---
title: "Trigonometry in the Right Triangle"
book: "Primary & Middle School Mathematics"
subject: math
language: en
chapter: 69
exercises: 9
source: https://one-course.com/books/math/1/en/chapter/69-trigonometry-in-the-right-triangle
---

# Chapter 69 — Trigonometry in the Right Triangle

In a [right triangle](https://one-course.com/books/math/1/en/chapter/41-triangles-and-quadrilaterals#def-g6-shapes-triangles), 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](#def-g9-trig-ratios), [sine](#def-g9-trig-ratios) and [tangent](#def-g9-trig-ratios). Together with the Pythagorean theorem, they let you compute every side and every angle of a [right triangle](https://one-course.com/books/math/1/en/chapter/41-triangles-and-quadrilaterals#def-g6-shapes-triangles) from very little information.

## 69.1 The Pythagorean theorem, again

**Theorem 69.1 (Pythagoras).**

In a triangle $ABC$ [right-angled](https://one-course.com/books/math/1/en/chapter/41-triangles-and-quadrilaterals#def-g6-shapes-triangles) at $A$, the square of the hypotenuse is the [sum](https://one-course.com/books/math/1/en/chapter/2-addition-first-steps#def-g1-addition-def) 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](https://one-course.com/books/math/1/en/chapter/41-triangles-and-quadrilaterals#def-g6-shapes-triangles) at $A$.

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

**Example 69.2.**

A [right triangle](https://one-course.com/books/math/1/en/chapter/41-triangles-and-quadrilaterals#def-g6-shapes-triangles) 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](https://one-course.com/books/math/1/en/chapter/41-triangles-and-quadrilaterals#def-g6-shapes-triangles), relative to an acute angle $\theta$, the three sides have names: the *hypotenuse* (opposite the [right angle](https://one-course.com/books/math/1/en/chapter/18-shapes-and-right-angles#def-g3-shapes-rightangle), 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](https://one-course.com/books/math/1/en/chapter/41-triangles-and-quadrilaterals#def-g6-shapes-triangles):

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

These ratios depend only on $\theta$, not on the size of the triangle (all [right triangles](https://one-course.com/books/math/1/en/chapter/41-triangles-and-quadrilaterals#def-g6-shapes-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).](https://one-course.com/images/onecourse/chapters/math-1/g9-trig/fig-b3579ad88883.svg)

*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 $\theta$:

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

**Proof.** Write $a$ for the adjacent side, $o$ for the opposite side, $h$ for the hypotenuse of a [right triangle](https://one-course.com/books/math/1/en/chapter/41-triangles-and-quadrilaterals#def-g6-shapes-triangles) 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:

1. mark the known angle and label the three sides (hypotenuse, opposite, adjacent) *from that angle* ;
2. choose the ratio (cos, sin or tan) that involves the known side and the wanted side;
3. write the [equation](https://one-course.com/books/math/1/en/chapter/57-literal-calculation-and-equations#def-g8-equations-def) and solve it;
4. sanity-check: the hypotenuse must come out longest.

**Example 69.6.**

In a [right triangle](https://one-course.com/books/math/1/en/chapter/41-triangles-and-quadrilaterals#def-g6-shapes-triangles), 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](#def-g9-trig-ratios):

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

For the adjacent side, use the [cosine](#def-g9-trig-ratios): $8\cos 35^\circ \approx 8 \times 0.819 \approx 6.6$. 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](#def-g9-trig-ratios), [sine](#def-g9-trig-ratios) or [tangent](#def-g9-trig-ratios) of the unknown angle, and apply the calculator’s [inverse](https://one-course.com/books/math/1/en/chapter/55-fractions-all-four-operations#def-g8-fractions-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.](https://one-course.com/images/onecourse/chapters/math-1/g9-trig/fig-6daa11ccbd02.svg)

*The ladder problem: the adjacent side ($1.4$ m) and the hypotenuse ($5$ m) are known, so the [cosine](#def-g9-trig-ratios) gives the angle.*

**Example 69.9 (Special angles).**

Two triangles give exact values. [Half](https://one-course.com/books/math/1/en/chapter/17-sharing-and-division#def-g3-division-half) a square of side 1 (a right [isosceles triangle](https://one-course.com/books/math/1/en/chapter/41-triangles-and-quadrilaterals#def-g6-shapes-triangles)) 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](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 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](https://one-course.com/books/math/1/en/chapter/41-triangles-and-quadrilaterals#def-g6-shapes-triangles) has legs $9$ and $12$. Compute its hypotenuse. Another has hypotenuse $17$ and one leg $8$: compute the other leg.

**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](https://one-course.com/books/math/1/en/chapter/41-triangles-and-quadrilaterals#def-g6-shapes-triangles)? Same question for sides $5$, $6$ and $8$.

**Solution of Exercise 69.2.**

$7^2 + 24^2 = 49 + 576 = 625 = 25^2$: yes, [right-angled](https://one-course.com/books/math/1/en/chapter/41-triangles-and-quadrilaterals#def-g6-shapes-triangles) (converse of Pythagoras), with the [right angle](https://one-course.com/books/math/1/en/chapter/18-shapes-and-right-angles#def-g3-shapes-rightangle) opposite the side $25$.

$5^2 + 6^2 = 61 \neq 64 = 8^2$: not [right-angled](https://one-course.com/books/math/1/en/chapter/41-triangles-and-quadrilaterals#def-g6-shapes-triangles).

**Exercise 69.3 ★.**

In a triangle $DEF$ [right-angled](https://one-course.com/books/math/1/en/chapter/41-triangles-and-quadrilaterals#def-g6-shapes-triangles) at $D$, 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 of Exercise 69.3.**

The hypotenuse is $[EF]$ (opposite the [right angle](https://one-course.com/books/math/1/en/chapter/18-shapes-and-right-angles#def-g3-shapes-rightangle) 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](https://one-course.com/books/math/1/en/chapter/41-triangles-and-quadrilaterals#def-g6-shapes-triangles), the hypotenuse measures $10$ and one acute angle is $28^\circ$. Compute the two legs ([round](https://one-course.com/books/math/1/en/chapter/21-large-numbers#def-g4-numbers-round) to the tenth; $\cos 28^\circ \approx 0.883$, $\sin 28^\circ \approx 0.469$).

**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](https://one-course.com/books/math/1/en/chapter/41-triangles-and-quadrilaterals#def-g6-shapes-triangles), the leg adjacent to the angle $\theta$ measures $6$ and the opposite leg measures $4.5$. Compute $\tan\theta$, then $\theta$ ([round](https://one-course.com/books/math/1/en/chapter/21-large-numbers#def-g4-numbers-round) to the degree; $\tan^{-1}(0.75) \approx 36.9^\circ$).

**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 of Exercise 69.6.**

In the [right triangle](https://one-course.com/books/math/1/en/chapter/41-triangles-and-quadrilaterals#def-g6-shapes-triangles) 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](https://one-course.com/books/math/1/en/chapter/67-linear-and-affine-functions#prop-g9-linfunc-affinegraph) must the ramp have ($\sin 5^\circ \approx 0.0872$)? Round up to the nearest tenth of a meter.

**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](#prop-g9-trig-properties): an acute angle $\theta$ satisfies $\cos\theta = 0.6$. Compute $\sin\theta$ without finding $\theta$, then $\tan\theta$.

**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](#ex-g9-trig-special): in the right [isosceles triangle](https://one-course.com/books/math/1/en/chapter/41-triangles-and-quadrilaterals#def-g6-shapes-triangles) with legs $1$, compute the hypotenuse; in the [equilateral triangle](https://one-course.com/books/math/1/en/chapter/41-triangles-and-quadrilaterals#def-g6-shapes-triangles) of side $1$, compute the height, and deduce the [cosine](#def-g9-trig-ratios) and [sine](#def-g9-trig-ratios) of $30^\circ$ and $60^\circ$.

**Solution of Exercise 69.9.**

*Right [isosceles triangle](https://one-course.com/books/math/1/en/chapter/41-triangles-and-quadrilaterals#def-g6-shapes-triangles)* 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](https://one-course.com/books/math/1/en/chapter/41-triangles-and-quadrilaterals#def-g6-shapes-triangles)* of side $1$: its height splits it into two [right triangles](https://one-course.com/books/math/1/en/chapter/41-triangles-and-quadrilaterals#def-g6-shapes-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: $\tan 35^\circ \approx 0.700$, $\sin 35^\circ \approx 0.574$, $\cos 35^\circ \approx 0.819$, $\tan 1^\circ \approx 0.0175$.)

1. 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?
2. 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?
3. The two acute angles of a [right triangle](https://one-course.com/books/math/1/en/chapter/41-triangles-and-quadrilaterals#def-g6-shapes-triangles) are complementary, and each one’s opposite side is the other’s adjacent side. Deduce the “co” identities: $$\sin(90^\circ - \theta) = \cos\theta,  \qquad  \cos(90^\circ - \theta) = \sin\theta .$$
4. An $8$ m ladder leans at $60^\circ$ to the ground. Using the exact values of [Example 69.9](#ex-g9-trig-special) , give the height reached and the distance of the foot from the wall — exactly, then to the centimeter.
5. 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](#prop-g9-trig-properties) ) 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. ($\tan 30^\circ \approx 0.577$, $\tan 40^\circ \approx 0.839$.)

6. Express $\tan 40^\circ$ and $\tan 30^\circ$ using $h$ , $x$ and $x + 200$ : two [equations](https://one-course.com/books/math/1/en/chapter/57-literal-calculation-and-equations#def-g8-equations-def) for two unknowns.
7. From each [equation](https://one-course.com/books/math/1/en/chapter/57-literal-calculation-and-equations#def-g8-equations-def) 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.
8. Compute $x$ as well, and sanity-check your two answers against the $30^\circ$ sighting from $A$ .
9. 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?
10. 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](https://one-course.com/books/math/1/en/chapter/64-arithmetic-divisors-and-prime-numbers#def-g9-arith-divisor) of $\sqrt3$ — then to the meter.

**Part III — To the horizon and up the wall.**

11. How far is the horizon? Standing with eyes $h$ meters above a spherical Earth of [radius](https://one-course.com/books/math/1/en/chapter/18-shapes-and-right-angles#def-g3-shapes-shapes) $R$ , your line of sight grazes the sphere: the sight line, the [radius](https://one-course.com/books/math/1/en/chapter/18-shapes-and-right-angles#def-g3-shapes-shapes) to the grazing point and the [radius](https://one-course.com/books/math/1/en/chapter/18-shapes-and-right-angles#def-g3-shapes-shapes) to your feet form a [right triangle](https://one-course.com/books/math/1/en/chapter/41-triangles-and-quadrilaterals#def-g6-shapes-triangles) . Show with Pythagoras ( [Theorem 69.1](#thm-g9-trig-pythagoras) ) 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).
12. 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.)
13. 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?
14. 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](#ex-g9-trig-special) , and with a steep attic ladder at $60^\circ$ : what riser would *it* need on a $29$ cm tread?
15. The [tangent](#def-g9-trig-ratios) ’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 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](https://one-course.com/books/math/1/en/chapter/41-triangles-and-quadrilaterals#def-g6-shapes-triangles) 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 *co*[sine](#def-g9-trig-ratios) is the [sine](#def-g9-trig-ratios) of the *co*mplement.

**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](https://one-course.com/books/math/1/en/chapter/38-decimal-numbers#def-g6-decimals-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](https://one-course.com/books/math/1/en/chapter/3-subtraction-first-steps#ex-g1-subtraction-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](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-perp) 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](https://one-course.com/books/math/1/en/chapter/17-sharing-and-division#def-g3-division-half)* 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](https://one-course.com/books/math/1/en/chapter/17-sharing-and-division#def-g3-division-remainder) grows beyond every bound. At $90^\circ$ exactly there is no triangle and no value — the graph of the [tangent](#def-g9-trig-ratios) climbs a vertical asymptote, first of many in the High School volume.