way you will re-prove Pythagoras’ theorem by a completely different route, and end with a ruler-and-compass machine that constructs $\sqrt2$, $\sqrt5$, $\sqrt{n}$ for every whole number $n$.

Throughout, $ABC$ is a triangle [right-angled](https://one-course.com/books/math/1/en/chapter/41-triangles-and-quadrilaterals#def-g6-shapes-triangles) at $A$, and $H$ is the foot of the [perpendicular](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-perp) from $A$ to the hypotenuse $[BC]$, so that $H$ lies between $B$ and $C$ and $AH \perp BC$.

**Part I — One angle, two triangles.**

1. Using the angle [sum](https://one-course.com/books/math/1/en/chapter/2-addition-first-steps#def-g1-addition-def) of a triangle ( [Theorem 51.3](https://one-course.com/books/math/1/en/chapter/51-triangles-and-angles#thm-g7-triangles-sum) ) in $ABH$ and in $ABC$ , show that $\widehat{BAH} = \widehat{ACB}$ : the altitude cuts the [right triangle](https://one-course.com/books/math/1/en/chapter/41-triangles-and-quadrilaterals#def-g6-shapes-triangles) into two smaller triangles carrying the *same angles* as the original.
2. The triangles $ABC$ ([right-angled](https://one-course.com/books/math/1/en/chapter/41-triangles-and-quadrilaterals#def-g6-shapes-triangles) at $A$) and $ABH$ ([right-angled](https://one-course.com/books/math/1/en/chapter/41-triangles-and-quadrilaterals#def-g6-shapes-triangles) at $H$) share the angle $\widehat B$. Write $\cos\widehat B$ in each of them, and deduce from [Definition 60.3](#def-g8-cosine-def) that $$AB^2 = BH \times BC .$$ (To pass from equal ratios to equal [products](https://one-course.com/books/math/1/en/chapter/10-multiplication-first-steps#def-g2-mult-def), multiply both sides by both [denominators](https://one-course.com/books/math/1/en/chapter/24-first-fractions#def-g4-fractions-def), [Theorem 57.5](https://one-course.com/books/math/1/en/chapter/57-literal-calculation-and-equations#thm-g8-equations-balance).)
3. State and prove the twin relation at the [vertex](https://one-course.com/books/math/1/en/chapter/35-cubes-and-boxes#def-g5-solids-def) $C$: $$AC^2 = CH \times CB .$$
4. Take the $3$ – $4$ – $5$ triangle: $AB = 3$ , $AC = 4$ , $BC = 5$ . Compute $BH$ and $CH$ from questions 2 and 3, and check that $BH + HC = BC$ .
5. In general, express $BH$ and $CH$ as [fractions](https://one-course.com/books/math/1/en/chapter/39-fractions-first-steps#def-g6-fractions-def) involving only the three sides. In what sense does the foot $H$ split the hypotenuse “proportionally to the squares of the legs”?

**Part II — Pythagoras again, and the altitude.**

6. Add the relations of questions 2 and 3 and use $BH + HC = BC$ to obtain $$AB^2 + AC^2 = BC^2 :$$ Pythagoras’ theorem ([Theorem 58.1](https://one-course.com/books/math/1/en/chapter/58-the-pythagorean-theorem#thm-g8-pythagoras-direct)), re-proved — with no [area](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-area) puzzle in sight (compare [Exercise 58.11](https://one-course.com/books/math/1/en/chapter/58-the-pythagorean-theorem#exo-g8-pythagoras-11)).
7. Apply Pythagoras in the small triangle $ABH$ to write $AH^2 = AB^2 - BH^2$, replace $AB^2$ by $BH \times BC$, and factor out $BH$ to prove *Euclid’s altitude relation*: $$AH^2 = BH \times HC .$$
8. Compute the [area](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-area) of $ABC$ in two ways — legs as base and height, then hypotenuse as base ([Theorem 53.3](https://one-course.com/books/math/1/en/chapter/53-areas-and-volumes#thm-g7-areas-triangle)) — and deduce the third relation: $$AB \times AC = BC \times AH .$$
9. Back to the $3$ – $4$ – $5$ triangle: compute $AH$ with question 8, then verify the altitude relation of question 7 numerically, using the values of $BH$ and $CH$ found in question 4.
10. The altitude’s foot splits the hypotenuse of some [right triangle](https://one-course.com/books/math/1/en/chapter/41-triangles-and-quadrilaterals#def-g6-shapes-triangles) into [segments](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-objects) $BH = 2$ cm and $HC = 8$ cm. Compute the altitude $AH$ .

**Part III — A machine for square roots.** Draw a [segment](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-objects) $[BC]$ made of two pieces laid end to end: $BH = p$ and $HC = q$. Draw the half-circle of [diameter](https://one-course.com/books/math/1/en/chapter/26-lines-and-polygons#def-g4-geometry-circle) $[BC]$, and let $A$ be the point where the [perpendicular](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-perp) to $(BC)$ at $H$ meets it.

11. Using [Theorem 60.1](#thm-g8-cosine-circle), explain why the triangle $ABC$ is [right-angled](https://one-course.com/books/math/1/en/chapter/41-triangles-and-quadrilaterals#def-g6-shapes-triangles) at $A$ — so question 7 applies and $$AH^2 = p \times q .$$ (The length $AH$ is called the *geometric mean* of $p$ and $q$.)
12. Take $p = 1$ and $q = 2$ . What is $AH^2$ ? Which famous length of [Example 58.9](https://one-course.com/books/math/1/en/chapter/58-the-pythagorean-theorem#ex-g8-pythagoras-diagonal) has your ruler-and-compass figure just constructed?
13. Describe the recipe that constructs a [segment](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-objects) of length $\sqrt n$ for any whole number $n \geq 1$ , and say which choice of $p$ and $q$ constructs $\sqrt 5$ .
14. Let $O$ be the center of the half-circle. Explain why $AH$ can never exceed the [radius](https://one-course.com/books/math/1/en/chapter/18-shapes-and-right-angles#def-g3-shapes-shapes) $\frac{p + q}{2}$ , and for which position of $H$ the two are equal.
15. Deduce the inequality: for all positive numbers $p$ and $q$, $$p \times q \leq \left(\frac{p + q}{2}\right)^2,$$ with equality exactly when $p = q$. Then reprove it with no geometry at all: expand $\left(\frac{p+q}{2}\right)^2 - p q$ and recognize a remarkable identity ([Problem 57.1](https://one-course.com/books/math/1/en/chapter/57-literal-calculation-and-equations#pb-g8-equations-1)).

**Solution of Problem 60.1.**

**1.** In the triangle $ABH$, [right-angled](https://one-course.com/books/math/1/en/chapter/41-triangles-and-quadrilaterals#def-g6-shapes-triangles) at $H$, the angle [sum](https://one-course.com/books/math/1/en/chapter/2-addition-first-steps#def-g1-addition-def) ([Theorem 51.3](https://one-course.com/books/math/1/en/chapter/51-triangles-and-angles#thm-g7-triangles-sum)) gives $\widehat B + 90^\circ + \widehat{BAH} = 180^\circ$, so $\widehat{BAH} = 90^\circ - \widehat B$. In the triangle $ABC$, [right-angled](https://one-course.com/books/math/1/en/chapter/41-triangles-and-quadrilaterals#def-g6-shapes-triangles) at $A$: $\widehat B + \widehat C + 90^\circ = 180^\circ$, so $\widehat C = 90^\circ - \widehat B$. Hence $\widehat{BAH} = \widehat{ACB}$: both small triangles repeat the angles $\widehat B$, $\widehat C$, $90^\circ$ of the original.

**2.** In $ABC$ ([right-angled](https://one-course.com/books/math/1/en/chapter/41-triangles-and-quadrilaterals#def-g6-shapes-triangles) at $A$), the side adjacent to $\widehat B$ is $AB$ and the hypotenuse is $BC$: $\cos\widehat B = \frac{AB}{BC}$. In $ABH$ ([right-angled](https://one-course.com/books/math/1/en/chapter/41-triangles-and-quadrilaterals#def-g6-shapes-triangles) at $H$), the side adjacent to $\widehat B$ is $BH$ and the hypotenuse is $AB$: $\cos\widehat B = \frac{BH}{AB}$. The ratio depends only on the angle ([Definition 60.3](#def-g8-cosine-def)), so

$$
\frac{AB}{BC} = \frac{BH}{AB} ;
$$

multiplying both sides by $BC \times AB$ ([Theorem 57.5](https://one-course.com/books/math/1/en/chapter/57-literal-calculation-and-equations#thm-g8-equations-balance)): $AB^2 = BH \times BC$.

**3.** The triangles $ABC$ ([right-angled](https://one-course.com/books/math/1/en/chapter/41-triangles-and-quadrilaterals#def-g6-shapes-triangles) at $A$) and $ACH$ ([right-angled](https://one-course.com/books/math/1/en/chapter/41-triangles-and-quadrilaterals#def-g6-shapes-triangles) at $H$) share the angle $\widehat C$. Adjacent side over hypotenuse in each:

$$
\cos\widehat C = \frac{AC}{CB} = \frac{CH}{AC}
\qquad\Longrightarrow\qquad
AC^2 = CH \times CB .
$$

**4.** $AB^2 = BH \times BC$ reads $9 = BH \times 5$, so $BH = 1.8$; and $16 = CH \times 5$ gives $CH = 3.2$. Check: $BH + HC = 1.8 + 3.2 = 5 = BC$ — as it must, since $H$ lies on the hypotenuse between $B$ and $C$.

**5.** Dividing each relation by $BC$:

$$
BH = \frac{AB^2}{BC},
\qquad
CH = \frac{AC^2}{BC} .
$$

So $BH$ and $CH$ are [proportional](https://one-course.com/books/math/1/en/chapter/44-proportionality-and-data#def-g6-propdata-def) to $AB^2$ and $AC^2$: the foot of the altitude splits the hypotenuse in the ratio of the squares of the legs ($1.8 : 3.2 = 9 : 16$ in question 4).

**6.** Adding the two relations and factoring out $BC$ (distributivity):

$$
AB^2 + AC^2 = BH \times BC + CH \times BC
= (BH + CH) \times BC = BC \times BC = BC^2 .
$$

This is [Theorem 58.1](https://one-course.com/books/math/1/en/chapter/58-the-pythagorean-theorem#thm-g8-pythagoras-direct), obtained from the cosine alone — a genuinely different proof from the four-triangle [area](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-area) puzzle of [Exercise 58.11](https://one-course.com/books/math/1/en/chapter/58-the-pythagorean-theorem#exo-g8-pythagoras-11).

**7.** Pythagoras in $ABH$ ([right-angled](https://one-course.com/books/math/1/en/chapter/41-triangles-and-quadrilaterals#def-g6-shapes-triangles) at $H$, hypotenuse $[AB]$): $AH^2 = AB^2 - BH^2$. Replacing $AB^2$ by $BH \times BC$ (question 2) and factoring out $BH$:

$$
AH^2 = BH \times BC - BH^2
= BH \times (BC - BH)
= BH \times HC .
$$

**8.** With the legs as base and height, the [area](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-area) of $ABC$ is $\frac{AB \times AC}{2}$; with the hypotenuse $[BC]$ as base, the height is exactly $AH$, so the [area](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-area) is also $\frac{BC \times AH}{2}$ ([Theorem 53.3](https://one-course.com/books/math/1/en/chapter/53-areas-and-volumes#thm-g7-areas-triangle)). Equating and doubling: $AB \times AC = BC \times AH$.

**9.** $AH = \frac{AB \times AC}{BC} = \frac{3 \times 4}{5}
= 2.4$. Verification of question 7: $AH^2 = 2.4^2 = 5.76$ and $BH \times HC = 1.8 \times 3.2 = 5.76$. Equal.

**10.** $AH^2 = BH \times HC = 2 \times 8 = 16$, so $AH = \sqrt{16} = 4$ cm.

**11.** The point $A$ lies on the circle of [diameter](https://one-course.com/books/math/1/en/chapter/26-lines-and-polygons#def-g4-geometry-circle) $[BC]$ (and differs from $B$ and $C$), so by point 2 of [Theorem 60.1](#thm-g8-cosine-circle) the triangle $ABC$ is [right-angled](https://one-course.com/books/math/1/en/chapter/41-triangles-and-quadrilaterals#def-g6-shapes-triangles) at $A$. By construction $(AH) \perp (BC)$ with $H$ between $B$ and $C$: the [segment](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-objects) $[AH]$ is precisely the altitude of question 7, and $AH^2 = BH \times HC = p \times q$.

**12.** $AH^2 = 1 \times 2 = 2$: the figure constructs, with ruler and compass alone, a [segment](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-objects) whose square is $2$ — the diagonal of the unit square, $\sqrt2 \approx 1.414$, of [Example 58.9](https://one-course.com/books/math/1/en/chapter/58-the-pythagorean-theorem#ex-g8-pythagoras-diagonal).

**13.** Recipe: lay end to end two [segments](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-objects) of lengths $p = 1$ and $q = n$; draw the half-circle whose [diameter](https://one-course.com/books/math/1/en/chapter/26-lines-and-polygons#def-g4-geometry-circle) is the whole [segment](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-objects) (length $n + 1$); raise the [perpendicular](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-perp) at the junction point; it meets the half-circle at a point whose distance to the junction is $\sqrt{1 \times n} = \sqrt n$. For $\sqrt5$: take $p = 1$ and $q = 5$ (a half-circle of [diameter](https://one-course.com/books/math/1/en/chapter/26-lines-and-polygons#def-g4-geometry-circle) $6$).

**14.** $A$ is on the circle of center $O$, so $OA = \frac{p+q}{2}$, the [radius](https://one-course.com/books/math/1/en/chapter/18-shapes-and-right-angles#def-g3-shapes-shapes). If $H \neq O$, the triangle $AHO$ is [right-angled](https://one-course.com/books/math/1/en/chapter/41-triangles-and-quadrilaterals#def-g6-shapes-triangles) at $H$ with hypotenuse $[OA]$, and a leg is shorter than the hypotenuse: $AH < OA$. If $H = O$, then $AH = OA$ exactly. In all cases $AH \leq \frac{p+q}{2}$, with equality precisely when $H$ is the center — that is, when $p = q$.

**15.** Squaring $AH \leq \frac{p+q}{2}$ (both sides positive) and using $AH^2 = pq$:

$$
p \times q \leq \left(\frac{p+q}{2}\right)^2 ,
$$

with equality exactly for $p = q$. Algebraic re-proof, with the remarkable identities ([Problem 57.1](https://one-course.com/books/math/1/en/chapter/57-literal-calculation-and-equations#pb-g8-equations-1)):

$$
\left(\frac{p+q}{2}\right)^2 - pq
= \frac{p^2 + 2pq + q^2 - 4pq}{4}
= \frac{p^2 - 2pq + q^2}{4}
= \left(\frac{p-q}{2}\right)^2 ,
$$

a square, hence never negative — and [zero](https://one-course.com/books/math/1/en/chapter/1-counting-to-20#def-g1-counting-zero) exactly when $p = q$. The [geometric mean](#pb-g8-cosine-1) of two numbers never exceeds their [half-sum](https://one-course.com/books/math/1/en/chapter/2-addition-first-steps#def-g1-addition-def).