Primary & Middle School Mathematics · Grades 1–9

69Trigonometry in the Right Triangle

In a right triangle, once you know one acute angle, the shape of the triangle is fixed — only its size can vary. The ratios of its sides therefore depend only on the angle: these ratios are the cosine, sine and tangent. Together with the Pythagorean theorem, they let you compute every side and every angle of a right triangle from very little information.

69.1 The Pythagorean theorem, again

Theorem 69.1 (Pythagoras)

In a triangle $ABC$ right-angled at $A$ , the square of the hypotenuse is the sum of the squares of the two legs:

BC^2 = AB^2 + AC^2 .

Conversely, if $BC^2 = AB^2 + AC^2$ in a triangle, then the triangle is right-angled at $A$ .

Proof. Admitted at this level. ∎

Example 69.2

A right triangle has legs $AB = 5$ and $AC = 12$ . Then

BC^2 = 5^2 + 12^2 = 25 + 144 = 169, \qquad\text{so}\quad BC = \sqrt{169} = 13 .

To find a leg: if $BC = 10$ and $AB = 6$ , then $AC^2 = BC^2 - AB^2 = 100 - 36 = 64$ , so $AC = 8$ .

69.2 Cosine, sine, tangent

In a right triangle, relative to an acute angle $\theta$ , the three sides have names: the hypotenuse (opposite the right angle, the longest side), the side opposite to $\theta$ , and the side adjacent to $\theta$ (the leg touching $\theta$ ).

Definition 69.3 (Trigonometric ratios)

For an acute angle $\theta$ of a right triangle:

\cos\theta = \frac{\text{adjacent}}{\text{hypotenuse}}, \qquad \sin\theta = \frac{\text{opposite}}{\text{hypotenuse}}, \qquad \tan\theta = \frac{\text{opposite}}{\text{adjacent}} .

These ratios depend only on $\theta$ , not on the size of the triangle (all right triangles with the same acute angle are scalings of one another, by Thales).

The three sides seen from the angle = B: the hypotenuse [BC], the adjacent side [BA], the opposite side [AC]. — The three sides seen from the angle $\theta = \widehat B$ : the hypotenuse $[BC]$ , the adjacent side $[BA]$ , the opposite side $[AC]$ .

Proposition 69.4 (First properties)

For every acute angle $\theta$ :

$0 < \cos\theta < 1$ and $0 < \sin\theta < 1$ (a leg is shorter than the hypotenuse);
$\cos^2\theta + \sin^2\theta = 1$ ;
$\tan\theta = \dfrac{\sin\theta}{\cos\theta}$ .

Proof. Write $a$ for the adjacent side, $o$ for the opposite side, $h$ for the hypotenuse of a right triangle with angle $\theta$ .

1. $0 < a < h$ and $0 < o < h$ , so the two ratios are strictly between $0$ and $1$ .

2. By Pythagoras, $a^2 + o^2 = h^2$ . Divide everything by $h^2$ :

\cos^2\theta + \sin^2\theta = \frac{a^2}{h^2} + \frac{o^2}{h^2} = \frac{a^2 + o^2}{h^2} = 1 .

3. $\dfrac{\sin\theta}{\cos\theta} = \dfrac{o/h}{a/h} = \dfrac oa = \tan\theta$ . ∎

Method 69.5 (Finding a side)

To compute an unknown side when one side and one acute angle are known:

mark the known angle and label the three sides (hypotenuse, opposite, adjacent) from that angle;
choose the ratio (cos, sin or tan) that involves the known side and the wanted side;
write the equation and solve it;
sanity-check: the hypotenuse must come out longest.

Example 69.6

In a right triangle, the hypotenuse measures $8$ and one angle is $35^\circ$ ; compute the side opposite to it. The ratio involving the opposite side and the hypotenuse is the sine:

\sin 35^\circ = \frac{\text{opposite}}{8} \quad\Longrightarrow\quad \text{opposite} = 8 \sin 35^\circ \approx 8 \times 0.574 \approx 4.6 .

For the adjacent side, use the cosine: $8\cos 35^\circ \approx 8 \times 0.819 \approx 6.6$ . Check with Pythagoras: $4.6^2 + 6.6^2 \approx 21 + 43 \approx 64 = 8^2$ .

Method 69.7 (Finding an angle)

When two sides are known, compute the ratio they form, identify it as a cosine, sine or tangent of the unknown angle, and apply the calculator’s inverse function ( $\cos^{-1}$ , $\sin^{-1}$ or $\tan^{-1}$ ) to the ratio.

Example 69.8

A ladder $5$ m long leans against a wall, its foot $1.4$ m from the wall. The angle $\theta$ between the ladder and the ground satisfies

\cos\theta = \frac{1.4}{5} = 0.28, \qquad\text{so}\quad \theta = \cos^{-1}(0.28) \approx 73.7^\circ .

The ladder problem: the adjacent side (1.4 m) and the hypotenuse (5 m) are known, so the cosine gives the angle. — The ladder problem: the adjacent side ( $1.4$ m) and the hypotenuse ( $5$ m) are known, so the cosine gives the angle.

Example 69.9 (Special angles)

Two triangles give exact values. Half a square of side 1 (a right isosceles triangle) has hypotenuse $\sqrt2$ and angles of $45^\circ$ :

\cos 45^\circ = \sin 45^\circ = \frac{1}{\sqrt2} = \frac{\sqrt2}{2}, \qquad \tan 45^\circ = 1 .

Half an equilateral triangle of side 1 gives the angles $30^\circ$ and $60^\circ$ :

\cos 60^\circ = \sin 30^\circ = \frac12, \qquad \cos 30^\circ = \sin 60^\circ = \frac{\sqrt3}{2}.

69.3 Exercises

Exercise 69.1 ★

A right triangle has legs $9$ and $12$ . Compute its hypotenuse. Another has hypotenuse $17$ and one leg $8$ : compute the other leg.

Solution

Solution of Exercise 69.1.

Hypotenuse: $\sqrt{9^2 + 12^2} = \sqrt{81 + 144} = \sqrt{225} = 15$ .

Other leg: $\sqrt{17^2 - 8^2} = \sqrt{289 - 64} = \sqrt{225} = 15$ .

Exercise 69.2 ★

A triangle has sides $7$ , $24$ and $25$ . Is it right-angled? Same question for sides $5$ , $6$ and $8$ .

Solution

Solution of Exercise 69.2.

$7^2 + 24^2 = 49 + 576 = 625 = 25^2$ : yes, right-angled (converse of Pythagoras), with the right angle opposite the side $25$ .

$5^2 + 6^2 = 61 \neq 64 = 8^2$ : not right-angled.

Exercise 69.3 ★

In a triangle $DEF$ right-angled at $D$ , with the angle at $E$ noted $\theta$ : which side is the hypotenuse? Which side is opposite to $\theta$ ? Write $\cos\theta$ , $\sin\theta$ and $\tan\theta$ as ratios of the sides $DE$ , $DF$ , $EF$ .

Solution

Solution of Exercise 69.3.

The hypotenuse is $[EF]$ (opposite the right angle at $D$ ). The side opposite to $\theta = \widehat E$ is $[DF]$ ; the adjacent side is $[DE]$ . Hence

\cos\theta = \frac{DE}{EF}, \qquad \sin\theta = \frac{DF}{EF}, \qquad \tan\theta = \frac{DF}{DE}.

Exercise 69.4 ★

In a right triangle, the hypotenuse measures $10$ and one acute angle is $28^\circ$ . Compute the two legs (round to the tenth; $\cos 28^\circ \approx 0.883$ , $\sin 28^\circ \approx 0.469$ ).

Solution

Solution of Exercise 69.4.

Opposite leg: $10 \sin 28^\circ \approx 10 \times 0.469 = 4.7$ . Adjacent leg: $10 \cos 28^\circ \approx 10 \times 0.883 = 8.8$ . (Check: $4.7^2 + 8.8^2 \approx 22 + 77 \approx 100$ .)

Exercise 69.5 ★

In a right triangle, the leg adjacent to the angle $\theta$ measures $6$ and the opposite leg measures $4.5$ . Compute $\tan\theta$ , then $\theta$ (round to the degree; $\tan^{-1}(0.75) \approx 36.9^\circ$ ).

Solution

Solution of Exercise 69.5.

$\tan\theta = \dfrac{4.5}{6} = 0.75$ , so $\theta = \tan^{-1}(0.75) \approx 37^\circ$ .

Exercise 69.6 ★★

A drone flies at an altitude of $120$ m. From an observer on the ground, it is seen at an angle of elevation of $32^\circ$ . At what horizontal distance from the observer is the drone ( $\tan 32^\circ \approx 0.625$ )?

Solution

Solution of Exercise 69.6.

In the right triangle formed by the observer, the point on the ground below the drone, and the drone: the altitude $120$ m is the side opposite the elevation angle, the horizontal distance $d$ is adjacent. So

\tan 32^\circ = \frac{120}{d}, \qquad d = \frac{120}{\tan 32^\circ} \approx \frac{120}{0.625} = 192 \text{ m}.

Exercise 69.7 ★★

An access ramp must rise $1.2$ m with an angle of at most $5^\circ$ to the horizontal. What minimum length along the slope must the ramp have ( $\sin 5^\circ \approx 0.0872$ )? Round up to the nearest tenth of a meter.

Solution

Solution of Exercise 69.7.

The rise ( $1.2$ m) is opposite the angle; the ramp length $L$ is the hypotenuse:

\sin 5^\circ = \frac{1.2}{L}, \qquad L = \frac{1.2}{\sin 5^\circ} \approx \frac{1.2}{0.0872} \approx 13.8 \text{ m}.

The ramp must be at least $13.8$ m long.

Exercise 69.8 ★★

Using Proposition 69.4: an acute angle $\theta$ satisfies $\cos\theta = 0.6$ . Compute $\sin\theta$ without finding $\theta$ , then $\tan\theta$ .

Solution

Solution of Exercise 69.8.

From $\cos^2\theta + \sin^2\theta = 1$ : $\sin^2\theta = 1 - 0.36 = 0.64$ , and $\sin\theta > 0$ for an acute angle, so $\sin\theta = 0.8$ . Then $\tan\theta = \dfrac{\sin\theta}{\cos\theta} = \dfrac{0.8}{0.6} = \dfrac43$ .

Exercise 69.9 ★★★

Justify the exact values of Example 69.9: in the right isosceles triangle with legs $1$ , compute the hypotenuse; in the equilateral triangle of side $1$ , compute the height, and deduce the cosine and sine of $30^\circ$ and $60^\circ$ .

Solution

Solution of Exercise 69.9.

Right isosceles triangle with legs $1$ : hypotenuse $\sqrt{1 + 1} = \sqrt2$ ; each acute angle is $45^\circ$ , and

\cos 45^\circ = \frac{1}{\sqrt2} = \frac{\sqrt2}{2} = \sin 45^\circ, \qquad \tan 45^\circ = \frac11 = 1 .

Equilateral triangle of side $1$ : its height splits it into two right triangles with hypotenuse $1$ , base $\frac12$ , and height $h = \sqrt{1 - \frac14} = \frac{\sqrt3}{2}$ (Pythagoras). The angles are $60^\circ$ (at the base) and $30^\circ$ (at the top). Reading the ratios:

\cos 60^\circ = \frac{1/2}{1} = \frac12, \quad \sin 60^\circ = \frac{\sqrt3/2}{1} = \frac{\sqrt3}{2}, \quad \cos 30^\circ = \frac{\sqrt3}{2}, \quad \sin 30^\circ = \frac12 .

69.4 Problem: Measuring what you cannot reach

Problem 69.1

Weekend problem — the surveyor’s two-station method for inaccessible heights, road grades, the distance to the horizon, and the tangent’s race to infinity

No tape measure reaches the top of a mountain, the summit of a tower across a river, or the horizon at sea — but an angle gauge and this chapter’s three ratios do. The centerpiece of this problem is the surveyors’ classic two-station method, which computes a height without ever approaching its base; around it: road signs, ladders, stairs, the curvature of the Earth, and a first glimpse of a function exploding to infinity.

Part I — One station. (Values: $\tan 35^\circ \approx 0.700$ , $\sin 35^\circ \approx 0.574$ , $\cos 35^\circ \approx 0.819$ , $\tan 1^\circ \approx 0.0175$ .)

From a point $60$ m from the foot of a lighthouse, the top is seen at an elevation of $35^\circ$ (measured from the horizontal, at eye level). How high is the top above eye level?
A road sign announces a $12\,\%$ grade: the road rises $12$ m per $100$ m of horizontal run. What angle does the road make with the horizontal? And what grade percentage would a $45^\circ$ road have?
The two acute angles of a right triangle are complementary, and each one’s opposite side is the other’s adjacent side. Deduce the “co” identities:

$\sin(90^\circ - \theta) = \cos\theta, \qquad \cos(90^\circ - \theta) = \sin\theta .$
An $8$ m ladder leans at $60^\circ$ to the ground. Using the exact values of Example 69.9, give the height reached and the distance of the foot from the wall — exactly, then to the centimeter.
Verify $\cos^2\theta + \sin^2\theta = 1$ numerically for $\theta = 35^\circ$ and exactly for $\theta = 60^\circ$ . Why is this identity (Proposition 69.4) a good habit for checking calculator work?

Part II — Two stations. A mountain’s summit is visible, its base unreachable. From a point $A$ , the summit’s elevation is $30^\circ$ ; walking $200$ m straight toward the mountain, to $B$ , the elevation becomes $40^\circ$ . Write $h$ for the summit’s height above eye level and $x$ for the horizontal distance from $B$ to the vertical of the summit. ( $\tan 30^\circ \approx 0.577$ , $\tan 40^\circ \approx 0.839$ .)

Express $\tan 40^\circ$ and $\tan 30^\circ$ using $h$ , $x$ and $x + 200$ : two equations for two unknowns.
From each equation express the horizontal distance in terms of $h$ , subtract, and derive the two-station formula:

$h = \frac{200} {\ \dfrac{1}{\tan 30^\circ} - \dfrac{1}{\tan 40^\circ}\ } .$

Compute $h$ to the nearest ten meters.
Compute $x$ as well, and sanity-check your two answers against the $30^\circ$ sighting from $A$ .
In one sentence: why is the two-station method indispensable — what single measurement, impossible here, would the one-station method of question 1 have required?
A tower is sighted at $30^\circ$ , and at $60^\circ$ after walking $200$ m toward it. Using $\tan 30^\circ = \frac{1}{\sqrt3}$ and $\tan 60^\circ = \sqrt3$ , compute the height exactly — the answer is a multiple of $\sqrt3$ — then to the meter.

Part III — To the horizon and up the wall.

How far is the horizon? Standing with eyes $h$ meters above a spherical Earth of radius $R$ , your line of sight grazes the sphere: the sight line, the radius to the grazing point and the radius to your feet form a right triangle. Show with Pythagoras (Theorem 69.1) that the distance $d$ to the horizon satisfies $d^2 = 2Rh + h^2 \approx 2Rh$ , and compute $d$ for eyes at $1.80$ m ( $R = 6.37 \times 10^6$ m).
Same question from the top of a $324$ m tower. (Sailors’ rule of thumb: the horizon in kilometers is about $3.6\sqrt{h}$ with $h$ in meters — check your two answers against it.)
Your thumb, about $2$ cm wide, held at arm’s length, about $60$ cm from your eye: what angle does it cover? The full Moon covers about $0.5^\circ$ : how much of your thumb hides the whole Moon?
Comfortable stairs have risers of $17$ cm and treads of $29$ cm. What angle do they climb? Compare with the $30^\circ$ of Example 69.9, and with a steep attic ladder at $60^\circ$ : what riser would it need on a $29$ cm tread?
The tangent’s race: compute (calculator) $\tan 10^\circ$ , $\tan 45^\circ$ , $\tan 80^\circ$ , $\tan 89^\circ$ . What happens as the angle approaches $90^\circ$ , and why (think adjacent side)? The wall is vertical, the ratio has no value, and the graph shoots to infinity — your first asymptote, a creature studied closely in the High School volume.

Solution

Solution of Problem 69.1.

1. The height above eye level is the side opposite $35^\circ$ , with adjacent side $60$ m: $h = 60 \tan 35^\circ \approx 60 \times 0.700 = 42$ m.

2. $\tan\theta = 0.12$ gives $\theta = \tan^{-1}(0.12) \approx 6.8^\circ$ — steep for a road, yet a modest angle. A $45^\circ$ road has $\tan 45^\circ = 1$ : a $100\,\%$ grade. (“One hundred percent” is far from vertical — a favorite ski-lift misunderstanding.)

3. In a right triangle with acute angles $\theta$ and $90^\circ - \theta$ , the side opposite one is adjacent to the other and the hypotenuse is shared. So $\sin(90^\circ - \theta) = \frac{\text{its opposite}}{h} = \frac{\text{adjacent of }\theta}{h} = \cos\theta$ , and symmetrically $\cos(90^\circ - \theta) = \sin\theta$ : the cosine is the sine of the complement.

4. Height: $8 \sin 60^\circ = 8 \times \frac{\sqrt3}{2} = 4\sqrt3 \approx 6.93$ m. Foot: $8 \cos 60^\circ = 4$ m exactly.

5. $0.574^2 + 0.819^2 = 0.329 + 0.671 = 1.000$ (to rounding); and $\left(\frac12\right)^2 + \left(\frac{\sqrt3}{2}\right)^2 = \frac14 + \frac34 = 1$ . Any pair of values failing the identity betrays a wrong reading or a calculator in the wrong mode — a free error detector.

6. $\tan 40^\circ = \dfrac hx$ and $\tan 30^\circ = \dfrac{h}{x + 200}$ .

7. $x = \dfrac{h}{\tan 40^\circ}$ and $x + 200 = \dfrac{h}{\tan 30^\circ}$ ; subtracting, $200 = h\left(\frac{1}{\tan 30^\circ} - \frac{1}{\tan 40^\circ}\right)$ , whence the formula. Numbers: $\frac{1}{0.577} \approx 1.732$ , $\frac{1}{0.839} \approx 1.192$ , difference $0.540$ : $h \approx \frac{200}{0.540} \approx 370$ m.

8. $x = \frac{h}{\tan 40^\circ} \approx \frac{370}{0.839} \approx 441$ m. Check from $A$ : $\frac{h}{x + 200} \approx \frac{370}{641} \approx 0.577 = \tan 30^\circ$ : consistent.

9. The one-station method needs the horizontal distance to the point directly below the summit — unmeasurable across a gorge or inside a mountain. The two-station method replaces it by a distance you can pace out yourself: the $200$ m between your own two viewpoints.

10. $h = \dfrac{200}{\sqrt3 - \frac{1}{\sqrt3}} = \dfrac{200}{\frac{2}{\sqrt3}} = 100\sqrt3 \approx 173$ m — the Eiffel Tower’s second floor, measured with two sightings and a stroll.

11. Sight line $d$ , radii $R$ (to the grazing point, perpendicular to the sight line) and $R + h$ (to the eye): Pythagoras gives $d^2 + R^2 = (R + h)^2 = R^2 + 2Rh + h^2$ , so $d^2 = 2Rh + h^2 \approx 2Rh$ ( $h$ is minuscule against $R$ ). Eyes at $1.80$ m: $d \approx \sqrt{2 \times 6.37 \times 10^6 \times 1.8} \approx 4\,800$ m — the sea horizon is barely five kilometers away.

12. $d \approx \sqrt{2 \times 6.37 \times 10^6 \times 324} \approx 64$ km. Sailors’ rule: $3.6\sqrt{1.8} \approx 4.8$ km and $3.6\sqrt{324} = 64.8$ km — both match.

13. $\tan\theta = \frac{2}{60}$ gives $\theta \approx 1.9^\circ$ . The Moon’s $0.5^\circ$ is about a quarter of a thumb: the Moon looks enormous and is hidden by a fingernail — the eye is a poor protractor.

14. Stairs: $\tan^{-1}\frac{17}{29} \approx 30^\circ$ — comfortable stairs climb at almost exactly the $30^\circ$ of the special values. At $60^\circ$ on a $29$ cm tread, the riser would be $29\tan 60^\circ = 29\sqrt3 \approx 50$ cm: a ladder, not a staircase.

15. $\tan 10^\circ \approx 0.18$ ; $\tan 45^\circ = 1$ ; $\tan 80^\circ \approx 5.7$ ; $\tan 89^\circ \approx 57.3$ . As $\theta \to 90^\circ$ the adjacent side shrinks to nothing while the opposite side holds: the quotient grows beyond every bound. At $90^\circ$ exactly there is no triangle and no value — the graph of the tangent climbs a vertical asymptote, first of many in the High School volume.