Trigonometry

The last chapter ended on Everest: its height was calculated in 1856, almost a hundred years before anyone first climbed to the summit. The surveyors didn’t go up to the mountain, drop a rope or set up a staff. They measured only angles, and got $8840$ m, only nine metres less than today’s measurements.

In this chapter we’ll travel the same road ourselves, as a surveying expedition. The kit is modest: a pole, a tape measure and a theodolite, an instrument for reading off the angle between two directions. First the pole and its shadow will teach us to turn angles into numbers. Then, without crossing over, we’ll measure the width of a river, the height of a mountain, and finally the size of the whole Earth, as Eratosthenes did more than two thousand years ago.

The pole and the shadow

The first camp is a flat spot in the sun. Stick a pole in the ground and measure its shadow. There’s a tree nearby; you can’t reach its top, but its shadow is on the ground and can be measured with the tape. In chapter 17 Thales measured a pyramid this way: sunlight falls in parallel rays, so the pole with its shadow and the tree with its shadow form similar right triangles, and the tree is as many times taller than the pole as its shadow is longer.

Let’s note the key thing in this argument. The ratio “height : shadow” is the same for all objects at the same moment, and it depends only on how high the Sun is, that is, on the angle of the rays. So this ratio can be worked out once for each angle and written in a table. After that the pole isn’t needed: it’s enough to measure the angle.

All right triangles with the same acute angle are similar to one another.

Set up several poles of different heights along one ray coming out of the point $A$ at the angle $\alpha$ to the ground: the top of each pole lies on the ray. Each pole, together with its stretch of ground and its stretch of ray, forms a right triangle. All these triangles have the angle $\alpha$ at $A$ and a right angle at the foot of the pole. Two angles of one triangle equal two angles of another, so by the similarity test (chapter 17) the triangles are similar. Similar triangles have proportional corresponding sides: if all the sides of the second are $k$ times those of the first, then the ratio of pole to hypotenuse in the second is $\frac{k \cdot a}{k \cdot c} = \frac ac$. The numbers in the picture confirm it. The ratio of the opposite leg to the hypotenuse is one number for all triangles with the angle $\alpha$. It depends only on the angle. The same argument works for any pair of sides.

Since the ratios depend only on the angle, they can be regarded as properties of the angle itself and given names.

The leg opposite the angle $\alpha$: the one that doesn’t touch the angle’s vertex. For the pole, the pole itself. The leg adjacent to the angle: the angle’s other arm, apart from the hypotenuse. For the pole, the shadow. The hypotenuse is the longest side, so the sine and cosine of an acute angle are less than one. The tangent can be any positive number. Example: in the Egyptian triangle $3$, $4$, $5$ the smaller acute angle has $\sin\alpha = \frac35 = 0.6$, $\cos\alpha = \frac45 = 0.8$, $\tan\alpha = \frac34 = 0.75$; the angle itself is about $36.87^\circ$. In some books the tangent is written $\operatorname{tg}$ and the cotangent $\operatorname{ctg}$.

Now the tree can be measured without its shadow. Walk $30$ m away from it and sight its top at an angle of $35^\circ$ to the horizontal. The tangent of $35^\circ$ from a table (or a calculator) is about $0.700$, so the top is $30 \cdot 0.700 \approx 21.0$ m above eye level. Add the height of the theodolite above the ground, say $1.6$ m, and the tree is about $22.6$ m tall.

A ladder $5$ m long stands at an angle of $70^\circ$ to the ground. At what height does its top touch the wall? Take $\sin 70^\circ \approx 0.94$.

The ladder is the hypotenuse, the height is the leg opposite the $70^\circ$ angle. So $h = 5 \cdot \sin 70^\circ \approx 5 \cdot 0.94 = 4.7$ m.

Three angles to know by heart

For most angles the sine is an irrational number taken from tables. But for $30^\circ$, $45^\circ$ and $60^\circ$ the values can be found exactly, from two familiar figures: half a square and half an equilateral triangle.

We need right triangles with these angles whose sides we know exactly.

Cut a square with side $1$ along a diagonal. We get a right triangle with legs $1$ and $1$; its acute angles are equal (the triangle is isosceles) and add up to $90^\circ$, so each is $45^\circ$. The hypotenuse, by Pythagoras’ theorem, is $\sqrt{1 + 1} = \sqrt2$. An equilateral triangle with side $2$ has all its angles $60^\circ$. The altitude divides it into two right triangles: hypotenuse $2$, the shorter leg half the side, $1$, and the angle opposite it half of $60^\circ$, that is, $30^\circ$. The other leg is the altitude: $\sqrt{4 - 1} = \sqrt3$. Divide the sides. In the first triangle $\sin 45^\circ = \cos 45^\circ = \frac{1}{\sqrt2} = \frac{\sqrt2}{2}$. In the second, the leg $1$ is opposite the $30^\circ$ angle, and the leg $\sqrt3$ opposite the $60^\circ$: $\sin 30^\circ = \frac12$, $\cos 30^\circ = \frac{\sqrt3}{2}$, $\sin 60^\circ = \frac{\sqrt3}{2}$, $\cos 60^\circ = \frac12$. We’ve got all the values in the theorem. The tangents are their quotients: $\tan 30^\circ = \frac{1}{\sqrt3} = \frac{\sqrt3}{3}$, $\tan 45^\circ = 1$, $\tan 60^\circ = \sqrt3$.

The columns for $0^\circ$ and $90^\circ$ are limiting cases: if an acute angle is squeezed towards zero, the opposite leg vanishes and the adjacent one becomes equal to the hypotenuse. The unit circle of the next section gives them a precise meaning. The row of sines is easy to remember as $\frac{\sqrt0}{2}, \frac{\sqrt1}{2}, \frac{\sqrt2}{2}, \frac{\sqrt3}{2}, \frac{\sqrt4}{2}$, and the cosines are the same row backwards.

The angle $60^\circ$ is twice the angle $30^\circ$. How many times larger is $\sin 60^\circ$ than $\sin 30^\circ$?

Exactly twice $\sqrt3 \approx 1.73$ times $1.5$ times They’re equal

$\sin 60^\circ : \sin 30^\circ = \frac{\sqrt3}{2} : \frac12 = \sqrt3$. The sine grows more slowly than the angle, and slows right down towards $90^\circ$: from $60^\circ$ to $90^\circ$ it gains only $1 - 0.866 \approx 0.13$.

The hypotenuse of a right triangle is $10$, and one of its acute angles is $30^\circ$. Find the leg adjacent to this angle. You can write the answer with a root, for example 5sqrt(3).

The adjacent leg is $b = c\cos 30^\circ = 10 \cdot \frac{\sqrt3}{2} = 5\sqrt3 \approx 8.66$. The opposite one is $10 \cdot \frac12 = 5$, and by Pythagoras’ theorem $25 + 75 = 100$.

Angles larger than a right angle

A triangle with acute angles is a good camp, but the expedition will have to go further. By a river or a mountain the triangle of observations can be obtuse, and we’ll need the cosine of $120^\circ$. But there’s no right triangle with a $120^\circ$ angle. A wheel, a clock hand, the Earth on its axis turn through angles larger than $180^\circ$ and even larger than $360^\circ$. We need a definition that works for any angle.

For an acute angle the new definition agrees with the old one. Drop a perpendicular from $P_\alpha$ to the $x$-axis: we get a right triangle with hypotenuse $1$ whose legs are exactly the coordinates of the point. The sine “leg divided by hypotenuse” becomes “leg divided by one”, that is, the $y$-coordinate. But now the angle can be anything, and the coordinates can be negative: for the angle $120^\circ$ the point $P$ lies to the left of the $y$-axis, and $\cos 120^\circ = -\frac12$.

Here it’s handy to measure angles in radians: the radian measure of an angle is the length of the arc of the unit circle from the point $(1, 0)$ to $P_\alpha$. A full turn is $2\pi$, a right angle $\frac\pi2$. The widget below shows it all at once: the point on the circle, its coordinates, and the graphs you get by plotting the arc travelled horizontally and the sine, cosine or tangent vertically.

There’s nothing special about the sine wave: it’s the shadow of a point running round a circle, spread out over time. The cosine is the same shadow seen from the side, so its graph is a sine wave shifted by a quarter turn. The tangent is a segment on the tangent line to the circle at the point $(1, 0)$, from that point to the ray $OP$; when the ray becomes vertical, the segment runs off to infinity.

It’s Pythagoras’ theorem for a triangle hidden in the unit circle.

The point $P_\alpha$ has coordinates $(\cos\alpha, \sin\alpha)$ and lies on the circle of radius $1$. Drop a perpendicular from it to the $x$-axis. We get a right triangle with a vertex at the centre: its legs are $|\cos\alpha|$ and $|\sin\alpha|$ (lengths are never negative), and its hypotenuse is the radius, $1$. If $P$ lies on an axis, the triangle degenerates, but the equation below can be checked directly: for example, $\cos^2 0 + \sin^2 0 = 1 + 0$. By Pythagoras’ theorem $|\cos\alpha|^2 + |\sin\alpha|^2 = 1^2$. The square of an absolute value equals the square of the number itself, so the signs don’t get in the way. $\sin^2\alpha + \cos^2\alpha = 1$ in every quadrant. Move the point: the triangle changes, the sum of squares doesn’t.

The identity lets you find one value from another. If $\sin\alpha = \frac35$ and the angle is acute, then $\cos\alpha = \sqrt{1 - \frac{9}{25}} = \frac45$. If the angle is obtuse, the point $P$ is to the left of the $y$-axis, and $\cos\alpha = -\frac45$: the sign tells you which quadrant the point is in.

The circle is symmetric, and each of its symmetries turns into a formula. Here are three the expedition will need.

Each formula is a reflection of the point $P_\alpha$ in one of three lines.

Reflect $P_\alpha$ in the $y$-axis. A reflection preserves distances, so the image also lies on the circle, and the angle from the ray $Ox$ to it is $180^\circ - \alpha$: the angle the ray $OP_\alpha$ makes with the axis $Ox$ becomes, after the reflection, the same angle with the opposite ray. The $y$-coordinate is kept and the $x$-coordinate changes sign: $\sin(180^\circ - \alpha) = \sin\alpha$, $\cos(180^\circ - \alpha) = -\cos\alpha$. Reflect $P_\alpha$ in the $x$-axis: we get the point rotated by $-\alpha$, the same angle but clockwise. The $x$-coordinate is kept and the $y$-coordinate changes sign: $\cos(-\alpha) = \cos\alpha$, $\sin(-\alpha) = -\sin\alpha$. Reflect $P_\alpha$ in the line $y = x$, which bisects the first quadrant. This reflection swaps the coordinates, and turns the angle from $Ox$ into the angle from $Oy$, that is, into $90^\circ - \alpha$ from $Ox$. So $\cos(90^\circ - \alpha) = \sin\alpha$ and $\sin(90^\circ - \alpha) = \cos\alpha$. All six formulas are proved for every position of the point: move $P$ all the way round the circle.

The third pair explains the word “cosine” itself: it’s the sine of the complementary angle, complementi sinus, the one that completes $\alpha$ to a right angle. And the first pair will come in handy by the river: an obtuse angle has the same sine as the acute angle adjacent to it.

One more consequence can be seen in the widget if you let the point run. After a full turn the point comes back to where it started, so $\sin(\alpha + 2\pi) = \sin\alpha$ and $\cos(\alpha + 2\pi) = \cos\alpha$, and the graph repeats.

Which is larger: $\sin 1$ or $\sin 2$? The angles are in radians.

$\sin 2$ $\sin 1$ They’re equal $\sin 2$ is negative

$1$ radian is about $57.3^\circ$, and $2$ radians about $114.6^\circ$. The sine of the second angle equals the sine of the angle adjacent to it, $180^\circ - 114.6^\circ = 65.4^\circ$, and that angle is closer to $90^\circ$ than $57.3^\circ$ is. So $\sin 2 \approx 0.909 > \sin 1 \approx 0.841$.

Two rotations in a row

The table knows $30^\circ$ and $45^\circ$, but what about $75^\circ = 45^\circ + 30^\circ$? You can’t add the sines: otherwise $\sin 90^\circ$ would be $\sin 45^\circ + \sin 45^\circ \approx 1.41$, and a sine is never more than one. The right answer comes from the same circle: a rotation by $\alpha + \beta$ is a rotation by $\beta$ followed by another by $\alpha$.

The trick is to rotate not one point but the whole picture, axes included.

Take the unit arrows $e_1$ and $e_2$ along the axes and the point $P_\beta$ on the circle. To get to it from the centre you can go $\cos\beta$ along $e_1$ and then $\sin\beta$ along $e_2$: those are its coordinates (if one of them is negative, go the opposite way). Rotate the whole picture by the angle $\alpha$. The point $P_\beta$ goes to $P_{\alpha + \beta}$: two rotations in a row add up. The arrow $e_1$ goes to $e_1' = P_\alpha = (\cos\alpha, \sin\alpha)$. The arrow $e_2 = P_{90^\circ}$ goes to $P_{90^\circ + \alpha}$, and its coordinates come from the reduction formulas applied twice: $\cos(90^\circ + \alpha) = \sin(-\alpha) = -\sin\alpha$ and $\sin(90^\circ + \alpha) = \cos(-\alpha) = \cos\alpha$. So $e_2' = (-\sin\alpha, \cos\alpha)$. A rotation preserves lengths and angles, so the route to the new point is again $\cos\beta$ along $e_1'$ and $\sin\beta$ along $e_2'$. Compute the $x$-coordinate of the end of the route. The stretch of length $\cos\beta$ along $e_1'$ moves us horizontally by $\cos\beta \cdot \cos\alpha$, the stretch of length $\sin\beta$ along $e_2'$ by $\sin\beta \cdot (-\sin\alpha)$. The sum is the $x$-coordinate of the point $P_{\alpha + \beta}$, that is, $\cos(\alpha + \beta)$. Likewise the $y$-coordinate: $\cos\beta \cdot \sin\alpha + \sin\beta \cdot \cos\alpha$. That’s $\sin(\alpha + \beta)$. $\cos(\alpha + \beta) = \cos\alpha\cos\beta - \sin\alpha\sin\beta$, $\ \sin(\alpha + \beta) = \sin\alpha\cos\beta + \cos\alpha\sin\beta$, for any angles, obtuse and negative ones included.

In the chapter on complex numbers the same formulas came out of multiplication, and that’s no coincidence: multiplying by a complex number of modulus $1$ is exactly a rotation.

The first angle. The second angle. Substituting $-\beta$ and using the reduction formulas gives the sine of a difference: $\sin(\alpha - \beta) = \sin\alpha\cos\beta - \cos\alpha\sin\beta$. Example: $\sin 75^\circ = \sin 45^\circ\cos 30^\circ + \cos 45^\circ\sin 30^\circ = \frac{\sqrt2}{2} \cdot \frac{\sqrt3}{2} + \frac{\sqrt2}{2} \cdot \frac12 = \frac{\sqrt6 + \sqrt2}{4} \approx 0.9659$.

With $\beta = \alpha$ the formulas become the double-angle formulas: $\sin 2\alpha = 2\sin\alpha\cos\alpha$ and $\cos 2\alpha = \cos^2\alpha - \sin^2\alpha$. Their proof is a single substitution into the addition formulas already proved.

Find $\cos 15^\circ$ exactly by writing $15^\circ = 45^\circ - 30^\circ$. A square root is written as sqrt(…).

$\cos(\alpha - \beta) = \cos\alpha\cos\beta + \sin\alpha\sin\beta$ (substitute $-\beta$ into the addition formula). So $\cos 15^\circ = \frac{\sqrt2}{2} \cdot \frac{\sqrt3}{2} + \frac{\sqrt2}{2} \cdot \frac12 = \frac{\sqrt6 + \sqrt2}{4} \approx 0.9659$, the same as $\sin 75^\circ$, and that’s no accident: $75^\circ = 90^\circ - 15^\circ$.

Across the river

The next camp is a riverbank. On the far bank stands a tree $T$, and we need the width of the river without crossing it. Measure a base line $AB$ along our bank with the tape, say $60$ m, and measure two angles with the theodolite: at $A$, between the base line and the direction to the tree, and at $B$. A side and the two angles on it determine a triangle uniquely: that’s the second congruence test (chapter 17). It remains to turn the angles into lengths.

The key is to find, for the side $a$, a right triangle in which it’s a leg and the hypotenuse is a diameter.

Draw the circumcircle of triangle $ABC$, with centre $O$ and radius $R$. The side $a = BC$ is a chord of it. Draw the diameter $BD$ through the centre. Its length is $2R$. The angles $BAC$ and $BDC$ are inscribed and stand on the same chord $BC$. By the inscribed angle theorem each is half the arc it stands on. If $A$ and $D$ lie on the same side of $BC$, the arc is the same, and $\angle BDC = \angle A$. If on opposite sides, the arcs add up to the whole circle, and $\angle BDC = 180^\circ - \angle A$. Either way the sines of these angles are equal, by the reduction formula. The angle $BCD$ stands on a diameter, so it’s a right angle (Thales’ theorem). In the right triangle $BCD$ the side $BC$ is the leg opposite the angle $D$, and the hypotenuse is $BD = 2R$. By the definition of the sine, $a = 2R\sin D = 2R\sin A$. (If $C$ coincides with $D$, then $a$ is the diameter itself, angle $A$ is right, and $a = 2R = 2R\sin 90^\circ$.) $\frac{a}{\sin A} = 2R$. For the sides $b$ and $c$ the argument is the same, only the diameter is drawn from another vertex. All three ratios equal the one number $2R$. Any side of the triangle. The angle opposite this side. An obtuse angle has the same sine as the acute angle adjacent to it. The diameter of the circumcircle, common to all three sides. Example by the river: $AB = 60$ m, $\angle A = 66^\circ$, $\angle B = 73^\circ$. The angle at the tree is $180^\circ - 66^\circ - 73^\circ = 41^\circ$, so $AT = \frac{60\sin 73^\circ}{\sin 41^\circ} \approx 87.5$ m, and the width of the river, the altitude of the triangle from $T$, is $AT\sin 66^\circ \approx 79.9$ m.

The widget explains the surveyors’ main rule. If the base line is short and the tree far away, the angles at $A$ and $B$ are close to $90^\circ$ and the angle at $T$ is tiny. Such a triangle is “thin”, and a theodolite error of half a degree turns the point into a long stretched-out band. Triangles close to equilateral are best: there the errors in the angles hardly swell at all.

The third side

There’s another kind of problem. From the camp the distances $a$ and $b$ to two landmarks have been measured, along with the angle $C$ between the directions to them. What’s the distance $c$ between the landmarks themselves? Two sides and the angle between them determine a triangle (the first congruence test), and the answer comes from a generalisation of Pythagoras’ theorem. In chapter 18 Euclid already derived it without cosines, with the correction $2bx$; now the correction gets a name.

Put the triangle on the coordinate plane so that the coordinates of the vertices are expressed through the cosine and sine.

Put the vertex $C$ at the origin and point the side $CA$ along the $x$-axis. Then $A = (b, 0)$. The point $B$ lies at distance $a$ from the origin, on the ray at the angle $C$ to the $x$-axis. If the picture is shrunk by a factor of $a$, $B$ lands on the unit circle at the point $P_C = (\cos C, \sin C)$. So $B = (a\cos C, a\sin C)$, and for an obtuse angle $C$ too: then $\cos C < 0$, and $B$ lies to the left of the $y$-axis. The distance between $A$ and $B$ is the hypotenuse of a right triangle whose legs are the differences of the coordinates: $|a\cos C - b|$ horizontally and $a\sin C$ vertically. By Pythagoras’ theorem $c^2 = (a\cos C - b)^2 + (a\sin C)^2$. Expand the brackets: $c^2 = a^2\cos^2 C - 2ab\cos C + b^2 + a^2\sin^2 C = a^2\left(\cos^2 C + \sin^2 C\right) + b^2 - 2ab\cos C$. By the Pythagorean identity the bracket is $1$. $c^2 = a^2 + b^2 - 2ab\cos C$. The number $a\cos C$ is exactly the $x$ of Euclid’s formula: the signed distance from $C$ to the foot of the altitude from $B$. One side at the angle $C$. The other side at the angle $C$. The side opposite the angle $C$: the one we’re looking for. The angle between $a$ and $b$. At $C = 90^\circ$ the cosine is zero and Pythagoras’ theorem is left; for an obtuse angle $\cos C < 0$, and the third side is longer than it would be for a right angle. Example: $a = 5$, $b = 8$, $C = 60^\circ$. Then $c^2 = 25 + 64 - 2 \cdot 5 \cdot 8 \cdot \frac12 = 49$, and $c = 7$.

The same picture gives the area too. In chapter 17 the area of a triangle was half the base times the height, and the altitude from vertex $B$ to side $CA$ is the $y$-coordinate of the point $B$, that is, $a\sin C$.

Take the side $CA = b$ as the base. The altitude from $B$ is the distance from $B$ to the line $CA$, that is, the absolute value of the $y$-coordinate of $B = (a\cos C, a\sin C)$. An angle of a triangle is less than $180^\circ$, so $\sin C > 0$ and the altitude equals $a\sin C$, even if angle $C$ is obtuse and the altitude falls beyond the point $C$. $S = \frac12 \cdot b \cdot a\sin C = \frac12 ab\sin C$.

Two sides of a triangle are $3$ and $5$, and the angle between them is $120^\circ$. Find the third side.

$\cos 120^\circ = -\frac12$, so $c^2 = 9 + 25 - 2 \cdot 3 \cdot 5 \cdot \left(-\frac12\right) = 34 + 15 = 49$, and $c = 7$. The area of this triangle is $\frac12 \cdot 3 \cdot 5 \cdot \sin 120^\circ = \frac{15\sqrt3}{4} \approx 6.5$.

The mountain

Now we can go back to Everest. The foot of the mountain can’t be reached, so the distance to it is unknown. But you can measure the angle at which the summit is seen from two points $A$ and $B$ in line with the mountain, a known distance $d$ apart. Let the summit be seen from the farther point $A$ at the angle $\alpha$, and from the nearer point $B$ at a larger angle $\beta$. In triangle $ABT$ the angle at $B$ is $180^\circ - \beta$, and the angle at the summit $T$ is $\beta - \alpha$. By the law of sines $BT = \frac{d\sin\alpha}{\sin(\beta - \alpha)}$, and the height of the summit above the level of the points is a leg of a right triangle with hypotenuse $BT$:

$$h = BT\sin\beta = \frac{d\sin\alpha\sin\beta}{\sin(\beta - \alpha)}.$$

For example, with $d = 1000$ m, $\alpha = 25^\circ$ and $\beta = 35^\circ$ we get $h \approx 1396$ m. The surveyors of the Great Trigonometrical Survey of India measured Everest from several stations on the plain, more than a hundred kilometres away. At such distances two more things have to be taken into account: the Earth is round, and the summit “hides” behind its curve; and rays of light bend slightly in the air. In 1856 the figure announced was $29{,}002$ feet, about $8840$ m. Today’s value, adopted in 2020, is $8848.86$ m.

From the point $A$ the top of a hill is seen at an angle of $30^\circ$, and from the point $B$, $100$ m closer to the hill, at an angle of $60^\circ$. How high above the points $A$ and $B$ is the top? Hint: look at the angle at the top.

The angle of triangle $ABT$ at $T$ is $60^\circ - 30^\circ = 30^\circ$, the same as at $A$. The triangle is isosceles, $BT = AB = 100$ m. The height is $h = BT\sin 60^\circ = 100 \cdot \frac{\sqrt3}{2} = 50\sqrt3 \approx 86.6$ m.

The size of the Earth

The expedition’s last camp is the whole Earth. Around 240 BC Eratosthenes of Cyrene, keeper of the Library of Alexandria, learned that in Syene (today’s Aswan), at noon on the summer solstice, the Sun stands directly overhead: the story goes that its reflection can be seen at the bottom of a deep well, and upright objects cast no shadow. On the same day in Alexandria, north of Syene, an upright pole did cast a shadow at noon, and its angle was, as Cleomedes writes in retelling the experiment, one fiftieth of a full circle: $7.2^\circ$.

The rest is geometry. The Sun is very far away, so its rays in Syene and in Alexandria are parallel. The ray in Syene runs along the plumb line, that is, along a radius of the Earth. The pole in Alexandria also stands along a radius. The angle between the ray and the pole in Alexandria and the angle between the two radii at the centre of the Earth are alternate angles for parallel lines, so they’re equal (chapter 17). So the arc from Syene to Alexandria is one fiftieth of the Earth’s circumference. The distance between the cities was taken to be $5000$ stadia, and the whole circumference came out at $250{,}000$ stadia.

How accurate the result is is hard to say: it’s not known which stadion Eratosthenes meant. If he used a stadion of about $157.5$ m (called the Egyptian one), we get $39{,}375$ km, astonishingly close to today’s $40{,}008$ km along a meridian. If the Attic one, about $185$ m, we get $46{,}250$ km, $16\%$ too much. Besides, Syene isn’t exactly due south of Alexandria, and $5000$ is a rounded number. But the method is flawless, and its accuracy is limited only by the accuracy of the measurements.

Note the “What if flat?” switch. A flat Earth also explains the same two measurements, if the Sun hangs fairly low, about $40$ thousand stadia above Syene. Two poles can’t tell one from the other. Eratosthenes relied on the Sun being very far away; by then Greek astronomers already knew it was many times farther than the Moon. Aristarchus of Samos estimated the ratio at $18$–$20$ times (in fact it’s about $400$).

Stretching out a meridian arc more accurately than “five thousand stadia” was made possible by the technique we used at the river, repeated many times over.

The first to measure a meridian arc this way was the Dutchman Willebrord Snellius; in 1617 he called the book about his measurement Eratosthenes Batavus, “The Dutch Eratosthenes”. The most famous structure of this kind is the Struve Geodetic Arc: from 1816 to 1855, under the direction of the astronomer Friedrich Georg Wilhelm von Struve, a chain of $258$ triangles was run for $2820$ km from Hammerfest in Norway to the Black Sea. Today $34$ of its surviving points in ten countries are on UNESCO’s World Heritage List.

The wave

The expedition is heading home, but the sine goes much further than triangles. Everything that oscillates steadily, a pendulum, a string, a loudspeaker cone, moves like the shadow of a point on a circle. A pure tone is air pressure changing according to $A\sin(2\pi f t)$, where $f$ is the frequency, the number of oscillations per second. The A above middle C, to which orchestras tune, is $440$ oscillations per second.

If two notes of almost the same pitch sound together, the pressures add, and you hear something strange: the sound swells and fades. These are beats, and they’re explained by the same addition formulas.

For two tones with frequencies $f_1$ and $f_2$ this means: the sum oscillates at the average frequency $\frac{f_1 + f_2}{2}$, but its amplitude $2\cos\left(\pi(f_1 - f_2)t\right)$ changes slowly. The loudness dies away $f_2 - f_1$ times a second. For an A at $440$ and $444$ oscillations per second, four times.

A complex sound, a voice, a violin, a chord, can also be broken down into sine waves, only many of them are needed. How to do it is told in the chapter on Fourier series.

The field log

The expedition is over; it’s time to fill in the log. The trainer has three levels: the values of sine, cosine and tangent for acute angles; any angles on the unit circle, in degrees and radians; triangles, with sides, angles and areas by the laws of sines and cosines.

For problems of your own there are solvers: solving a triangle from any three pieces of data, and trigonometric equations.

Where next

We’ve learned to measure the inaccessible without leaving the spot. But on the road you don’t just measure, you also move. A motorboat goes at $4$ km/h in still water, its bow pointed straight across the river, and the current carries it downstream at $3$ km/h. Where, and how fast, will it actually go? The speeds can’t simply be added here: each has a direction, and they have to be added “as arrows”. How to calculate with arrows is told in the chapter on vectors and coordinates.