Primary & Middle School Mathematics · Grades 1–9

60The Right Triangle and the Cosine

Two beautiful facts tie the right triangle to the circle: a right triangle fits exactly in a half-circle, its hypotenuse being a diameter. And one number, the cosine of an angle, encodes the shape of every right triangle with that angle — the first trigonometric ratio, ahead of its siblings sine and tangent (Chapter 69).

60.1 The right triangle and its circle

Theorem 60.1 (Circle theorem)

If a triangle $ABC$ is right-angled at $A$ , then $A$ lies on the circle whose diameter is the hypotenuse $[BC]$ .
Conversely, if $A$ lies on a circle of diameter $[BC]$ (with $A \neq B, C$ ), then the triangle $ABC$ is right-angled at $A$ .

Equivalently: in a right triangle, the midpoint of the hypotenuse is at equal distance from the three vertices — the median from the right angle measures half the hypotenuse.

Proof of 1. Let $O$ be the midpoint of $[BC]$ and $D$ the symmetric of $A$ about $O$ . The diagonals of $ABDC$ cut at their common midpoint $O$ , so $ABDC$ is a parallelogram (Definition 52.7) — with a right angle 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 $[BC]$ . (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. — Wherever $A$ sits on the circle, the angle $\widehat{BAC}$ is right — the conjecture of Exercise 41.10, now a theorem. The red median $[OA]$ is a radius: half the hypotenuse.

Example 60.2

A triangle has a hypotenuse of $10$ cm. Without knowing anything else, the median from the right angle measures $5$ cm, and the circumscribed circle of the triangle has radius $5$ cm, centered at the midpoint of the hypotenuse.

60.2 The cosine of an acute angle

Definition 60.3 (Cosine)

In a right triangle, 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 with the same acute angle are enlargements of one another, and enlargements preserve ratios of lengths (Chapter 59 began this story; Chapter 68 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 . — Two right 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, when the known and wanted quantities are an acute angle, its adjacent side, and the hypotenuse:

write the definition: $\cos\theta = \frac{\text{adjacent}}{\text{hypotenuse}}$ with the known values in place;
solve the small equation for the unknown (multiply or divide);
for an unknown angle, apply the calculator’s $\cos^{-1}$ to the computed ratio;
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, 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 has a hypotenuse of $13$ cm. What is the length of the median from the right-angle vertex? What is the radius of the circle through the three vertices?

Solution

Solution of Exercise 60.1.

Median from the right angle: half the hypotenuse, $6.5$ cm (Theorem 60.1). The circumscribed circle is centered at the midpoint of the hypotenuse with radius $6.5$ cm.

Exercise 60.2 ★

Draw a circle of diameter $8$ cm with a diameter $[BC]$ , 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

Solution of Exercise 60.2.

The angle at $A$ is right ( $[BC]$ is a diameter: point 2 of Theorem 60.1). The measures satisfy $AB^2 + AC^2 \approx 64 = BC^2$ , up to measuring precision.

Exercise 60.3 ★

In a triangle $DEF$ right-angled 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

Solution of Exercise 60.3.

Hypotenuse: $[EF]$ (opposite the right angle $D$ ). Adjacent to $\widehat E$ : $[ED]$ . Opposite: $[DF]$ . So $\cos\widehat E = \dfrac{ED}{EF}$ .

Exercise 60.4 ★

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

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

Solution

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, 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

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.

Solution

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: the quotient 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

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). Then check with a calculator.

Solution

Solution of Exercise 60.8.

The cosine decreases as the angle grows (Proposition 60.4):

\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

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 of a circle of center $O$ and radius $4.5$ cm, and $A$ is a point of the circle with $AB = 5.4$ cm.

Why is $ABC$ right-angled at $A$ ?
Compute $AC$ , then $\cos \widehat B$ .

Solution

Solution of Exercise 60.10.

1. $A$ is on a circle of diameter $[BC]$ : right angle at $A$ (Theorem 60.1).

2. $BC = 9$ cm (twice the radius). 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 an equilateral triangle of side $1$ (cut along a height), justify the landmark $\cos 60^\circ = \frac12$ of Proposition 60.4. (Where does the foot of the height fall on the base? The full trigonometric version is Example 69.9.)

Solution

Solution of Exercise 60.11.

In the equilateral triangle of side $1$ , the height from one vertex lands on the midpoint of the opposite side (axis of symmetry). Half the triangle is right-angled with hypotenuse $1$ (a full side), an angle of $60^\circ$ at the base vertex, and adjacent side $\frac12$ (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 of a right triangle, the perpendicular to the hypotenuse: this short segment — 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), 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 at $A$ , and $H$ is the foot of the perpendicular from $A$ to the hypotenuse $[BC]$ , so that $H$ lies between $B$ and $C$ and $AH \perp BC$ .

Part I — One angle, two triangles.

Using the angle sum of a triangle (Theorem 51.3) in $ABH$ and in $ABC$ , show that $\widehat{BAH} = \widehat{ACB}$ : the altitude cuts the right triangle into two smaller triangles carrying the same angles as the original.
The triangles $ABC$ (right-angled at $A$ ) and $ABH$ (right-angled at $H$ ) share the angle $\widehat B$ . Write $\cos\widehat B$ in each of them, and deduce from Definition 60.3 that

$AB^2 = BH \times BC .$

(To pass from equal ratios to equal products, multiply both sides by both denominators, Theorem 57.5.)
State and prove the twin relation at the vertex $C$ :

$AC^2 = CH \times CB .$
Take the $3$ – $4$ – $5$ triangle: $AB = 3$ , $AC = 4$ , $BC = 5$ . Compute $BH$ and $CH$ from questions 2 and 3, and check that $BH + HC = BC$ .
In general, express $BH$ and $CH$ as fractions 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.

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), re-proved — with no area puzzle in sight (compare Exercise 58.11).
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:

$AH^2 = BH \times HC .$
Compute the area of $ABC$ in two ways — legs as base and height, then hypotenuse as base (Theorem 53.3) — and deduce the third relation:

$AB \times AC = BC \times AH .$
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.
The altitude’s foot splits the hypotenuse of some right triangle into segments $BH = 2$ cm and $HC = 8$ cm. Compute the altitude $AH$ .

Part III — A machine for square roots. Draw a segment $[BC]$ made of two pieces laid end to end: $BH = p$ and $HC = q$ . Draw the half-circle of diameter $[BC]$ , and let $A$ be the point where the perpendicular to $(BC)$ at $H$ meets it.

Using Theorem 60.1, explain why the triangle $ABC$ is right-angled at $A$ — so question 7 applies and

$AH^2 = p \times q .$

(The length $AH$ is called the geometric mean of $p$ and $q$ .)
Take $p = 1$ and $q = 2$ . What is $AH^2$ ? Which famous length of Example 58.9 has your ruler-and-compass figure just constructed?
Describe the recipe that constructs a segment of length $\sqrt n$ for any whole number $n \geq 1$ , and say which choice of $p$ and $q$ constructs $\sqrt 5$ .
Let $O$ be the center of the half-circle. Explain why $AH$ can never exceed the radius $\frac{p + q}{2}$ , and for which position of $H$ the two are equal.
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).

Solution

Solution of Problem 60.1.

1. In the triangle $ABH$ , right-angled at $H$ , the angle sum (Theorem 51.3) gives $\widehat B + 90^\circ + \widehat{BAH} = 180^\circ$ , so $\widehat{BAH} = 90^\circ - \widehat B$ . In the triangle $ABC$ , right-angled 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 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 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), so

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

multiplying both sides by $BC \times AB$ (Theorem 57.5): $AB^2 = BH \times BC$ .

3. The triangles $ABC$ (right-angled at $A$ ) and $ACH$ (right-angled 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 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, obtained from the cosine alone — a genuinely different proof from the four-triangle area puzzle of Exercise 58.11.

7. Pythagoras in $ABH$ (right-angled 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 of $ABC$ is $\frac{AB \times AC}{2}$ ; with the hypotenuse $[BC]$ as base, the height is exactly $AH$ , so the area is also $\frac{BC \times AH}{2}$ (Theorem 53.3). 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 $[BC]$ (and differs from $B$ and $C$ ), so by point 2 of Theorem 60.1 the triangle $ABC$ is right-angled at $A$ . By construction $(AH) \perp (BC)$ with $H$ between $B$ and $C$ : the segment $[AH]$ 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 whose square is $2$ — the diagonal of the unit square, $\sqrt2 \approx 1.414$ , of Example 58.9.

13. Recipe: lay end to end two segments of lengths $p = 1$ and $q = n$ ; draw the half-circle whose diameter is the whole segment (length $n + 1$ ); raise the perpendicular 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 $6$ ).

14. $A$ is on the circle of center $O$ , so $OA = \frac{p+q}{2}$ , the radius. If $H \neq O$ , the triangle $AHO$ is right-angled 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):

\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 exactly when $p = q$ . The geometric mean of two numbers never exceeds their half-sum.