Lectures 19–21 | measuring what you cannot reach
Chapter 4 treated sine and cosine as coordinates on a circle, which is what they are and what makes them good at repetition. This chapter goes back to where they came from historically: the triangle, and the problem of measuring things that are too far away, too tall, or too large to put a ruler against.
That problem is older than algebra and it was solved, remarkably well, by people with no equipment beyond a stick and a shadow. By the end of the chapter we will have measured a building, the Earth, and the distance to the Moon, and then we will find a triangle whose angles add to more than \(180^{\circ}\) — at which point the geometry you have been using since primary school quietly stops being the whole story.
Seeing is a triangle. Light leaves the top of a tree, travels in a straight line to your eye, and light from the base does the same; those two rays plus the tree make a triangle, and so do you, your shadow, and the Sun. Every act of looking at something distant sets one up.
This is why geometry was the mathematics of sight, as Lecture 1 claimed. A triangle is the smallest rigid shape — fix three side lengths and the shape cannot flex, which is untrue of every other polygon — and rigidity is exactly what lets you deduce what you cannot reach from what you can.
Two triangles are similar if they have the same three angles. They may be wildly different in size. The key fact, known to the Greeks and provable from parallel lines, is that similar triangles have proportional sides: if one is three times wider it is three times taller too.
In a right triangle with acute angle \(\theta\), label the sides opposite, adjacent and hypotenuse. Then $$\sin\theta=\frac{\text{opp}}{\text{hyp}},\qquad \cos\theta=\frac{\text{adj}}{\text{hyp}},\qquad \tan\theta=\frac{\text{opp}}{\text{adj}}.$$ These are well defined precisely because similar triangles have proportional sides: any right triangle with that angle gives the same three numbers.
A right triangle has legs 3 and 4 and hypotenuse 5. Let \(\theta\) be the angle opposite the side of length 3. What is \(\sin\theta\)?
decimals, fractions, pi and sqrt( ) are all fine
This agrees with Chapter 4. Put the triangle inside the unit circle with \(\theta\) at the origin and the hypotenuse equal to 1, and the opposite side is the height of the terminal point, the adjacent side is its horizontal coordinate. One definition, two costumes. The circle version handles angles beyond \(90^{\circ}\), which no triangle can contain; the triangle version is what you use with a tape measure.
In a right triangle with legs \(a,b\) and hypotenuse \(c\), $$a^{2}+b^{2}=c^{2}.$$
It is the identity \(\cos^{2}\theta+\sin^{2}\theta=1\) of Chapter 4, multiplied through by \(c^{2}\) — or, read the other way, that identity is Pythagoras applied to a triangle whose hypotenuse is 1.
The three ratios are worth memorising by their initials, because you will reach for them constantly:
| SOH | \(\textbf{S}\)ine \(=\) \(\textbf{O}\)pposite \(\div\) \(\textbf{H}\)ypotenuse |
| CAH | \(\textbf{C}\)osine \(=\) \(\textbf{A}\)djacent \(\div\) \(\textbf{H}\)ypotenuse |
| TOA | \(\textbf{T}\)angent \(=\) \(\textbf{O}\)pposite \(\div\) \(\textbf{A}\)djacent |
Which side counts as "opposite" and "adjacent" depends on which acute angle you are standing at. The hypotenuse never changes — it is always the side facing the right angle — but swap to the other acute angle and opposite and adjacent trade places. That is the single most common source of wrong answers in this lecture.
Same 3-4-5 triangle. Now stand at the other acute angle, \(\varphi\). What is \(\cos\varphi\)?
decimals, fractions, pi and sqrt( ) are all fine
A right triangle has six parts: three sides and three angles, one of which is the right angle. Completing it means finding all six from the few you are given. There are only three situations, and each has a standard opening move.
A right triangle has an angle of \(28^{\circ}\) and the side adjacent to it is 15 cm. Complete the triangle.
Third angle: \(90-28=62^{\circ}\).
Opposite side: the ratio connecting opposite to adjacent is TOA, so \(\tan28^{\circ}=\dfrac{\text{opp}}{15}\) and opp \(=15\tan28^{\circ}=7.98\) cm.
Hypotenuse: CAH connects adjacent to hypotenuse, so \(\cos28^{\circ}=\dfrac{15}{\text{hyp}}\) and hyp \(=\dfrac{15}{\cos28^{\circ}}=16.99\) cm.
Check with Pythagoras: \(15^{2}+7.98^{2}=225+63.7=288.7\), and \(16.99^{2}=288.7\). ✓
Notice that the hypotenuse came from the given side rather than from the one we had just computed. Where you have a choice, always build on the numbers you were handed, so that one slip does not poison everything after it.
A wheelchair ramp rises 0.45 m over a horizontal run of 5.4 m. Complete the triangle, and check it against the common building requirement that a ramp be no steeper than 1 in 12.
Hypotenuse: \(\sqrt{0.45^{2}+5.4^{2}}=\sqrt{0.2025+29.16}=5.419\) m.
Angle at the bottom: opposite and adjacent are known, so use TOA backwards: $$\tan\theta=\frac{0.45}{5.4}=0.08333,\qquad \theta=\tan^{-1}(0.08333)=4.76^{\circ}.$$
The remaining angle is \(85.24^{\circ}\).
A slope of 1 in 12 means a rise of 1 for a run of 12, which is \(\tan^{-1}(1/12)=4.76^{\circ}\). This ramp is exactly at the limit. The rule is written as a ratio rather than an angle precisely because a ratio can be checked on site with a tape measure and no instrument at all.
Most angles give ugly ratios that only a calculator can supply. Two do not, and they come from shapes you already know.
| \(\theta\) | \(\sin\theta\) | \(\cos\theta\) | \(\tan\theta\) |
|---|---|---|---|
| \(30^{\circ}\) | \(\tfrac12\) | \(\tfrac{\sqrt3}{2}\) | \(\tfrac{1}{\sqrt3}\) |
| \(45^{\circ}\) | \(\tfrac{1}{\sqrt2}\) | \(\tfrac{1}{\sqrt2}\) | \(1\) |
| \(60^{\circ}\) | \(\tfrac{\sqrt3}{2}\) | \(\tfrac12\) | \(\sqrt3\) |
Every entry is read straight off the two pictures above. Nothing has to be memorised separately — if you can draw a square and an equilateral triangle, you can rebuild this table in twenty seconds.
Without a calculator: what is \(10\sin30^{\circ}+4\cos60^{\circ}\)?
decimals, fractions, pi and sqrt( ) are all fine
A guy wire runs from the top of a 12 m mast to the ground at \(60^{\circ}\) to the horizontal. Find the exact length of the wire and its exact distance from the base.
The mast is opposite the \(60^{\circ}\) angle, so \(\sin60^{\circ}=\dfrac{12}{L}\), giving $$L=\frac{12}{\sqrt3/2}=\frac{24}{\sqrt3}=8\sqrt3\ \text{m}\approx13.86\ \text{m}.$$
The base distance uses TOA: \(d=\dfrac{12}{\tan60^{\circ}}=\dfrac{12}{\sqrt3}=4\sqrt3\approx6.93\) m.
Note that \(L=2d\), which is the 30-60-90 triangle announcing itself: the hypotenuse is always twice the shortest side. Spotting that lets you write both answers down without touching a calculator.
Chapter 4 defined sine and cosine as coordinates of a point on the unit circle. This lecture defined them as ratios of sides. Those look like two different definitions. They are one picture, and here it is.
Take any right triangle and place it in standard position: the vertex of the angle \(\theta\) at the origin, the adjacent side lying along the positive \(x\)-axis, the opposite side going straight up. Now scale the whole triangle until the hypotenuse has length exactly 1 — which similarity says changes no angle and no ratio.
With the hypotenuse equal to 1, the ratios lose their denominators. The adjacent side is \(\cos\theta\) and the opposite side is \(\sin\theta\), so the far corner of the triangle sits at the point \((\cos\theta,\sin\theta)\) — exactly the terminal point of Chapter 4.
And there is a second thing to notice, which is the reason radians exist. The corner can be described two ways: by the angle turned, or by the distance travelled along the circle from \((1,0)\). On a circle of radius 1 those two numbers are equal, because arc length is \(r\theta\) and \(r=1\). Turning and travelling carry the same information.
That is the bridge. Triangles are what you use when you have a ruler and a distant object. The circle is what you use when the thing keeps coming back around. Projecting the moving point of the circle onto the axes gives you a triangle at every instant, and reading the triangle at every instant gives you back the circle.
A point on the unit circle has travelled an arc of length \(\tfrac{\pi}{4}\) from \((1,0)\). What are its coordinates?
2 numbers, separated by a comma: x, then y (sqrt(2)/2 is fine)
A point on the unit circle has travelled an arc of length \(\tfrac{\pi}{3}\) from \((1,0)\). Draw the right triangle it makes with the \(x\)-axis, give its three sides, and state the coordinates of the point.
An arc of \(\tfrac{\pi}{3}\) on the unit circle is an angle of \(\tfrac{\pi}{3}\) radians, which is \(60^{\circ}\).
Dropping a vertical line from the point to the \(x\)-axis makes a 30-60-90 triangle with hypotenuse 1. From the special triangle, the side adjacent to \(60^{\circ}\) is \(\tfrac12\) and the opposite side is \(\tfrac{\sqrt3}{2}\).
So the point is \(\left(\tfrac12,\tfrac{\sqrt3}{2}\right)\) — which is exactly the entry in the unit-circle table of Lecture 15, now derived from a cut-up equilateral triangle rather than quoted.
A 1.8 m person casts a 2.5 m shadow. At the same moment a building casts a 31 m shadow. How tall is the building, and what is the Sun's angle of elevation?
The Sun is so far away that its rays arrive parallel, so both triangles have the same angles — they are similar. Therefore $$\frac{h}{31}=\frac{1.8}{2.5}=0.72,\qquad h=22.3\ \text{m}.$$
The elevation is \(\tan\theta=0.72\), so \(\theta=35.8^{\circ}\).
Notice that the Sun's angle was never needed. The ratio did all the work, and the ratio came from a stick you could reach. Thales is said to have measured the Great Pyramid this way around 600 BCE, by waiting for the moment when his own shadow equalled his height — at which instant the pyramid's shadow equals the pyramid's height, and no arithmetic is required at all.
A cable runs from the top of a 24 m mast to a point on the ground, making a \(52^{\circ}\) angle with the ground. Find the length of the cable and its distance from the base.
The mast is opposite the angle and the cable is the hypotenuse, so \(\sin 52^{\circ}=\dfrac{24}{c}\), giving $$c=\frac{24}{\sin52^{\circ}}=\frac{24}{0.7880}=30.5\ \text{m}.$$
The base distance is adjacent: \(\tan52^{\circ}=\dfrac{24}{d}\), so \(d=\dfrac{24}{1.2799}=18.8\) m.
Check with Pythagoras: \(24^{2}+18.8^{2}=576+353=929\), and \(30.5^{2}=930\). ✓ Always run that check; it catches a sine typed where a cosine was meant.
From a cliff-top viewpoint 85 m above the sea, a boat is seen at an angle of depression of \(11^{\circ}\). How far is the boat from the base of the cliff? It then moves to an angle of depression of \(19^{\circ}\); how far has it travelled?
The angle of depression from the top equals the angle of elevation from the boat, since the two horizontals are parallel. So $$\tan 11^{\circ}=\frac{85}{d_1},\qquad d_1=\frac{85}{0.1944}=437\ \text{m}.$$
At \(19^{\circ}\): \(d_2=\dfrac{85}{0.3443}=247\) m. The boat has come \(190\) m closer.
Worth noticing: the same \(8^{\circ}\) change would correspond to a very different distance if the boat were further out. Near the horizon, small angle changes mean enormous distance changes — which is why judging distance at sea by eye is so unreliable, and why the tangent function shoots off to infinity.
A total solar eclipse happens because the Moon, seen from the Earth, is almost exactly the same size as the Sun. Not actually the same size — the Sun is 400 times wider. But it is also about 400 times further away, and similar triangles say that is what matters.
The angular size of an object of width \(D\) at distance \(d\) is the angle it takes up in your field of view. For \(D\) much smaller than \(d\), $$\theta\approx\frac{D}{d}\ \text{radians}.$$
This is the small-angle approximation of Lecture 15 doing real work: it is why the Moon and a coin held at arm's length can block the same amount of sky.
A 1 m object is 100 m away. Roughly what is its angular size, in radians?
decimals, fractions, pi and sqrt( ) are all fine
Compute the angular size of the Sun and of the Moon in degrees. Then repeat the Moon's calculation at its closest approach, 363,300 km, and at its furthest, 405,500 km. What do you conclude?
Sun: \(\theta=\dfrac{1{,}392{,}000}{149{,}600{,}000}=0.009305\) rad \(=0.533^{\circ}\).
Moon on average: \(\dfrac{3{,}475}{384{,}400}=0.009040\) rad \(=0.518^{\circ}\) — very slightly smaller than the Sun.
At its closest: \(0.548^{\circ}\), comfortably larger. At its furthest: \(0.491^{\circ}\), noticeably smaller.
So it depends on where the Moon is in its orbit. When it is near, it covers the Sun completely and you get a total eclipse with a visible corona. When it is far, a ring of Sun remains uncovered — an annular eclipse, from the Latin for ring, and one you must never look at directly.
The whole phenomenon hangs on a coincidence of two ratios, and it is temporary: the Moon recedes about 3.8 cm a year, so in roughly 600 million years total eclipses will stop happening. We are living in the era when they work.
Around 240 BCE the librarian at Alexandria, Eratosthenes, measured the circumference of the planet to within a couple of per cent, using a stick, a well and a single angle. It is the best example there is of how far a triangle can be made to reach.
It is worth being clear that the shape of the Earth was not in doubt among educated Greeks by then, and not because anyone had guessed. There were measurements.
That last observation is not merely evidence that the Earth is curved. Measure the difference and it tells you how much it curves — which is the whole method.
Suppose the shadow angle at Alexandria had been \(3.6^{\circ}\) instead, with the cities still 5,000 stadia apart. What circumference, in stadia, would that give?
decimals, fractions, pi and sqrt( ) are all fine
At noon on the summer solstice the Sun is directly overhead at Syene: a vertical pole casts no shadow and sunlight reaches the bottom of a deep well. At the same moment in Alexandria, 5,000 stadia to the north, an identical pole casts a shadow, and the Sun's rays make an angle of \(7.2^{\circ}\) with the pole. Find the circumference.
The Sun is so distant that its rays arrive parallel. A pole points straight at the centre of the Earth, so the angle between the ray and the pole at Alexandria equals the angle between the two cities measured at the centre — alternate angles across parallel lines, the same fact that made similar triangles work.
That angle is \(7.2^{\circ}\), and $$\frac{7.2}{360}=\frac{1}{50},$$ so the arc between the cities is one fiftieth of the whole way round: $$C=50\times5{,}000=250{,}000\ \text{stadia}.$$
A stadion was roughly 157.5 m, giving about 39,400 km against the true 40,075 km — an error under 2%. From the circumference the radius follows immediately: \(r=C/2\pi\approx6{,}270\) km.
Every step is in this lecture. What the calculation actually required was noticing that a shadow in one city and no shadow in another, at the same instant, is a measurement of the planet.
With the size of the Earth known, the Moon falls next — and it takes two observations and a ratio.
During a lunar eclipse the Earth's shadow at the Moon's distance is measured to be about \(2.5\) Moon-diameters wide, and the shadow there is about \(0.72\) of the Earth's own diameter (it has narrowed on the way out). The Moon's angular size is \(0.52^{\circ}\). Taking the Earth's diameter as 12,742 km, find the Moon's diameter and its distance.
Size. The shadow at that distance is \(0.72\times12{,}742=9{,}174\) km across, and the Moon is 2.5 times narrower than that: $$D_{\text{Moon}}=\frac{9{,}174}{2.5}=3{,}670\ \text{km},$$ against the true 3,475 km. Close, from a stopwatch and an eclipse.
Distance. Now use angular size, which is the small-angle version of a very thin triangle: \(\theta\approx D/d\), with \(\theta\) in radians. Here \(0.52^{\circ}=0.00908\) rad, so $$d=\frac{D}{\theta}=\frac{3{,}475}{0.00908}=382{,}700\ \text{km},$$ against the true average of 384,400 km.
Stand back from what just happened. Two thousand two hundred years ago, with no telescope, no clock worth the name and no algebra, people established the size of the Earth, the size of the Moon and the distance between them — and got all three right to within a few per cent. The only equipment was a shadow, a stopwatch of sorts, and the fact that similar triangles have proportional sides.
Aristarchus of Samos, around 270 BCE, tried to measure the distance to the Sun. At the exact moment the Moon is half lit, the Earth–Moon–Sun angle at the Moon is a right angle, so the triangle is determined by the single angle at the Earth. He measured it as \(87^{\circ}\), giving a Sun roughly 19 times further than the Moon.
The true angle is \(89.85^{\circ}\) and the true ratio is about 390. His method was flawless and his answer was off by a factor of twenty, because near \(90^{\circ}\) the cosine is brutally sensitive: a fifth of a degree of measurement error moves the answer by a factor of two. Good mathematics plus an instrument that is not quite good enough gives a confident wrong answer, which is a lesson worth more than the number.
Everything so far needed a right angle. Most triangles do not have one, and the fix is two formulas that between them solve any triangle at all.
First, a definitional loose end. A triangle can contain an obtuse angle, and \(\sin 130^{\circ}\) means nothing in the opposite-over-hypotenuse picture — no right triangle has a \(130^{\circ}\) angle in it. The circle definition of Chapter 4 has no such difficulty: put the angle in standard position, read off the coordinates of the terminal point, done.
So we use the circle definition and the triangle one becomes a special case. The consequence worth memorising: \(\sin(180^{\circ}-\theta)=\sin\theta\), because reflecting a terminal point across the vertical axis leaves its height unchanged, while \(\cos(180^{\circ}-\theta)=-\cos\theta\), because the horizontal coordinate flips. Sine cannot tell an angle from its supplement. That fact will cause us trouble shortly, and it is honest trouble.
Without a calculator: what is \(\sin150^{\circ}\)?
decimals, fractions, pi and sqrt( ) are all fine
In any triangle with sides \(a,b,c\) opposite angles \(A,B,C\), $$\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}.$$
Use it when a side and its opposite angle are known as a pair.
A triangle has \(A=30^{\circ}\), \(a=5\) and \(B=90^{\circ}\). What is \(b\)?
decimals, fractions, pi and sqrt( ) are all fine
Use it for three sides, or for two sides and the angle between them. When \(C=90^{\circ}\) the cosine is zero and this collapses to Pythagoras — so the Law of Cosines is Pythagoras with a correction term for the angle not being right.
A triangle has \(A=41^{\circ}\), \(B=77^{\circ}\) and \(a=13\) cm. Solve it completely.
\(C=180-41-77=62^{\circ}\).
By the Law of Sines, \(b=\dfrac{a\sin B}{\sin A}=\dfrac{13(0.9744)}{0.6561}=19.3\) cm and \(c=\dfrac{13(0.8829)}{0.6561}=17.5\) cm.
Sanity check: the largest angle is \(B\) and the largest side came out as \(b\). In any triangle the biggest side faces the biggest angle, and checking that ordering catches most arithmetic slips instantly.
Two roads leave a junction at an angle of \(60^{\circ}\). One town is 5 km along the first, another 7 km along the second. How far apart are they, and what angle does the connecting road make with the first?
The angle is between the two known sides, so use the Law of Cosines: $$d^{2}=5^{2}+7^{2}-2(5)(7)\cos60^{\circ}=25+49-35=39,$$ so \(d=6.245\) km.
For the angle at the first town, use the Law of Sines: \(\dfrac{\sin\theta}{7}=\dfrac{\sin60^{\circ}}{6.245}\), giving \(\sin\theta=0.9707\) and \(\theta=76.1^{\circ}\) (the obtuse alternative \(103.9^{\circ}\) is impossible here, since the remaining angles must sum to \(120^{\circ}\)).
Note how much shorter the direct road is than going back through the junction: 6.2 km against 12. That saving is the triangle inequality, and it is the same fact that made the diagonal across the park worth taking in Lecture 1.
A triangular plot has sides 8, 11 and 14 m. Find all three angles and the area.
Rearrange the Law of Cosines to find the angle opposite the side of length 8: $$\cos A=\frac{11^{2}+14^{2}-8^{2}}{2(11)(14)}=\frac{253}{308}=0.8214,\qquad A=34.8^{\circ}.$$
Similarly \(\cos B=\dfrac{8^{2}+14^{2}-11^{2}}{2(8)(14)}=0.6206\), so \(B=51.6^{\circ}\), and \(C=180-34.8-51.6=93.6^{\circ}\).
Area \(=\tfrac12ab\sin C=\tfrac12(8)(11)\sin93.6^{\circ}=43.9\) m².
Always find the largest angle with the cosine rule, not the sine rule. Cosine distinguishes acute from obtuse by its sign; sine does not, and here \(C\) is obtuse by a hair — the sine rule would have quietly offered you \(86.4^{\circ}\).
There is only one area anybody really knows how to find, and it is the rectangle: lay out \(b\) columns of \(h\) unit squares and count them. Every other area in mathematics is obtained by reducing it to that one, and the tool for the reduction is the triangle.
Cutting a hexagon into triangles from one corner gives how many triangles?
decimals, fractions, pi and sqrt( ) are all fine
That last line is the interesting one, and the circle is the case everybody wants.
Slice a circle into \(n\) equal wedges from the centre. Each wedge is nearly a triangle: two straight sides of length \(r\), and a third side that is very slightly curved. Pretend it is straight and you can find its area exactly. Add up all \(n\) of them and you have an estimate of the circle's area — an underestimate, because each chord cuts off a thin sliver of the circle that you left out.
Joining the \(n\) points around a circle of radius \(r\) makes a polygon of \(n\) triangles, each with two sides \(r\) and an included angle \(\dfrac{2\pi}{n}\). By the area formula of this lecture, $$A_n=n\cdot\frac12 r^{2}\sin\!\left(\frac{2\pi}{n}\right).$$ Since the true area is \(\pi r^{2}\), the quantity \(\dfrac{n}{2}\sin\!\left(\dfrac{2\pi}{n}\right)\) is an estimate of \(\pi\) — one you can compute with nothing but the sine of a small angle.
Use \(n=4\) wedges (an inscribed square). What estimate of \(\pi\) does \(\dfrac{n}{2}\sin\!\left(\dfrac{2\pi}{n}\right)\) give?
decimals, fractions, pi and sqrt( ) are all fine
Compute the inscribed estimate of \(\pi\) for \(n=6\), \(12\), \(24\) and \(96\), and compare with Archimedes’ published bounds of \(3\tfrac{10}{71}\) and \(3\tfrac17\).
| \(n\) | \(\tfrac{n}{2}\sin(2\pi/n)\) | error |
|---|---|---|
| 6 | 2.59808 | −0.543 |
| 12 | 3.00000 | −0.142 |
| 24 | 3.10583 | −0.036 |
| 96 | 3.13935 | −0.0022 |
Archimedes did exactly this around 250 BCE, starting from a hexagon and doubling four times to reach 96 sides. He also did the same with polygons drawn outside the circle, which overestimate, and so trapped \(\pi\) between two numbers: $$3\tfrac{10}{71}=3.1408\ <\ \pi\ <\ 3\tfrac17=3.1429.$$
He had no decimals, no trigonometric tables and no algebra. The doubling was done with square roots computed by hand. And notice what the method gives that a single estimate never can: a guarantee, with the answer caught between a number known to be too small and a number known to be too big.
Look at what the estimate is doing. Each wedge has area \(\tfrac12r^{2}\sin\theta\) where \(\theta=2\pi/n\), and we saw in Lecture 15 that \(\sin\theta\approx\theta\) for small angles. So for large \(n\), $$A_n\approx n\cdot\tfrac12 r^{2}\cdot\frac{2\pi}{n}=\pi r^{2}.$$ The \(n\) cancels. The formula for the area of a circle is not a separate fact to memorise — it is what the triangle formula turns into when the wedges get thin.
But it is only exactly \(\pi r^{2}\) in the limit. For any finite \(n\), however large, the slivers are still there and the answer is still too small. This is the fourth time this course has run the same argument: shrink the pieces, watch the estimate settle, and name the destination. Rectangles under a curve in Lecture 10 did it, polynomial approximations in Lecture 11 did it, secant lines in Lecture 8 did it, and here it is again with triangles in a circle. Calculus is the subject that makes the last step legal.
Given two sides and an angle not between them, the triangle may not be determined at all. Sine cannot distinguish \(\theta\) from \(180^{\circ}-\theta\), and sometimes both genuinely work.
A triangle has \(A=35^{\circ}\), \(a=6\) and \(b=8\). Solve it.
\(\sin B=\dfrac{b\sin A}{a}=\dfrac{8(0.5736)}{6}=0.7648\).
That gives \(B=49.9^{\circ}\) or \(B=130.1^{\circ}\), and both leave room for a third angle:
| \(B\) | \(C\) | \(c\) | |
|---|---|---|---|
| first triangle | 49.9° | 95.1° | 10.42 |
| second triangle | 130.1° | 14.9° | 2.69 |
Two different triangles, both with a \(35^{\circ}\) angle, a side of 6 opposite it, and a side of 8. The data does not determine the shape, and no amount of care with the algebra will make it.
This is why surveyors prefer to measure the angle between two known sides. Three sides, or two sides with the included angle, always give exactly one triangle. Two sides and a stray angle may give two — the same ambiguity that makes SSA useless as a congruence rule in geometry class.
A phone is 4.0 km from tower P and 6.5 km from tower Q, which are 8.0 km apart. Find the angle at P, and explain why two towers are not enough to locate the phone.
Three sides are known, so use the Law of Cosines at P: $$\cos P=\frac{4.0^{2}+8.0^{2}-6.5^{2}}{2(4.0)(8.0)}=\frac{37.75}{64}=0.5898,\qquad P=53.8^{\circ}.$$
But the angle could open to either side of the line PQ: the phone might be 53.8° above it or 53.8° below it. Two distances place you on a circle around each tower, and two circles meet in two points.
A third tower breaks the tie, because a third circle passes through only one of them. That is the whole principle of trilateration, and it is what GPS does with satellites instead of towers — using four rather than three, because the clock in your phone is not accurate enough and the fourth measurement is spent working out what time it actually is.
Triangulation is how large distances have been measured for four hundred years. Pick two points you can reach, measure the distance between them precisely once — this is the hard part, and the one measurement everything else rests on — then measure angles to a distant landmark from both ends. The Law of Sines gives the distance. Now use that landmark as one end of a new baseline and repeat, chaining triangles across a country.
The Great Trigonometrical Survey of India ran this from 1802 for seventy years, measured the subcontinent, and in the process found that a peak catalogued as XV was the highest in the world. The same idea, with the Earth's orbit as the baseline, gives the distance to nearby stars. One measured length plus a pile of angles is enough to map anything.
Everything in the last two lectures assumed the triangle was drawn on a flat surface. Almost nothing important is. The Earth is round, and once the triangle gets large enough, every rule you have just learned becomes false — not slightly inaccurate, but false in an interesting and measurable way.
You cannot draw a straight line on a sphere; nothing in the surface is straight. What you can do is ask for the shortest path between two points, staying on the surface, and that turns out to be a piece of a great circle — a circle whose centre is the centre of the sphere. The equator is one. Every line of longitude is one. Lines of latitude, apart from the equator, are not.
A geodesic is the straightest available path in a curved space: the route you get by never turning left or right, only following the surface. On a plane the geodesics are straight lines; on a sphere they are great circles.
Replacing "straight line" with "geodesic" is the move that lets all of geometry be rebuilt on curved surfaces.
There is a spherical version of the Law of Cosines. For a triangle on a unit sphere with sides \(a,b,c\) measured as angles at the centre, $$\cos c=\cos a\cos b+\sin a\sin b\cos C.$$ Use it to find the distance from New York (\(40.7^{\circ}\)N, \(74.0^{\circ}\)W) to Madrid (\(40.4^{\circ}\)N, \(3.7^{\circ}\)W), taking the Earth's radius as 6,371 km.
Use the North Pole as the third vertex. The two sides from the pole are the co-latitudes, \(90-40.7=49.3^{\circ}\) and \(90-40.4=49.6^{\circ}\), and the angle at the pole is the difference in longitude, \(74.0-3.7=70.3^{\circ}\).
$$\cos c=\cos49.3^{\circ}\cos49.6^{\circ}+\sin49.3^{\circ}\sin49.6^{\circ}\cos70.3^{\circ}=0.4225+0.1949=0.6174$$So \(c=51.89^{\circ}=0.9056\) rad, and the distance is \(6371\times0.9056=5{,}769\) km.
Compare the flat version: it had the same shape, \(c^{2}=a^{2}+b^{2}-2ab\cos C\), with sides multiplying instead of adding. That is a general pattern — spherical formulas look like flat ones with the sides wrapped in sines and cosines, and they collapse back to the flat versions when the triangle is small.
Start at the North Pole. Travel straight south to the equator. Turn left through a right angle and travel a quarter of the way round the equator. Turn left again through a right angle and travel straight north. You arrive back at the pole, having walked a triangle with three right angles.
On a sphere of radius \(r\), a triangle with angles \(A,B,C\) measured in radians has $$\text{area}=(A+B+C-\pi)\,r^{2}.$$
The quantity \(E=A+B+C-\pi\) is the excess. It is zero only for a triangle of zero area, so on a sphere every triangle has too much angle, and the bigger the triangle the worse it gets.
The pole-to-equator triangle has three right angles. What is its excess \(E\), in radians?
decimals, fractions, pi and sqrt( ) are all fine
Estimate the angle excess for an equilateral triangle on the Earth with sides of 100 km, 1,000 km and 5,000 km, using excess \(\approx\) area \(/\,r^{2}\) with \(r=6{,}371\) km.
A flat equilateral triangle of side \(s\) has area \(\tfrac{\sqrt3}{4}s^{2}\).
| side | area (km²) | excess (rad) | excess (degrees) |
|---|---|---|---|
| 100 km | 4,330 | 0.00011 | 0.006° |
| 1,000 km | 433,000 | 0.0107 | 0.61° |
| 5,000 km | 10,825,000 | 0.267 | 15.3° |
At 100 km the excess is a fiftieth of a degree — smaller than most surveying instruments of the nineteenth century could see, so plane trigonometry was fine. At 1,000 km it is over half a degree, which no serious survey can ignore. At 5,000 km it is enormous.
So the flat rules are not wrong, they are local. They are the small-triangle approximation to the real thing, in exactly the way a tangent line is the small-interval approximation to a curve. You have seen this shape of statement several times now, and it is the same shape every time.
Here is the thing that makes this more than a curiosity about globes. Imagine a creature living in a surface, unable to leave it or look at it from outside. It cannot see that its world is curved. But it can do experiments.
It can walk a triangle and add up the angles. It can pace out a circle of radius \(r\) and measure the circumference, comparing it with the \(2\pi r\) that flatness demands — on a sphere it comes up short, on a saddle it comes out long. Every one of these tests is performed entirely inside the surface, with a ruler and a protractor, and every one of them detects curvature.
That is what makes curvature intrinsic. It is not a statement about how the surface sits in space; it is a property of distances within the surface itself, and it can be measured by inhabitants who have no idea there is an outside.
The intrinsic curvature of a surface is unchanged by bending, as long as nothing is stretched, compressed or torn.
Gauss proved this in 1827 and called it the theorema egregium — the remarkable theorem. He was not a man given to exclamation, and he named it that because it says something no one expected: curvature, which looks like a fact about how a surface sits in space, is really a fact about distances measured within it.
Three consequences, all of which you have already met or can check over lunch.
A surface that looks flat if you stand close enough, but need not be flat overall, is called a manifold. You have been standing on one all your life and it took humanity a surprisingly long time to notice.
The mathematics of these objects — geodesics instead of straight lines, curvature instead of flatness, a triangle's excess telling you how much space is bending underneath it — is differential geometry, and it is the language in which general relativity is written. In that theory gravity is not a force pulling planets into orbit; it is curvature in the geometry of space and time, and a planet moving "straight ahead" along a geodesic traces an ellipse for the same reason a plane flying straight from New York to Madrid bows north on the map.
You are four ideas away from that: limits, derivatives, several variables, and tensors. But the picture — the actual conceptual content — is the octant triangle you just looked at. Nothing about it required anything beyond this chapter.
For two thousand years, mathematicians tried to prove Euclid's fifth postulate — essentially that parallel lines never meet — from the other four. The effort was enormous and it always failed. In the 1820s Bolyai in Hungary and Lobachevsky in Russia independently stopped trying to prove it and asked what happens if you assume it is false. What happens is: a complete, consistent geometry, describing a saddle-shaped world where triangles have too little angle.
Gauss had reached the same conclusion years earlier and published nothing, fearing what he called the outcry of the Boeotians. Bolyai's father, who had spent his own life on the problem, wrote to his son begging him to abandon it: it may take all your time, and deprive you of your health and peace of mind. The son ignored him, and geometry stopped being one thing.
Ninety years later Einstein needed a geometry in which space itself bends. It was already built and waiting.
Similar triangles let you trade a measurement you can make for one you cannot, which is how a stick measures a building and a shadow measures a planet. The Law of Sines and the Law of Cosines finish the job for triangles with no right angle, with one honest warning about the case where the data does not determine the answer.
And then the sphere, where the angles come out wrong, geodesics replace straight lines, and curvature turns out to be something the inhabitants of a surface can measure without ever leaving it. That is the door to the geometry of curved spaces, and you have walked up to it using nothing beyond a protractor and the contents of Chapter 4.
Chapter 6 goes back inside the circle to collect the identities — the many different ways of writing the same quantity that the circle’s symmetry makes possible.
Next: Chapter 6 — Trigonometric identities.