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Topos Uranos

Summary

This lesson defines oriented angles, which measure a rotation together with its direction and may exceed a full turn, and their two usual measures: the sexagesimal degree, fixed by an ancient convention, and the radian, which is the length of the arc that the angle intercepts on the unit circle. It admits the circumference of the circle, the proportionality of arcs to radii and the preservation of lengths under reflections, defines the number π\pi as half the circumference of the unit circle and states the axiom of traversal by arc length, which assigns to each real number a point of that circle; from this it deduces the behavior of the traversal in its second half turn, the rule for converting between degrees and radians, the length of an arc, the area of a sector and the inscribed angle theorem. It then places angles in standard position, proves that every angle reduces in exactly one way to an angle of the first turn and that one turn does not pass twice through the same point, characterizes coterminal angles and relates angular velocity to linear velocity. The problems measure the Earth as Eratosthenes did, compute a circular segment, transmit a rotation through a chain, maximize the area of a sector of fixed perimeter and count how many times the hands of a clock overlap.

Learning objectives

  1. Distinguish the initial side and the terminal side of an oriented angle, measure it in degrees, minutes and seconds, and recognize its sign according to the direction of rotation.
  2. State the axiom of traversal by arc length, define the radian with it and convert measures between degrees and radians, in particular those of the special angles.
  3. Prove and apply the formulas for the length of an arc and the area of a sector, the inscribed angle theorem and the relation between angular velocity and linear velocity.
  4. Reduce any angle to an angle of the first turn, proving the existence and uniqueness of the reduction, and decide whether two angles are coterminal.

Oriented angles

In elementary geometry, an angle is the figure formed by two rays with a common endpoint, and its measure lies between 0∘0^\circ and 180∘180^\circ. To study rotations, such as that of a wheel, of the hands of a clock or of the Earth about its axis, this notion is not enough: what matters is how far something has turned and in which direction, and a rotation may make several turns. For this reason the angle is regarded as the result of a rotation.

An oriented angle is determined by a ray, called the initial side, which turns about its endpoint, called the vertex, until it reaches a final position, called the terminal side, together with the rotation performed. The angle is positive if the rotation is made in the direction opposite to that of the hands of a clock (counterclockwise), and negative if it is made in the direction of the hands of a clock (clockwise). The zero angle is the one that does not turn.

One and the same pair of sides may correspond to many oriented angles: the terminal side reaches the same position if it turns a quarter turn counterclockwise, three quarters of a turn clockwise, or a quarter turn plus a full turn. These angles are different, because they describe different rotations, even though they end on the same side; the third section makes precise how they are related.

The sexagesimal system

The oldest measure of angles divides the full turn into 360360 equal arcs, and calls degree the angle that intercepts each of them: an angle measures gg degrees if its arc is gg times the arc of one degree, so that the measure in degrees is proportional to the length of the arc. Each degree is divided into 6060 minutes, and each minute into 6060 seconds. The system comes from Babylonian astronomy, which counted in base sixty, as explained in the Course I lesson on positional numeral systems; the number 360360 has the advantage of being divisible by 22, 33, 44, 55, 66, 88, 99, 1010 and 1212, among others. A right angle measures 90∘90^\circ, a straight angle measures 180∘180^\circ and a full turn, 360∘360^\circ; a negative angle is measured with the minus sign, so that a quarter turn clockwise measures −90∘-90^\circ. The angle of 40 degrees, 15 minutes and 36 seconds, which is abbreviated by writing after each figure the degree sign, one prime for the minutes and two for the seconds, measures the following in degrees, that is, 40.26 degrees:

40+1560+363600=40+14+1100=20135040 + \frac{15}{60} + \frac{36}{3600} = 40 + \frac{1}{4} + \frac{1}{100} = \frac{2013}{50}

The number 360360 is, however, a convention: nothing in the geometry of the circle distinguishes it from any other number. The formulas for the length of an arc and the area of a sector, and later those of the trigonometric functions, become simpler if angles are measured in a way that arises from the circle itself.

The radian

The idea is to measure an angle by the length of the arc it intercepts on a circle of radius 11 centered at its vertex. This requires knowing what the length of an arc is, and here a difficulty appears: an arc is not a union of line segments, and its length is defined as the value approached by the lengths of the polygonal lines inscribed in it as their sides become shorter and shorter, that is, by means of a limit. Its study belongs to the course Calculus II, where lengths of curves and areas of regions bounded by curves are computed. Here the following facts are admitted.

Admitted fact (lengths and areas in the circle). Every arc of a circle has a length, and every region bounded by line segments and arcs has an area. Congruent figures have the same length or the same area; if a figure is decomposed into two that do not overlap, its measure is the sum of the measures of the pieces; and if one region contains another, its area is not smaller, and it is larger if the difference contains some disk. Moreover, if two circles have radii rr and RR, the arcs that one and the same central angle intercepts on them are proportional to the radii, and the area of a disk of radius rr is πr2\pi r^2.

This fact is admitted without proof, because the lengths of arcs and the areas of curved regions are defined by means of limits and computed with the integral, in the course Calculus II. Its last clause presupposes the number π\pi, which is now defined: the number π\pi is half the circumference of the unit circle, that is, of the circle of radius 11. By the proportionality of arcs to radii, applied to the angle of a full turn, the circle of radius rr has circumference 2πr2\pi r, and the ratio of the circumference of any circle to its diameter is always π\pi. It is also admitted that π\pi is irrational, and that its decimal expansion begins with 3.14159; in particular, 3<π<43 < \pi < 4.

To measure angles one must also traverse the circle by arc length: start from a point and advance along it a given length, in one direction or the other, and know which point is reached. Everything this unit says about angles and trigonometric functions rests on that traversal, so it is worth stating precisely what is admitted about it. It is stated in the Cartesian plane, for the unit circle x2+y2=1x^2 + y^2 = 1 and its point A=(1,0)A = (1, 0); since lengths do not change when figures are moved, it holds equally for any circle of radius 11 and any starting point.

Axiom (traversal by arc length). For each real number ss there is a unique point P(s)P(s) of the unit circle which is reached by starting from A=(1,0)A = (1, 0) and traversing along it a length ∣s∣\abs{s}, counterclockwise if s>0s > 0 and clockwise if s<0s < 0, making as many turns as necessary; for s=0s = 0 we have P(0)=AP(0) = A. This traversal has the four properties below.
  1. Concatenation: if from the point P(s)P(s) one traverses a length t≥0t \geq 0 counterclockwise, one reaches P(s+t)P(s + t); if one traverses it clockwise, one reaches P(s−t)P(s - t).
  2. Periodicity: a full turn, in either direction, traverses the length 2π2\pi of the circle and returns to the starting point; that is, P(s+2π)=P(s)P(s + 2\pi) = P(s) for every real number ss.
  3. Half turn: the counterclockwise traversal first passes through the upper semicircle, from right to left. Precisely, P(π)=(−1,0)P(\pi) = (-1, 0), and if 0≤s<t≤π0 \leq s < t \leq \pi, the points P(s)P(s) and P(t)P(t) have nonnegative ordinate and the abscissa of P(s)P(s) is greater than that of P(t)P(t).
  4. Covering: every point of the unit circle is P(s)P(s) for some real number ss.

The axiom is admitted without proof for the same reason as the fact stated before it: the length of an arc is defined by means of a limit, and its properties are proved in the course Calculus II; covering, which asserts that in traversing the circle no point is skipped, rests moreover on the intermediate value property of continuous functions, which is proved in the course Calculus I. The half-turn property has a simple geometric reading. If (x,y)(x, y) lies on the unit circle, the square of its distance to the point AA is as follows:

(x−1)2+y2=x2+y2−2x+1=2−2x(x - 1)^2 + y^2 = x^2 + y^2 - 2x + 1 = 2 - 2x

Hence saying that the abscissa decreases strictly along the upper semicircle is equivalent to saying that the chord joining AA to the point reached increases strictly with the arc, until it becomes a diameter after a half turn. When, in the lesson on trigonometric ratios, the cosine of ss is defined as the abscissa of P(s)P(s), this property will say that the cosine is strictly decreasing on [0,π]\icc{0}{\pi}: that monotonicity is not deduced; it is the half-turn clause itself read in coordinates. Now, from this axiom, from the facts about lengths and areas and from the fact about reflections admitted below, everything else that is used about the traversal is deduced: in this lesson, the behavior of the second half turn and the fact that one turn does not pass twice through the same point; in the lesson on the inverse trigonometric functions, the monotonicity of the sine and of the tangent.

Admitted fact (reflections and the half-turn rotation). A reflection in a line preserves distances and, with them, the lengths of arcs, but it reverses the direction of rotation: it transforms a counterclockwise traversal of a circle centered at a point of that line into a clockwise traversal of the same length, and conversely. The half-turn rotation about a point, which is the composition of the reflections in two perpendicular lines through it, preserves distances, lengths and also the direction of rotation.

This fact belongs to the elementary geometry of plane motions, which this course does not develop. In coordinates, the reflection in the xx-axis is the transformation (x,y)↦(x,−y)(x, y) \mapsto (x, -y); it maps the unit circle onto itself, because x2+(−y)2=x2+y2x^2 + (-y)^2 = x^2 + y^2, and leaves the point A=(1,0)A = (1, 0) fixed. With it, the second half turn of the traversal is obtained from the first.

PropositionThe second half turn

For every real number uu, the point P(−u)P(-u) is the reflection of P(u)P(u) in the xx-axis. Moreover, if π≤s<t≤2π\pi \leq s < t \leq 2\pi, the points P(s)P(s) and P(t)P(t) have nonpositive ordinate and the abscissa of P(s)P(s) is less than that of P(t)P(t): in its second half turn, the counterclockwise traversal passes through the lower semicircle, from left to right.

The traversal is reflected in the xx-axis, and periodicity carries the first half turn, traversed backward, to the second. We write (xw,yw)(x_w, y_w) for the coordinates of P(w)P(w), and σ\sigma for the reflection in the xx-axis.

Proof

  1. σ(x,y)=(x,−y),σ(A)=A\sigma(x, y) = (x, \resaltar{-y}), \quad \sigma(A) = A

    The reflection σ\sigma changes the sign of the ordinate, maps the unit circle onto itself and leaves the point AA fixed.

  2. P(−u)=σ(P(u))=(xu,−yu)\dato{r}{P(-u) = \resaltar{\sigma(P(u))} = \left( x_u, -y_u \right)}

    Let u≥0u \geq 0. By the admitted fact, σ\sigma transforms the counterclockwise traversal of length uu that starts at AA and ends at P(u)P(u) into a clockwise traversal of the same length that starts at AA; by the uniqueness of the point in the axiom, that traversal ends at P(−u)P(-u). If u<0u < 0, the above is applied to −u-u and one reflects again, because σ\sigma applied twice leaves every point where it was.

  3. P(s)=P(−u)=(xu,−yu),P(t)=(xv,−yv)P(s) = P(\resaltar{-u}) = \dato{st}{\resaltar{\left( x_u, -y_u \right)}}, \quad P(t) = \left( x_v, -y_v \right)

    Let π≤s<t≤2π\pi \leq s < t \leq 2\pi, and let u=2π−su = 2\pi - s and v=2π−tv = 2\pi - t, which satisfy 0≤v<u≤π0 \leq v < u \leq \pi. By periodicity, P(s)=P(s−2π)=P(−u)P(s) = P(s - 2\pi) = P(-u), which is the reflection of P(u)P(u); likewise, P(t)P(t) is the reflection of P(v)P(v).

  4. xs=xu<xv=xt,ys=−yu≤0,yt=−yv≤0x_s = \resaltar{x_u < x_v} = x_t, \quad y_s = -y_u \leq 0, \quad y_t = -y_v \leq 0

    By the half-turn property, applied to v<uv < u in [0,π]\icc{0}{\pi}, the ordinates of P(u)P(u) and of P(v)P(v) are not negative and the abscissa of P(v)P(v) is greater than that of P(u)P(u). The reflection preserves abscissas and changes the sign of ordinates: the statement follows.

Consider an oriented angle with vertex at the center OO of the unit circle, and let AA be the point where the initial side meets that circle. As the side turns, the point AA traverses an arc of the circle, as in the axiom stated above. The radian measure of the angle is the length of that arc, with positive sign if the rotation is counterclockwise and negative if it is clockwise; if the rotation makes several turns, each full turn contributes the length 2π2\pi of the circle.
ExampleSome angles measured in radians

The definition makes it possible to measure at once the angles that are simple fractions of a turn.

Proof

  1. L=2πL = 2\pi

    The unit circle has its center at the vertex OO and radius 11; by the definition of π\pi, its circumference is 2π2\pi. The initial side meets it at the point AA.

  2. θ=s\theta = s

    As the side turns counterclockwise, the point AA traverses an arc; the length ss of that arc is the measure of the angle.

  3. s=1→θ=1s = 1 \rightarrow \theta = 1

    One radian is the angle whose arc has the same length as the radius. Since 2π2\pi is somewhat greater than 66, one radian is somewhat less than one sixth of a turn.

  4. θ=2π4=π2\theta = \frac{\resaltar{2\pi}}{4} = \resaltar{\frac{\pi}{2}}

    A quarter turn traverses one quarter of the circle: the right angle measures π2\frac{\pi}{2} radians.

  5. θ=π,θ=2π\theta = \pi, \quad \theta = 2\pi

    A half turn traverses half the circle, whose length is π\pi by definition; the full turn measures 2π2\pi.

  6. θ=−π2\theta = -\frac{\pi}{2}

    If the side turns a quarter turn clockwise, the arc traversed has the same length, but the angle is negative.

An angle measured in radians is a dimensionless real number: it is the ratio of an arc length to the length of the radius, which equals 11. That is why one simply writes θ=2\theta = 2 for an angle of two radians; when one wishes to stress the unit, one writes 2 rad. By contrast, a measure in degrees is always indicated with its symbol, as in 30∘30^\circ. Conversely, every real number is the measure of some oriented angle: by the axiom of traversal, for each θ\theta the point AA can traverse the circle a length ∣θ∣\abs{\theta} in the direction indicated by the sign of θ\theta, making as many full turns as necessary, and the ray from the vertex to the point reached is the terminal side.

Converting between degrees and radians

Since the full turn measures 360∘360^\circ and also 2π2\pi radians, and since both measures are proportional to the length of the arc that the angle intercepts on the unit circle (the degree measure by the definition of the degree, and the radian measure because it is that length), the two measures of an angle are proportional.

TheoremConversion between degrees and radians

If an angle measures gg degrees and θ\theta radians, then θ=πg180\theta = \frac{\pi g}{180} and g=180θπg = \frac{180 \theta}{\pi}.

The rule is obtained from the proportion between the angle and the full turn, and is then applied to the special marks of the circle, the angles that appear in elementary figures.

Proof

  1. 360∘=2π\dato{v}{360^\circ = 2\pi}

    A full turn measures 360360 degrees and 2π2\pi radians.

  2. θ2π=g360→θ=2πg360=πg180\frac{\theta}{2\pi} = \frac{g}{360} \rightarrow \dato{c}{\theta = \resaltar{\frac{2\pi g}{360}} = \resaltar{\frac{\pi g}{180}}}

    The ratio of an angle to the full turn is the same in both measures, because both are proportional to the length of the arc; each additional turn adds 360∘360^\circ in one and 2π2\pi in the other. Solving for θ\theta gives the rule.

  3. 1∘=π180,1=(180π)∘1^\circ = \frac{\pi}{180}, \quad 1 = \left( \frac{180}{\pi} \right)^\circ

    With g=1g = 1 and with θ=1\theta = 1: one degree is a small fraction of a radian, and one radian measures somewhat more than 5757 degrees, because 180π\frac{180}{\pi} is approximately 57.3.

  4. 30∘=30π180=π6,45∘=π4,60∘=π330^\circ = \resaltar{\frac{30\pi}{180}} = \resaltar{\frac{\pi}{6}}, \quad 45^\circ = \frac{\pi}{4}, \quad 60^\circ = \frac{\pi}{3}

    The angles of 3030, 4545 and 6060 degrees, which appear in the equilateral triangle and in the square, are converted by simplifying the fraction.

  5. 90∘=π2,180∘=π,270∘=3π290^\circ = \frac{\pi}{2}, \quad 180^\circ = \pi, \quad 270^\circ = \frac{3\pi}{2}

    The right angle, the straight angle and three quarters of a turn mark the semi-axes.

  6. 120∘=2π3,135∘=3π4,150∘=π−π6=5π6120^\circ = \frac{2\pi}{3}, \quad 135^\circ = \frac{3\pi}{4}, \quad 150^\circ = \resaltar{\pi - \frac{\pi}{6}} = \resaltar{\frac{5\pi}{6}}

    The other special angles of the turn are obtained by adding or subtracting; for example, 150∘150^\circ is 180∘180^\circ minus 30∘30^\circ.

In practice, to pass from degrees to radians one multiplies by π180\frac{\pi}{180}, and to pass from radians to degrees, by 180π\frac{180}{\pi}. Thus 225∘=5π4225^\circ = \frac{5\pi}{4}, −60∘=−π3-60^\circ = -\frac{\pi}{3} and 7π6=210∘\frac{7\pi}{6} = 210^\circ; the angle of 22 radians measures 360π\frac{360}{\pi} degrees, approximately 114.6 degrees.

Arc length and area of a sector

On a circle of radius rr, a central angle θ\theta intercepts an arc, and the two sides of the angle, together with that arc, bound a circular sector. Radian measure gives their formulas a particularly simple form.

TheoremArc length and area of the sector

On a circle of radius rr, a central angle of θ\theta radians, with 0<θ≤2π0 < \theta \leq 2\pi, intercepts an arc of length s=rθs = r\theta, and the sector it determines has area A=12r2θ=12rsA = \frac{1}{2} r^2 \theta = \frac{1}{2} rs.

The length of the arc follows from the admitted proportionality. For the area, the formula is proved first when the angle is a rational fraction of the turn, and then extended to every angle by the density of the rationals, proved in the Course I lesson on the real line.

Proof

  1. s1=θs_1 = \theta

    On the unit circle with the same center, the angle intercepts an arc of length θ\theta, by the definition of the radian.

  2. s=rs1=rθs = r \resaltar{s_1} = \dato{s}{\resaltar{r\theta}}

    By the proportionality of arcs to radii, the corresponding arc on the circle of radius rr is rr times larger.

  3. A=pqπr2=12r2θA = \resaltar{\frac{p}{q} \pi r^2} = \dato{f}{\resaltar{\frac{1}{2} r^2 \theta}}

    If θ=2πpq\theta = \frac{2\pi p}{q}, with pp and qq natural numbers and p≤qp \leq q, the disk is divided into qq congruent sectors of angle 2πq\frac{2\pi}{q}, each of area πr2q\frac{\pi r^2}{q}, and the sector of angle θ\theta is the union of pp of them.

  4. 2Ar2<φ<θ→A<12r2φ=Aφ≤A\frac{2A}{r^2} < \varphi < \theta \rightarrow A < \resaltar{\frac{1}{2} r^2 \varphi} = A_\varphi \leq A

    Now let θ\theta be arbitrary, and suppose that the area AA of its sector were less than 12r2θ\frac{1}{2} r^2 \theta. By the density of the rationals there is an angle φ\varphi, a rational fraction of the turn, between 2Ar2\frac{2A}{r^2} and θ\theta. Its sector is contained in that of θ\theta, and its area is given by the formula of the preceding step: a contradiction is reached.

  5. A=12r2θ=12rsA = \frac{1}{2} r^2 \theta = \frac{1}{2} r \resaltar{s}

    Likewise, with a φ\varphi between θ\theta and 2Ar2\frac{2A}{r^2}, which does not exceed the turn because the sector is not larger than the disk, the case where AA is greater than 12r2θ\frac{1}{2} r^2 \theta is ruled out. Therefore the area is the one stated, and with rθ=sr\theta = s one obtains a formula analogous to that of the area of a triangle with base ss and height rr.

If the angle were measured in degrees, the formulas would carry the conversion factor: the arc would be πrg180\frac{\pi r g}{180} and the area of the sector πr2g360\frac{\pi r^2 g}{360}. The simplicity of s=rθs = r\theta is the main reason for preferring the radian. Moreover, the formula gives the measure of an angle on any circle, not only on the unit circle: θ=sr\theta = \frac{s}{r} is the ratio of the arc to the radius, and it does not depend on the radius chosen.

The inscribed angle

An angle with its vertex at the center of a circle is called a central angle. If AA, BB and CC are three distinct points of a circle, the angle formed at CC by the chords CACA and CBCB is called an inscribed angle, and it is said to intercept the arc with endpoints AA and BB that does not contain CC. By the theorem proved above, to an arc of length ss on a circle of radius rr corresponds the central angle of sr\frac{s}{r} radians, even if that number exceeds π\pi. To compare the inscribed angle with the central one, the following facts of elementary geometry are admitted, which this course does not prove because it does not develop synthetic geometry. The first two measure the angles of a triangle; the last two are facts of position, which say how angles and arcs are shared out when a ray passes through the interior of an angle.

Admitted fact (angles of the triangle and positions on the circle). The four statements below are admitted, in which non-oriented angles are measured in radians, between 00 and π\pi.
  1. Sum of the angles: the three angles of a triangle add up to a straight angle, that is, π\pi radians.
  2. Isosceles triangle: in a triangle with two equal sides, the angles opposite those sides are equal.
  3. Sum of adjacent angles: if a ray starting from the vertex of an angle passes through its interior, it divides it into two angles whose measures add up to that of the given angle.
  4. Position of the center: let AA, BB and CC be three distinct points of a circle with center OO, and let DD be the other endpoint of the diameter starting from CC. If OO lies on neither of the chords CACA and CBCB, one of two things happens. Either OO lies in the interior of the angle formed at CC by those chords, and then the ray from CC to DD passes through that interior and DD lies on the arc with endpoints AA and BB that does not contain CC. Or OO lies outside that angle, and then one of the points AA and BB, which may be called AA, lies on the arc with endpoints DD and BB that does not contain CC, and the ray from CC to AA passes through the interior of the angle formed at CC by the chords CDCD and CBCB.
TheoremAn inscribed angle measures half the arc it intercepts

Let AA, BB and CC be three distinct points of a circle with center OO and radius rr, and let ss be the length of the arc with endpoints AA and BB that does not contain CC. The inscribed angle γ\gamma formed at CC by the chords CACA and CBCB is half the central angle corresponding to that arc.

γ=s2r\gamma = \frac{s}{2r}

The proof first treats the case in which one side of the inscribed angle passes through the center, and reduces to it the other two cases of the admitted fact on the position of the center by means of the diameter starting from CC.

Proof

  1. α=γ\dato{i}{\alpha = \gamma}

    Suppose first that the center OO lies on the chord CBCB, which is then a diameter (the case where it lies on the chord CACA is the same). Let α\alpha be the angle at AA of the triangle with vertices OO, CC and AA, which is a genuine triangle because AA, distinct from CC and from BB, does not lie on the line joining them; that triangle is isosceles, because OAOA and OCOC are radii, and its angles at CC and at AA are equal.

  2. δ=π−(π−α−γ)=α+γ=2γ<π\delta = \pi - \left( \pi - \alpha - \gamma \right) = \resaltar{\alpha + \gamma} = \dato{d}{\resaltar{2\gamma}} < \pi

    Let δ\delta be the central angle formed by the radii OAOA and OBOB. Since CC, OO and BB are collinear, δ\delta and the angle at OO of the triangle add up to π\pi; since the three angles of the triangle also add up to π\pi, δ\delta is the sum of the other two. Moreover, the angle at OO is positive, so that α+γ\alpha + \gamma is less than π\pi, and so is δ\delta.

  3. s=rδ=2rγ→γ=s2rs = r \resaltar{\delta} = \resaltar{2r\gamma} \rightarrow \dato{c}{\gamma = \frac{s}{2r}}

    The radii OAOA and OBOB divide the circle into two arcs; since δ<π\delta < \pi, by the arc theorem, the one intercepted by δ\delta measures rδr\delta, less than the semicircle, and the other one, more. The point CC, the endpoint of the diameter opposite BB, is a semicircle away from BB in both directions, so it does not lie on the minor arc: this is the arc intercepted by γ\gamma, and s=rδs = r\delta.

  4. γ=s12r+s22r=s1+s22r=s2r\gamma = \frac{s_1}{2r} + \frac{s_2}{2r} = \frac{\resaltar{s_1 + s_2}}{2r} = \frac{\resaltar{s}}{2r}

    If the center lies in the interior of the angle, let DD be the other endpoint of the diameter starting from CC. By the admitted fact on the position of the center, the ray from CC to DD divides the angle into two, each with a side passing through the center, whose measures add up to γ\gamma by the sum of adjacent angles; and DD divides the intercepted arc into two arcs, of lengths s1s_1 and s2s_2, which add up to ss. By the case already treated, applied to each of the two angles, one obtains the following.

  5. γ=s22r−s12r=s2−s12r=s2r\gamma = \frac{s_2}{2r} - \frac{s_1}{2r} = \frac{\resaltar{s_2 - s_1}}{2r} = \frac{\resaltar{s}}{2r}

    If the center lies outside the angle, let DD again be the other endpoint of the diameter starting from CC. By the admitted fact on the position of the center, it may be assumed that AA lies on the arc from DD to BB not passing through CC, and that the ray from CC to AA divides the angle with sides CDCD and CBCB into the one with sides CDCD and CACA and γ\gamma. That arc, of length s2s_2, is made up of the arc from DD to AA, of length s1s_1, and of the arc intercepted by γ\gamma; and γ\gamma is the difference of two angles with a side passing through the center.

CorollaryInscribed angles on the same chord

Two inscribed angles with the same endpoints AA and BB are equal if their vertices lie on the same one of the two arcs that the chord ABAB determines, and supplementary if they lie on different arcs. In particular, every angle inscribed in a semicircle, that is, whose endpoints AA and BB are those of a diameter, is a right angle.

It suffices to read, in each case, which arc each angle intercepts, and to apply the theorem. Let CC and C′C' be the vertices, γ\gamma and γ′\gamma' the angles, and ss the length of the arc with endpoints AA and BB that does not contain CC.

Proof

  1. γ=s2r=γ′\dato{g}{\gamma = \frac{s}{2r}} = \gamma'

    If C′C' lies on the same arc as CC, both angles intercept the other arc, of length ss, and by the theorem they have the same measure.

  2. γ+γ′=s2r+2πr−s2r=π\gamma + \gamma' = \frac{s}{2r} + \frac{\resaltar{2\pi r - s}}{2r} = \resaltar{\pi}

    If C′C' lies on the other arc, it intercepts the arc that contains CC, whose length is what ss lacks to complete the circle; the two measures add up to a straight angle.

  3. s=πr→γ=πr2r=π2s = \pi r \rightarrow \gamma = \frac{\resaltar{\pi r}}{2r} = \resaltar{\frac{\pi}{2}}

    If ABAB is a diameter, both arcs are semicircles, of length πr\pi r, and the angle is a right angle; this particular case is known as Thales's theorem.

Points that do not lie on the circle see the chord under other angles. The following result is also admitted from elementary geometry: it follows from the theorem above and from proposition 21 of book I of Euclid's Elements, according to which a point inside a triangle sees one of its sides under an angle greater than the opposite vertex does; but its proof requires knowing where the line segments starting from the point meet the circle, a question of position that this course does not develop. It will be used in the last solved problem of the lesson on the inverse trigonometric functions.

Admitted fact (angles from interior and exterior points). Let AA and BB be two points of a circle, and γ\gamma the common measure of the inscribed angles with endpoints AA and BB whose vertex lies on one of the two arcs these points determine. If QQ is a point not on the circle, situated on the same side of the line ABAB as that arc, the angle formed at QQ by the line segments QAQA and QBQB is less than γ\gamma if QQ is outside the disk, and greater than γ\gamma if it is inside.

Standard position and coterminal angles

In the Cartesian plane, studied in the Course I lesson on the equation of a line and Cartesian coordinate systems, every angle has a privileged position.

An oriented angle is in standard position if its vertex is the origin and its initial side is the positive xx-semi-axis. Its terminal side is then the ray that starts at the origin and passes through the point P(θ)P(\theta) of the axiom of traversal, that is, the point of the unit circle x2+y2=1x^2 + y^2 = 1 reached by traversing, from (1,0)(1, 0), an arc of length ∣θ∣\abs{\theta} in the direction indicated by the sign of θ\theta. Two angles are coterminal if, in standard position, they have the same terminal side.

The axes divide the plane into four quadrants, numbered counterclockwise starting from the one that contains the points with positive coordinates. An angle in standard position lies in the quadrant that contains its terminal side; if that side falls on an axis, the angle is called quadrantal. The coordinates of the point P(θ)P(\theta) are the cosine and the sine of θ\theta, which will be studied in the lesson that follows this one, on trigonometric ratios.

A full turn, in either direction, returns the side to its position, by the periodicity of the traversal: the angle θ+2π\theta + 2\pi is coterminal with θ\theta, and so are θ−2π\theta - 2\pi, θ+4π\theta + 4\pi and, in general, θ+2kπ\theta + 2k\pi with kk an integer. To know where an arbitrary angle ends, then, it suffices to discount the full turns, as the multiples of the divisor are discounted in the Course I lesson on divisibility and division with remainder. The theorem below is the real version of that division: the divisor is 2π2\pi, the quotient is the number of turns and the remainder, the angle that is left.

TheoremReduction to the first turn

For every real number θ\theta there exist a unique integer kk and a unique real number θ0\theta_0 such that θ=θ0+2kπ\theta = \theta_0 + 2k\pi and 0≤θ0<2π0 \leq \theta_0 < 2\pi.

Existence rests on the Archimedean property, proved in the Course I lesson on the decimal expansion of rational numbers, and uniqueness on the fact that an integer lying strictly between −1-1 and 11 is zero.

Proof

  1. k≤θ2π<k+1\dato{k}{k \leq \frac{\theta}{2\pi} < k + 1}

    Let θ\theta be a real number. By the Archimedean property and the well-ordering principle, as in the proof of the density of the rationals in the lesson on the real line, there exists the least integer mm greater than θ2π\frac{\theta}{2\pi}; let k=m−1k = m - 1, which is no longer greater than that number.

  2. 2kπ≤θ<2kπ+2π→0≤θ−2kπ<2π\resaltar{2k\pi \leq \theta < 2k\pi + 2\pi} \rightarrow \resaltar{0 \leq \theta - 2k\pi < 2\pi}

    We multiply by 2π2\pi, which is positive, and subtract 2kπ2k\pi from all three members.

  3. θ=θ0+2kπ,θ0∈[0,2π)\dato{e}{\theta = \theta_0 + 2k\pi, \quad \theta_0 \in \ico{0}{2\pi}}

    Let θ0=θ−2kπ\theta_0 = \theta - 2k\pi. Then θ\theta consists of kk full turns followed by the angle θ0\theta_0, which lies in the first turn: θ\theta and θ0\theta_0 are coterminal.

  4. 2π(k−j)=θ1−θ0,−2π<θ1−θ0<2π\dato{d}{2\pi(k - j) = \theta_1 - \theta_0}, \quad -2\pi < \theta_1 - \theta_0 < 2\pi

    For uniqueness, suppose that also θ=θ1+2jπ\theta = \theta_1 + 2j\pi, with jj an integer and 0≤θ1<2π0 \leq \theta_1 < 2\pi. Subtracting the two expressions, the difference θ1−θ0\theta_1 - \theta_0 is an integer multiple of 2π2\pi; and since both numbers lie in the interval [0,2π)\ico{0}{2\pi}, that difference lies between −2π-2\pi and 2π2\pi.

  5. ∣k−j∣<1→k=j→θ1=θ0\abs{k - j} < 1 \rightarrow k = j \rightarrow \theta_1 = \theta_0

    Dividing by 2π2\pi, the integer k−jk - j lies strictly between −1-1 and 11: it is zero. Therefore j=kj = k and, by the equality obtained above, θ1=θ0\theta_1 = \theta_0.

  6. θ0=−23π6+4π=π6\theta_0 = \resaltar{-\frac{23\pi}{6} + 4\pi} = \resaltar{\frac{\pi}{6}}

    For example, for θ=−23π6\theta = -\frac{23\pi}{6} we have θ2π=−2312\frac{\theta}{2\pi} = -\frac{23}{12}, which lies between −2-2 and −1-1: thus k=−2k = -2, and the angle makes almost two turns clockwise to end on the side of π6\frac{\pi}{6}.

The number θ0\theta_0 is called the reduction of θ\theta to the first turn. In degrees, the same reasoning, with 360360 instead of 2π2\pi, reduces any angle to one between 0∘0^\circ and 360∘360^\circ, the latter excluded: for example, 1000∘=280∘+2⋅360∘1000^\circ = 280^\circ + 2 \cdot 360^\circ. With the reduction one recognizes all the angles that end on the same side; but first one needs to know that one turn does not pass twice through the same point, which is deduced from the axiom of traversal.

LemmaOne turn does not repeat points

If 0≤α<β<2π0 \leq \alpha < \beta < 2\pi, then P(α)≠P(β)P(\alpha) \neq P(\beta).

The cases are separated according to the half turn in which α\alpha and β\beta lie: in each, the abscissa is strictly monotonic; from one to the other, the sign of the ordinate changes. We write (xs,ys)(x_s, y_s) for the coordinates of P(s)P(s).

Proof

  1. 0<s<π∨π<s<2π→−1<xs<10 < s < \pi \vee \pi < s < 2\pi \rightarrow \dato{x}{-1 < x_s < 1}

    By the half-turn property, as ss increases from 00 to π\pi the abscissa of P(s)P(s) decreases strictly from 11 to −1-1; by the proposition on the second half turn, as ss increases from π\pi to 2π2\pi it increases strictly from −1-1 to the abscissa of P(2π)P(2\pi), which by periodicity is P(0)=(1,0)P(0) = (1, 0).

  2. ys2=1−xs2>0→π<s<2π→ys<0y_s^2 = \resaltar{1 - x_s^2} > 0 \rightarrow \dato{y}{\pi < s < 2\pi \rightarrow y_s < 0}

    In those cases the ordinate is not zero, because its square is 1−xs2>01 - x_s^2 > 0: it is positive in the first half turn, because it is not negative, and negative in the second, because it is not positive.

  3. xα≠xβ→P(α)≠P(β)x_\alpha \neq x_\beta \rightarrow P(\alpha) \neq P(\beta)

    If α\alpha and β\beta both lie in [0,π]\icc{0}{\pi}, or both in [π,2π)\ico{\pi}{2\pi}, the two points have different abscissas, by the strict monotonicity in each half turn.

  4. yβ<0≤yα→P(α)≠P(β)\resaltar{y_\beta < 0 \leq y_\alpha} \rightarrow P(\alpha) \neq P(\beta)

    If α<π<β\alpha < \pi < \beta, the point P(β)P(\beta) has negative ordinate, whereas that of P(α)P(\alpha), in the first half turn, is not negative. In all cases the two points are distinct.

LemmaCoterminal angles

Two angles θ\theta and φ\varphi are coterminal if and only if their difference is an integer multiple of 2π2\pi, that is, if there exists an integer kk such that θ−φ=2kπ\theta - \varphi = 2k\pi. In particular, they are coterminal if and only if they have the same reduction to the first turn.

Each angle is replaced by its reduction, and the lemma proved just before compares the reductions.

Proof

  1. P(θ)=P(θ0),P(φ)=P(φ0)\dato{p}{P(\theta) = P(\resaltar{\theta_0}), \quad P(\varphi) = P(\resaltar{\varphi_0})}

    By the reduction theorem, θ=θ0+2kπ\theta = \theta_0 + 2k\pi and φ=φ0+2jπ\varphi = \varphi_0 + 2j\pi, with θ0\theta_0 and φ0\varphi_0 in [0,2π)\ico{0}{2\pi}; by the periodicity of the traversal, full turns do not change the point, and the terminal side of each angle is that of its reduction.

  2. P(θ)=P(φ)↔θ0=φ0\dato{c}{P(\theta) = P(\varphi) \leftrightarrow \theta_0 = \varphi_0}

    If θ0≠φ0\theta_0 \neq \varphi_0, by the lemma proved just before the points P(θ0)P(\theta_0) and P(φ0)P(\varphi_0) are distinct, and so are the terminal sides; if they are equal, the sides coincide. Therefore the angles are coterminal if and only if they have the same reduction.

  3. θ0=φ0→θ−φ=2(k−j)π,θ−φ=2mπ→θ=φ0+2(j+m)π→θ0=φ0\theta_0 = \varphi_0 \rightarrow \theta - \varphi = \resaltar{2(k - j)\pi}, \qquad \theta - \varphi = 2m\pi \rightarrow \theta = \varphi_0 + \resaltar{2(j + m)\pi} \rightarrow \theta_0 = \varphi_0

    If the reductions are equal, the difference is an integer multiple of 2π2\pi. Conversely, if θ−φ=2mπ\theta - \varphi = 2m\pi, then θ=φ0+2(j+m)π\theta = \varphi_0 + 2(j + m)\pi, and by the uniqueness of the reduction, θ0=φ0\theta_0 = \varphi_0.

It follows that the angles coterminal with θ\theta are exactly those of the form θ+2kπ\theta + 2k\pi, with kk an integer, and that the quadrant of an angle is decided by its reduction: it lies in the first quadrant if 0<θ0<π20 < \theta_0 < \frac{\pi}{2}, in the second if π2<θ0<π\frac{\pi}{2} < \theta_0 < \pi, in the third if π<θ0<3π2\pi < \theta_0 < \frac{3\pi}{2} and in the fourth if 3π2<θ0<2π\frac{3\pi}{2} < \theta_0 < 2\pi; it is quadrantal if θ0\theta_0 is 00, π2\frac{\pi}{2}, π\pi or 3π2\frac{3\pi}{2}. Thus −23π6-\frac{23\pi}{6} lies in the first quadrant, and 1000∘1000^\circ, whose reduction is 280∘280^\circ, in the fourth.

Finally, every point of the unit circle is the point P(θ)P(\theta) of some angle: this is the covering property of the axiom of traversal, which says that in traversing the circle from (1,0)(1, 0) one passes through all its points. If this is so, by the reduction to the first turn, periodicity and the lemma on coterminal angles, each point of the unit circle is P(θ)P(\theta) for a unique θ\theta in [0,2π)\ico{0}{2\pi}, and the angles corresponding to it are exactly those coterminal with that θ\theta.

Angular velocity and linear velocity

When a body rotates about an axis, all its points describe arcs of circles centered on the axis, and all of them turn through the same angle in the same time. If in a time tt the body turns, at a constant rate, through an angle θ\theta, its angular velocity is the ratio ω=θt\omega = \frac{\theta}{t}, which is measured in radians per second; it is also expressed in revolutions per minute, and since one revolution is 2π2\pi radians and one minute, 6060 seconds, one revolution per minute is equivalent to π30\frac{\pi}{30} radians per second.

A point at distance rr from the axis traverses, in that time, an arc of length s=rθs = r\theta, by the arc length theorem. Its linear velocity, the ratio of the distance traveled to the time, is therefore the product of the radius and the angular velocity:

v=st=rθt=rωv = \frac{s}{t} = \frac{r\theta}{t} = r\omega

It follows that, in a rotating body, the angular velocity is the same for all points, whereas the linear velocity grows in proportion to the distance from the axis: the rim of a spinning disk moves faster than the points near the center. This relation, valid only if angles are measured in radians, is the starting point of the study of circular motion in the course Foundations of Mechanics. Two wheels joined by a belt or a chain that does not slip, or that mesh with each other, have the same linear velocity at the point of contact; therefore ω1r1=ω2r2\omega_1 r_1 = \omega_2 r_2, and the smaller wheel turns faster, as will be seen in solved problem 3.

Solved problems

The problems measure the Earth from an angle, compute the area of a circular segment, transmit a rotation through a chain, look for the sector of greatest area with a given perimeter and count the overlaps of the hands of a clock. Each one calls for an idea of its own, which is worth looking for before reading the solution.

Solved problem 1The measurement of the Earth according to Eratosthenes

Around the year 240 before the common era, Eratosthenes observed that at noon on the summer solstice the Sun lit up the bottom of the wells of Syene, whereas in Alexandria, located to the north and, as he assumed, on the same meridian, the rays formed with the vertical an angle of 7 degrees and 12 minutes. Taking 800 km as the distance between the two cities, compute the radius of the Earth and the length of the meridian, the length of the arc of meridian that corresponds to one minute, and the distance, along the meridian, from the equator to a point of latitude 33∘33^\circ.

Solution

  1. α=7∘+(1260)∘=(365)∘\alpha = 7^\circ + \resaltar{\left( \frac{12}{60} \right)^\circ} = \resaltar{\left( \frac{36}{5} \right)^\circ}

    The angle, expressed in degrees, is 77 plus twelve sixtieths; it is the fiftieth part of the turn, because 365⋅50=360\frac{36}{5} \cdot 50 = 360.

  2. θ=365π180=π25\theta = \frac{\resaltar{\frac{36}{5}} \pi}{180} = \dato{t}{\resaltar{\frac{\pi}{25}}}

    Since the rays of the Sun arrive parallel and at Syene they fall in the direction of the Earth's radius, the angle at Alexandria between the ray and the vertical is equal to the angle formed at the center of the Earth by the radii through the two cities: they are alternate angles between parallels. In radians, by the conversion rule, it is as follows.

  3. R=sθ=800π25=20000πR = \frac{s}{\theta} = \frac{\resaltar{800}}{\resaltar{\frac{\pi}{25}}} = \dato{r}{\resaltar{\frac{20000}{\pi}}}

    The 800 km are the arc of meridian intercepted by that central angle. From s=Rθs = R\theta one solves for the radius, approximately 6366 km.

  4. 2πR=2π⋅20000π=400002\pi R = 2\pi \cdot \resaltar{\frac{20000}{\pi}} = \resaltar{40000}

    The length of the meridian is that of the circle of radius RR; it is also 5050 times the distance between the two cities.

  5. s=20000π⋅π10800=5027s = \frac{20000}{\pi} \cdot \resaltar{\frac{\pi}{10800}} = \resaltar{\frac{50}{27}}

    A minute is the sixtieth part of a degree, that is, π10800\frac{\pi}{10800} radians. Its arc measures about 1.852 km: this is the nautical mile, which was defined precisely in this way.

  6. s=20000π⋅11π60=110003s = \frac{20000}{\pi} \cdot \resaltar{\frac{11\pi}{60}} = \resaltar{\frac{11000}{3}}

    The latitude of a point is the angle that its radius forms with the plane of the equator; 33∘33^\circ is 11π60\frac{11\pi}{60} radians. The required distance is about 3667 km.

Eratosthenes expressed the distance in stadia, whose exact value is disputed; with the figure adopted here, the meridian measures 40000 km, and not by chance: at the end of the eighteenth century the meter was defined as the ten-millionth part of the quarter meridian. On the other hand, Alexandria is not exactly north of Syene but some three degrees farther west, so the arc joining them is not quite an arc of meridian.

Solved problem 2The circular segment of sixty degrees

A chord with endpoints AA and BB, on a circle with center OO and radius rr, subtends a central angle of π3\frac{\pi}{3}. Compute the area of the circular segment bounded by the chord and the minor arc with endpoints AA and BB, and the fraction of the disk that it occupies.

Solution

  1. As=12r2⋅π3=πr26A_s = \frac{1}{2} r^2 \cdot \resaltar{\frac{\pi}{3}} = \dato{s}{\resaltar{\frac{\pi r^2}{6}}}

    The circular segment is what remains of the sector with vertex OO and endpoints AA and BB when the triangle with vertices OO, AA and BB is removed from it. The area of the sector is given by the sector theorem.

  2. π−π32=π3\frac{\resaltar{\pi - \frac{\pi}{3}}}{2} = \resaltar{\frac{\pi}{3}}

    That triangle has two sides equal to the radius: it is isosceles, and its angles at AA and at BB are equal. Since the three angles of a triangle add up to π\pi, by the fact admitted in this lesson, each of them measures π3\frac{\pi}{3}, and the triangle is equilateral, with side rr.

  3. h=32r,At=12r⋅32r=34r2h = \frac{\sqrt{3}}{2} r, \quad A_t = \frac{1}{2} r \cdot \resaltar{\frac{\sqrt{3}}{2} r} = \dato{t}{\resaltar{\frac{\sqrt{3}}{4} r^2}}

    The altitude from OO meets the base at its midpoint; by the Pythagorean theorem, its square is r2−r24r^2 - \frac{r^2}{4}. The area of the triangle is half the base times the altitude.

  4. A=πr26−34r2=2π−3312r2A = \resaltar{\frac{\pi r^2}{6} - \frac{\sqrt{3}}{4} r^2} = \dato{a}{\resaltar{\frac{2\pi - 3\sqrt{3}}{12} r^2}}

    The area of the circular segment is the difference, with common denominator 1212; it is positive, because 2π>6>332\pi > 6 > 3\sqrt{3}.

  5. Aπr2=2π−3312π\frac{A}{\pi r^2} = \frac{2\pi - 3\sqrt{3}}{12\pi}

    Dividing by the area of the disk, the circular segment occupies approximately 2.9 percent of the disk; the sector occupies one sixth of it, and the triangle fills about 83 percent of the sector.

Solved problem 3The chain of a bicycle

On a bicycle, the chain connects the chainring, of radius 1010 cm, driven by the pedals, with the sprocket, of radius 44 cm, fixed to the axle of the rear wheel, whose radius is 3535 cm. If the cyclist makes one turn of the pedals per second, compute the angular velocity of the wheel and the speed of the bicycle, assuming that the chain does not slip and that the wheel rolls without skidding.

Solution

  1. ω1=2π\omega_1 = \dato{w}{2\pi}

    One turn per second is 2π2\pi radians per second: that is the angular velocity of the chainring.

  2. v=r1ω1=10⋅2π=20πv = r_1 \omega_1 = 10 \cdot \resaltar{2\pi} = \dato{v}{\resaltar{20\pi}}

    The whole chain moves with one and the same speed, which is the linear velocity of the rim of the chainring, in centimeters per second.

  3. ω2=vr2=20π4=5π\omega_2 = \frac{v}{r_2} = \frac{\resaltar{20\pi}}{\resaltar{4}} = \dato{p}{\resaltar{5\pi}}

    The chain drives the rim of the sprocket with the same linear velocity; from ω1r1=ω2r2\omega_1 r_1 = \omega_2 r_2 one obtains the angular velocity of the sprocket: two and a half turns per second.

  4. vr=35⋅5π=175πv_r = 35 \cdot \resaltar{5\pi} = \dato{b}{\resaltar{175\pi}}

    The wheel is fixed to the axle of the sprocket and turns with the same angular velocity, but its rim is much farther from the axle, and its linear velocity is much greater, in centimeters per second.

  5. 175π⋅3600100000=63π10175\pi \cdot \resaltar{\frac{3600}{100000}} = \resaltar{\frac{63\pi}{10}}

    If the wheel does not skid, in each turn the bicycle advances the circumference of the wheel, which is laid down successively on the ground: its speed is that of the rim of the wheel. One centimeter per second is 3600100000\frac{3600}{100000} kilometers per hour, so the bicycle travels at about 19.8 kilometers per hour.

Solved problem 4The sector of fixed perimeter and maximum area

With a wire of length PP one forms the boundary of a circular sector: two radii and the arc that joins them. Find the radius and the angle of the sector of greatest area, and the value of that area.

Solution

  1. P=2r+s→s=P−2rP = 2r + s \rightarrow \dato{s}{s = P - 2r}

    If the radius is rr and the angle θ\theta, the boundary consists of two radii and an arc of length s=rθs = r\theta.

  2. A=12rs=12r(P−2r)=Pr2−r2A = \frac{1}{2} r \resaltar{s} = \frac{1}{2} r \resaltar{(P - 2r)} = \dato{a}{\frac{Pr}{2} - r^2}

    By the sector theorem, the area is half the radius times the arc: it is a quadratic function of the radius.

  3. P2+2π≤r<P2\dato{d}{\frac{P}{2 + 2\pi} \leq r < \frac{P}{2}}

    The arc must be positive and not greater than the whole circle, that is, 0<θ≤2π0 < \theta \leq 2\pi; from P−2r>0P - 2r > 0 and P−2r≤2πrP - 2r \leq 2\pi r one obtains the interval of possible radii.

  4. A=P216−(r−P4)2≤P216A = \resaltar{\frac{P^2}{16} - \left( r - \frac{P}{4} \right)^2} \leq \resaltar{\frac{P^2}{16}}

    Completing the square, the area is a fixed number minus a square. Its graph, as a function of rr, is the parabola (r−P4)2=−(A−P216)\left( r - \frac{P}{4} \right)^2 = -\left( A - \frac{P^2}{16} \right), which opens downward; by the vertex theorem of the Course I lesson on characterizing parabolas, its highest point is the vertex.

  5. r=P4,A=P216r = \frac{P}{4}, \quad A = \frac{P^2}{16}

    The abscissa of the vertex, P4\frac{P}{4}, belongs to the interval of possible radii, because 2+2π>42 + 2\pi > 4. Therefore the maximum area is attained exactly with that radius.

  6. θ=sr=P−P2P4=2\theta = \frac{s}{r} = \frac{\resaltar{P - \frac{P}{2}}}{\frac{P}{4}} = \resaltar{2}

    The corresponding arc measures P2\frac{P}{2}, and the angle is the ratio of the arc to the radius: 22 radians, about 114.6 degrees, whatever PP may be. Curiously, the maximum area is that of the square of perimeter PP.

Solved problem 5The hands of the clock

Compute the angle formed by the hands of a clock at 3:40. Then prove that, between 12:00 and the following 12:00 (the latter excluded), the hands overlap exactly eleven times, separated by equal intervals of 1211\frac{12}{11} hours.

Solution

  1. αm=360t,αh=30t\dato{p}{\alpha_m = 360t, \quad \alpha_h = 30t}

    We measure the position of each hand by the angle, in degrees and clockwise, through which it has turned since 12:00. After tt hours, the minute hand has made tt turns, and the hour hand, which makes one turn every 12 hours, has turned 30∘30^\circ per hour.

  2. αm=360⋅113=1320∘=240∘+3⋅360∘\alpha_m = 360 \cdot \frac{11}{3} = \resaltar{1320^\circ} = \resaltar{240^\circ + 3 \cdot 360^\circ}

    At 3:40, 3+4060=1133 + \frac{40}{60} = \frac{11}{3} hours have passed. The minute hand has made three full turns, which do not change its position, and it stands at the reduction of its angle.

  3. αh=30⋅113=110∘\alpha_h = 30 \cdot \resaltar{\frac{11}{3}} = \resaltar{110^\circ}

    The hour hand has not completed any turn.

  4. 240∘−110∘=130∘=13π18240^\circ - 110^\circ = \resaltar{130^\circ} = \resaltar{\frac{13\pi}{18}}

    The difference of the reduced positions is less than 180∘180^\circ, so it is the angle formed by the hands, in degrees and in radians.

  5. 360t−30t=330t=360k↔t=12k11,k∈Z360t - 30t = 330t = 360k \leftrightarrow \dato{t}{t = \frac{12k}{11}}, \quad k \in \mathbb{Z}

    The hands overlap when their angles are coterminal, that is, by the lemma on coterminal angles, when their difference is an integer multiple of a turn.

  6. 0≤12k11<12↔0≤k<110 \leq \frac{12k}{11} < 12 \leftrightarrow 0 \leq k < 11

    The condition 0≤t<120 \leq t < 12 is equivalent to 0≤k<110 \leq k < 11: there are exactly eleven overlaps, for k=0,1,…,10k = 0, 1, \ldots, 10, and two consecutive ones are 1211\frac{12}{11} hours apart, approximately one hour, 5 minutes and 27 seconds. The positions of the hour hand, 360k11\frac{360k}{11} degrees, are the vertices of a regular hendecagon.

Proposed problems

The problems follow the ideas of the lesson: turning a sector into a cone, estimating the length of a spiral by adding up circles, decomposing a band into straight portions and arcs whose central angles add up, and reducing angles to the first turn by means of coterminal angles in order to count positions and instants. Each one has a hint, which is best opened only after having attempted the problem; those that ask for a proof also have a worked solution.

Proposed problem 1A paper cone

A circular sector of radius 1212 cm and angle 120∘120^\circ is cut out of a sheet, and its two radii are glued together to form the lateral surface of a cone. Compute the radius of the base and the height of the cone.

Hint

The arc of the sector becomes the circumference of the base, and the radius of the sector becomes the distance from the apex of the cone to any point of that circumference. The height is then obtained with the Pythagorean theorem.

Answer

The radius of the base is 44 cm and the height, 828\sqrt{2} cm.

Proposed problem 2The track of a compact disc

On a compact disc, the information is recorded along a single spiral track, which begins 2.3 cm from the center and ends at 5.8 cm; two consecutive turns of the spiral are 1.6 micrometers apart. The reading head follows the track with a constant linear velocity of 1.2 meters per second. Estimate the length of the track and the duration of a complete reading, approximating each turn of the spiral by a circle centered on the axis of the disc.

Hint

If rr and RR are the initial and final radii and pp is the separation between turns, there are N=R−rpN = \frac{R - r}{p} turns, of radii rr, r+pr + p, r+2pr + 2p, and so on. Their lengths form an arithmetic progression, whose sum was computed in the Course I lesson on summations, progressions and induction; the result is very close to the area of the annulus divided by the pitch. Take care with the units.

Answer

The spiral makes 2187521875 turns, and its length is approximately as follows, about 5.57 km; the complete reading lasts about 4640 seconds, that is, about 77 minutes.

L≈π(R2−r2)pL \approx \frac{\pi \left( R^2 - r^2 \right)}{p}
Proposed problem 3The band that ties three pipes

Three equal cylindrical pipes, of radius rr, are placed parallel, each resting on the other two, and are tied with a taut band that goes around them. Prove that the band measures 6r+2πr6r + 2\pi r, and compute its length for pipes of radius 5 cm.

Hint

In a cross-section perpendicular to the pipes, each straight portion of the band forms a rectangle with the radii drawn to its points of tangency and with a side of the triangle of the centers. For the arcs, compute the central angle that the band wraps around on each pipe, and use the sum of the angles of a triangle.

Answer

The band consists of three straight portions of length 2r2r and of three arcs that together form a complete circle. For r=5r = 5 it measures 30+10π30 + 10\pi cm, about 61.4 cm.

Worked solution

Solution

  1. d⁡(A,B)=d⁡(B,C)=d⁡(C,A)=2r\operatorname{d}(A, B) = \operatorname{d}(B, C) = \operatorname{d}(C, A) = \dato{l}{2r}

    The bundle is cut by a plane perpendicular to the pipes: three circles of radius rr are seen, with centers AA, BB and CC. Two externally tangent circles have their centers at a distance equal to the sum of the radii, so the triangle of the centers is equilateral, with side 2r2r.

  2. d⁡(T,T′)=d⁡(A,B)=2r→3⋅2r=6r\operatorname{d}(T, T') = \resaltar{\operatorname{d}(A, B)} = \resaltar{2r} \rightarrow \dato{t}{3 \cdot 2r = 6r}

    Each straight portion of the band touches two pipes, for example those with centers AA and BB, at two points TT and T′T'. The radius drawn to the point of tangency is perpendicular to the tangent, according to the Course I lesson on the circle; the radii ATAT and BT′BT' are perpendicular to the same line, equal and on the same side of it, and the quadrilateral with vertices AA, BB, T′T' and TT is a rectangle. Each straight portion has the same length as a side of the triangle.

  3. π2+α+π2+φA=2π→φA=π−α\frac{\pi}{2} + \alpha + \frac{\pi}{2} + \varphi_A = 2\pi \rightarrow \dato{f}{\varphi_A = \pi - \alpha}

    On the pipe with center AA, the band wraps around the arc lying between the points of tangency of the two straight portions that reach it. The radii drawn to those points form right angles with the sides ABAB and ACAC; together with the angle α\alpha of the triangle at AA and the central angle φA\varphi_A of the arc, they complete a full turn around AA, by the sum of adjacent angles.

  4. r(π−α)+r(π−β)+r(π−γ)=r(3π−(α+β+γ))=r(3π−π)=2πrr(\pi - \alpha) + r(\pi - \beta) + r(\pi - \gamma) = r\left( 3\pi - \resaltar{(\alpha + \beta + \gamma)} \right) = r\left( 3\pi - \resaltar{\pi} \right) = \dato{a}{2\pi r}

    The same happens on the other two pipes, with the angles β\beta and γ\gamma of the triangle. By the arc length theorem, the three arcs add up to the following, and since the angles of the triangle add up to π\pi, together they form a complete circle; here each one is a third of it, because α=β=γ=π3\alpha = \beta = \gamma = \frac{\pi}{3}.

  5. L=6r+2πr,r=5→L=30+10πL = \resaltar{6r} + \resaltar{2\pi r}, \qquad r = 5 \rightarrow L = 30 + 10\pi

    The band measures the sum of the straight portions and of the arcs. The computation of the arcs does not use the fact that the triangle is equilateral: the curved part of a taut band around three equal pipes always adds up to a complete circle. For r=5r = 5, the band measures about 61.4 cm.

Proposed problem 4A wheel that advances by jumps

A toothed wheel turns, at each jump, through an angle of 150∘150^\circ counterclockwise. Prove that it returns for the first time to its initial position after 1212 jumps, and that during those jumps it occupies 1212 distinct positions. Generalize: if each jump is a fraction pq\frac{p}{q} of the turn, with pp and qq coprime, how many distinct positions does it occupy?

Hint

By the lemma on coterminal angles, jumps jj and kk leave the wheel in the same position if and only if 150(k−j)150(k - j) is a multiple of 360360, that is, if 1212 divides 5(k−j)5(k - j). Use the fact that 55 and 1212 are coprime, as in the Course I lesson on the greatest common divisor and the least common multiple.

Answer

Two jumps give the same position if and only if their difference is a multiple of 1212; in general, the wheel occupies qq distinct positions, and returns to the initial one for the first time at jump qq.

Worked solution

Solution

  1. 150(k−j)=360m↔5(k−j)=12m,m∈Z150(k - j) = 360m \leftrightarrow \dato{c}{5(k - j) = 12m}, \quad m \in \mathbb{Z}

    After kk jumps, the wheel has turned 150k150k degrees. By the lemma on coterminal angles, jumps jj and kk leave it in the same position if and only if the difference of the angles is an integer multiple of a turn; dividing by 3030, the condition is that 5(k−j)5(k - j) be a multiple of 1212.

  2. 60∣5(k−j)↔12∣k−j\resaltar{60 \mid 5(k - j)} \leftrightarrow \dato{d}{\resaltar{12 \mid k - j}}

    The integer 5(k−j)5(k - j) is then a multiple of 1212 and of 55, which are coprime; by the lemma on common multiples of the Course I lesson on the greatest common divisor, it is a multiple of 6060, and k−jk - j is a multiple of 1212. Conversely, if k−j=12tk - j = 12t, the angles differ by 5t5t turns.

  3. 0≤j<k≤11→0<k−j<12,150⋅12=5⋅3600 \leq j < k \leq 11 \rightarrow 0 < k - j < 12, \qquad 150 \cdot \resaltar{12} = \resaltar{5 \cdot 360}

    Between jumps 00 and 1111, no difference is a multiple of 1212: the wheel occupies 1212 distinct positions. At jump 1212 it has turned 1800∘1800^\circ, five full turns, and returns for the first time to the initial position; from there on it repeats the same positions in the same order.

  4. q∣p(k−j)↔q∣k−jq \mid p(k - j) \leftrightarrow \resaltar{q \mid k - j}

    In general, if each jump is the fraction pq\frac{p}{q} of the turn, jumps jj and kk give the same position if and only if p(k−j)q\frac{p(k - j)}{q} is an integer, that is, if qq divides p(k−j)p(k - j). With pp and qq coprime, the same lemma gives that pqpq divides p(k−j)p(k - j), and qq divides k−jk - j: the wheel occupies exactly qq positions and returns to the initial one for the first time at jump qq. The fraction 150360=512\frac{150}{360} = \frac{5}{12} gives the case treated before.

Proposed problem 5The hands at a right angle

Prove that, between 12:00 and the following 12:00 (the latter excluded), the hands of a clock form a right angle exactly 2222 times, and find those instants.

Hint

With the notation of solved problem 5, the hands form a right angle when the difference 330t330t is coterminal with 90∘90^\circ or with 270∘270^\circ, that is, when 330t=90+180j330t = 90 + 180j for some integer jj.

Answer

The instants are the following, in hours after 12:00, with j=0,1,…,21j = 0, 1, \ldots, 21; the first is around 12:16, and among them are 3:00 and 9:00.

t=3(2j+1)11t = \frac{3(2j + 1)}{11}
Worked solution

Solution

  1. αm−αh=360t−30t=330t\alpha_m - \alpha_h = 360t - 30t = \dato{d}{330t}

    As in solved problem 5, after tt hours the minute hand has turned 360t360t degrees and the hour hand 30t30t, clockwise from 12:00; the difference of their positions is 330t330t degrees.

  2. 330t=90+360m∨330t=270+360m↔330t=90+180j,j∈Z330t = 90 + 360m \vee 330t = 270 + 360m \leftrightarrow \dato{e}{330t = 90 + 180j}, \quad j \in \mathbb{Z}

    The hands form a right angle if and only if that difference is coterminal with 90∘90^\circ or with 270∘270^\circ, that is, if it is 90+360m90 + 360m or 270+360m270 + 360m for some integer mm. The two families together are the numbers 90+180j90 + 180j: the even values of jj give the first and the odd ones, the second.

  3. t=90+180j330=3(2j+1)11t = \frac{\resaltar{90 + 180j}}{\resaltar{330}} = \dato{t}{\frac{3(2j + 1)}{11}}

    Solving for tt and simplifying the fraction by 3030, the instants are obtained.

  4. 0≤3(2j+1)11<12↔0≤j≤21,j=5→t=3,j=16→t=90 \leq \frac{3(2j + 1)}{11} < 12 \leftrightarrow \resaltar{0 \leq j \leq 21}, \qquad j = 5 \rightarrow t = 3, \quad j = 16 \rightarrow t = 9

    The condition 0≤t<120 \leq t < 12 is equivalent to 0≤2j+1<440 \leq 2j + 1 < 44, that is, to 0≤j≤210 \leq j \leq 21: there are exactly 2222 instants, distinct because tt increases with jj. With j=0j = 0 one obtains t=311t = \frac{3}{11}, about 16 minutes and 22 seconds; with j=5j = 5 and j=16j = 16, 3:00 and 9:00. The corresponding positions of the hour hand, 90(2j+1)11\frac{90(2j + 1)}{11} degrees, are regularly distributed around the dial.

With this lesson, angles are measured and placed in the plane. In the lesson that follows, on trigonometric ratios, the coordinates of the point P(θ)P(\theta) of the unit circle will define the cosine and the sine of any angle, and the reduction to the first turn and coterminal angles will explain their periodicity.