Chapter 15 Measuring an Angle

Chapter 15

Measuring an Angle

So far, the equations we have studied have an algebraic Cosmo S character, involving the variables x and y, arithmetic op- motors around erations and maybe extraction of roots. Restricting our Q the attention to such equations would limit our ability to de- circle R scribe certain natural phenomena. An important example P 20 feet involves understanding motion around a circle, and it can be motivated by analyzing a very simple scenario: Cosmo the dog, tied by a 20 foot long tether to a post, begins Figure 15.1: Cosmo the dog walking around a circle. A number of very natural ques- walking a circular path. tions arise:

Natural Questions 15.0.1. How can we measure the angles ∠SPR, ∠QPR, and ∠QPS? How can we measure the arc lengths arc(RS), arc(SQ) and arc(RQ)? How can we measure the rate Cosmo is moving around the circle? If we know how to measure angles, can we compute the coordinates of R, S, and Q? Turning this around, if we know how to compute the coordinates of R, S, and Q, can we then measure the angles ∠SPR, ∠QPR, and ∠QPS ? Finally, how can we specify the direction Cosmo is traveling?

We will answer all of these questions and see how the theory which evolves can be applied to a variety of problems. The deﬁnition and basic properties of the circular functions will emerge as a central theme in this Chapter. The full problem-solving power of these functions will become apparent in our discussion of sinusoidal functions in Chapter 19. The xy-coordinate system is well equipped to study straight line motion between two locations. For problems of this sort, the important tool is the distance formula. However, as Cosmo has illustrated, not all two-dimensional motion is along a straight line. In this section, we will describe how to calculate length along a circular arc, which requires a quick review of angle measurement. 191 192 CHAPTER 15. MEASURING AN ANGLE 15.1 Standard and Central Angles

An angle is the union of two rays emanating from a common point called the vertex of the angle. A typical angle can be dynamically generated by rotating a single ray from one position to another, sweeping counterclockwise or clockwise: See Figure 15.2. We often insert a curved arrow to indicate the direction in which we are sweeping out the angle. The ray ℓ1 is called the initial side and ℓ2 the terminal side of the angle ∠AOB.

(terminal side) l2

B SWEEP CLOCKWISE

vertex vertex A (initial side) l1 l1 l1 O A O A (initial side) O (terminal side) START SWEEP COUNTERCLOCKWISE B l2

Figure 15.2: Angle ∠AOB.

Working with angles, we need to agree on a standard frame of refer- ence for viewing them. Within the usual xy-coordinate system, imagine a model of ∠AOB in Figure 15.2 constructed from two pieces of rigid wire, welded at the vertex. Sliding this model around inside the xy-plane will not distort its shape, only its position relative to the coordinate axis. So, we can slide the angle into position so that the initial side is coincident with the positive x-axis and the vertex is the origin. Whenever we do this, we say the angle is in standard position. Once an angle is in standard position, we can construct a circle centered at the origin and view our standard angle as cutting out a particular “pie shaped wedge” of the corresponding disc. Notice, the shaded regions in Figure 15.3 depend on whether we sweep the angle counterclockwise or clockwise from the initial side. The portion of the “pie wedge” along the circle edge, which is an arc, is called the arc subtended by the angle. We can keep track of this arc using the notation arc(AB). A central angle is any angle with vertex at the center of a circle, but its initial side may or may not be the positive x-axis. For example, ∠QPS in the Figure beginning this Chapter is a central angle which is not in standard position. 15.2. AN ANALOGY 193

y-axis l2 l2 y-axis B B arc subtended arc(AB) arc(AB)

vertex vertex x-axis x-axis O l1 O l A A 1

COUNTERCLOCKWISE CLOCKWISE

Figure 15.3: Standard angles and arcs.

15.2 An Analogy

To measure the dimensions of a box you would use a ruler. In other words, you use an instrument (the ruler) as a standard against which you measure the box. The ruler would most likely be divided up into either English units (inches) or metric units (centimeters), so we could express the dimensions in a couple of different ways, depending on the units desired. By analogy, to measure the size of an angle, we need a standard against which any angle can be compared. In this section, we will describe two standards commonly used: the degree method and the radian method of angle measurement. The key idea is this: Beginning with a circular region, describe how to construct a “basic” pie shaped wedge whose interior angle becomes the standard unit of angle measurement.

15.3 Degree Method

Begin by drawing a circle of radius r, call it Cr, centered at the origin. Divide this circle into 360 equal sized pie shaped wedges, beginning with the the point (r,0) on the circle; i.e. the place where the circle crosses the x-axis. We will refer to the arcs along the outside edges of these wedges as one-degree arcs. Why 360 equal sized arcs? The reason for doing so is historically tied to the fact that the ancient Babylonians did so as they developed their study of astronomy. (There is actually an alternate system based on dividing the circular region into 400 equal sized wedges.) Any central angle which subtends one of these 360 equal sized arcs is 194 CHAPTER 15. MEASURING AN ANGLE

circle Cr

etc. a total of 360 this angle is equal sized r DEFINED to pie shaped typical have measure wedges inside (r,0) wedge 1 degree this disk r

etc.

***NOT TO SCALE***

Figure 15.4: Wedges as 1◦ arcs.

defined to have a measure of one degree, denoted 1◦. We can now use one-degree arcs to measure any angle: Begin by sliding the angle ∠AOB into standard central position, as in Figure 15.3. Piece together consecutive one-degree arcs in a counterclockwise or clockwise direction, beginning from the initial side and working toward the terminal side, approximating the angle ∠AOB to the nearest degree. If we are allowed to divide a one-degree arc into a fractional portion, then we could precisely determine the number m of one-degree arcs which consecutively fit together into the given arc. If we are sweeping counterclockwise from the initial side of the angle, m is defined to be the degree measure of the angle. If we sweep in a clockwise direction, then −m is defined to be the degree measure of the angle. So, in Figure 15.3, the left-hand angle has positive degree measure while the right-hand angle has negative degree measure. Simple examples would be angles like the ones in Figure 15.5. Notice, with our conventions, the rays determining an angle with measure −135◦ sit inside the circle in the same position as those for an angle of measure 225◦; the minus sign keeps track of sweeping the positive x-axis clockwise (rather than counterclockwise). We can further divide a one-degree arc into 60 equal arcs, each called a one minute arc. Each one-minute arc can be further divided into 60 equal arcs, each called a one second arc. This then leads to angle measures of one minute, denoted 1′ and one second, denoted 1′′:

1◦ = 60 minutes = 3600 seconds. 15.3. DEGREE METHOD 195

180◦ 90◦ 270◦

45◦ 315◦

−135◦

Figure 15.5: Examples of common angles.

For example, an angle of measure θ = 5 degrees 23 minutes 18 seconds is usually denoted 5◦ 23′ 18′′. We could express this as a decimal of degrees:

In degrees! 23 18 ◦ 5◦ 23′ 18′′ = 5 + + 60 3600 ← = 5.3883◦.

As another example, suppose we have an angle with measure 75.456◦ and we wish to convert this into degree/minute/second units. First, since 75.456◦ = 75◦ +0.456◦, we need to write 0.456◦ in minutes by the calculation: minutes 0.456 degree 60 = 27.36′. × degree

This tells us that 75.456◦ = 75◦27.36′ = 75◦27′ + 0.36′. Now we need to write 0.36′ in seconds via the calculation:

0.36 minutes 60 seconds/minute = 21.6′′. ×

In other words, 75.456◦ = 75◦ 27′ 21.6′′. Degree measurement of an angle is very closely tied to direction in the plane, explaining its use in map navigation. With some additional work, it is also possible to relate degree measure and lengths of circular arcs. To do this carefully, ﬁrst go back to Figure 15.3 and recall the situation 196 CHAPTER 15. MEASURING AN ANGLE

where an arc arc(AB) is subtended by the central angle ∠AOB. In this situation, the arc length of arc(AB), commonly denoted by the letter s, is the distance from A to B computed along the circular arc; keep in mind, this is NOT the same as the straight line distance between the points A and B. For example, consider the six angles pictured above, of measures 90◦, 180◦, 270◦, 45◦, −135◦, and 315◦. If the circle is of radius r and we want to compute the lengths of the arcs subtended by these six angles, then this can be done using the formula for the circumference of a circle (on the back of this text) and the following general principle:

Important Fact 15.3.1.

(length of a part) = (fraction of the part) (length of the whole) × For example, the circumference of the entire circle of radius r is 2πr; this is the “length of the whole” in the general principle. The arc sub- 90 1 tended by a 90◦ angle is 360 = 4 of the entire circumference; this is the “fraction of the part” in the general principle. The boxed formula implies:

1 πr s = arc length of the 90◦ arc = 2πr = . 4 2 s = distance along the arc Similarly, a 180◦ angle subtends an arc of length s = πr, 315 7πr r a 315◦ angle subtends an arc of length s = 360 2πr = 4 , etc. In general, we arrive at this formula: θ degrees

r Important Fact 15.3.2 (Arc length in degrees). Start with a central angle of measure θ degrees inside a circle of radius r. Then this angle will subtend an arc of length Cr Figure 15.6: The deﬁnition 2π of arc length. s = rθ 360 15.4 Radian Method

The key to understanding degree measurement was the description of a “basic wedge” which contained an interior angle of measure 1◦; this was straightforward and familiar to all of us. Once this was done, we could proceed to measure any angle in degrees and compute arc lengths as in Fact 15.3.2. However, the formula for the length of an arc subtended by an angle measured in degrees is sort of cumbersome, involving the curi- 2π ous factor 360 . Our next goal is to introduce an alternate angle measurement scheme called radian measure that begins with a different “basic wedge”. As will become apparent, a big selling point of radian measure is that arc length calculations become easy. 15.4. RADIAN METHOD 197

r r r r equilateral wedge

r (r,0) r

this angle is DEFINED to have circle Cr measure 1 radian

Figure 15.7: Constructing an equilateral wedge.

As before, begin with a circle Cr of radius r. Construct an equilateral wedge with all three sides of equal length r; see Figure 15.7. We define the measure of the interior angle of this wedge to be 1 radian. Once we have defined an angle of measure 1 radian, we can define an angle of measure 2 radians by putting together two equilateral wedges. 1 Likewise, an angle of measure 2 radian is obtained by symmetrically dividing an equilateral wedge in half, etc. Reasoning in this way, we can piece together equilateral wedges or fractions of such to compute the radian measure of any angle. It is important to notice an important relationship between the radian measure of an angle and arc length calculations. In the five angles pictured above, 1 1 1 radian, 2 radian, 3 radian, 2 radian and 4 radian, the length of the arcs 1 1 subtended by these angles θ are r, 2r, 3r, 2 r, and 4 r. In other words, a pattern emerges that gives a very simple relationship between the length s of an arc and the radian measure of the subtended angle:

Important Fact 15.4.1 (Arc length in radians). s = distance along the arc Start with a central angle of measure θ radians inside a circle of radius r. Then this angle will subtend an arc of r length s = θr. θ radians

These remarks allow us to summarize the deﬁnition of r the radian measure θ of ∠AOB inside a circle of radius r by the formula: Cr s r if angle is swept counterclockwise Figure 15.9: Deﬁning arc θ = s − r if angle is swept clockwise length when angles are measured in radians. 198 CHAPTER 15. MEASURING AN ANGLE

3r 2r

r 3 radians = θ 2 radians = θ r r r

r r

1 radian = θ 1 radian = θ r 4 1 = 2 radian θ

r r 2 r r 4 r r

Figure 15.8: Measuring angles in radians.

y-axis length of The units of θ are sometimes abbreviated as rad. It is arc(AB) = s important to appreciate the role of the radius of the circle B Cr when using radian measure of an angle: An angle of radian measure θ will subtend an arc of length |θ| on the θ radian unit circle. In other words, radian measure of angles is O A x-axis exactly the same as arc length on the unit circle; we couldn’t hope for a better connection! circle radius r The difficulty with radian measure versus degree mea- Figure 15.10: Arc length sure is really one of familiarity. Let’s view a few common after imposing a coordinate angles in radian measure. It is easiest to start with the system. case of angles in central standard position within the unit circle. Examples of basic angles would be fractional parts 1 of one complete revolution around the unit circle; for example, 12 revolu- 1 1 1 1 3 tion, 8 revolution, 6 revolution, 4 revolution, 2 revolution and 4 revolution. One revolution around the unit circle describes an arc of length 2π and so the subtended angle (1 revolution) is 2π radians. We can now easily 1 find the radian measure of these six angles. For example, 12 revolution 1 π would describe an angle of measure ( 12 )2π rad= 6 rad. Similarly, the other π π π five angles pictured below have measures 4 rad, 3 rad, 2 rad, π rad and 3π 2 rad. All of these examples have positive radian measure. For an angle with π negative radian measure, such as θ = − 2 radians, we would locate B 1 by rotating 4 revolution clockwise, etc. From these calculations and our 15.5. AREAS OF WEDGES 199 eplacements

π π π 4 6 3

1 1 1 12 revolution 8 revolution 6 revolution

π π 3π 2 2

1 1 3 4 revolution 2 revolution 4 revolution

Figure 15.11: Common angles measured in radians.

previous examples of degree measure we ﬁnd that

180 degrees = π radians. (15.1)

Solving this equation for degrees or radians will provide conversion formulas relating the two types of angle measurement. The formula π 2π also helps explain the origin of the curious conversion factor 180 = 360 in Fact 15.3.2.

15.5 Areas of Wedges

The beauty of radian measure is that it is rigged so that we can easily compute lengths of arcs and areas of circular sectors (i.e. “pie-shaped regions”). This is a key reason why we will almost always prefer to work with radian measure.

Example 15.5.1. If a 16 inch pizza is cut into 12 equal slices, what is the area of a single slice?

This can be solved using a general principle:

(Area of a part) = (area of the whole) (fraction of the part) × 200 CHAPTER 15. MEASURING AN ANGLE

So, for our pizza: (area one slice) = (area whole pie) (fraction of pie) × 1 = (82π) 12 16π = . 3 Let’s apply the same reasoning to ﬁnd the area of a

B circular sector. We know the area of the circular disc 2 bounded by a circle of radius r is πr . Let Rθ be the “pie shaped wedge” cut out by an angle ∠AOB with positive R θ θ O measure θ radians. Using the above principle

area(Rθ)=(area of disc bounded by Cr) A (portion of disc accounted for by R ) × θ Cr θ 1 =(πr2) = r2θ. 2π 2 Figure 15.12: Finding the area of a “pie shaped wedge”. π For example, if r = 3 in. and θ = 4 rad, then the area of 9 the pie shaped wedge is 8 π sq. in. Important Fact 15.5.2 (Wedge area). Start with a central angle with positive measure θ radians inside a circle of radius r. The area of the “pie 1 2 shaped region” bounded by the angle is 2 r θ. Example 15.5.3. A water drip irrigation arm 1200 feet long rotates around a pivot P once every 12 hours. How much area is covered by the arm in one hour? in 37 minutes? How much time is required to drip irrigate 1000 square feet? Solution. The irrigation arm will complete one revolution in 12 hours. The angle swept out by one complete revolution is 2π radians, so after t hours the arm sweeps out an angle θ(t) given by 2π radians π θ(t) = t hours = t radians. 12 hours × 6 Consequently, by Important Fact 15.5.2 the area A(t) of the irrigated region after t hours is 1 1 π A(t) = (1200)2θ(t) = (1200)2 t = 120,000πt square feet. 2 2 6 After 1 hour, the irrigated area is A(1) = 120,000π = 376,991 sq. ft. Like- 37 wise, after 37 minutes, which is 60 hours, the area of the irrigated region 37 37 is A( 60 ) = 120,000π( 60) = 232,500 square feet. To answer the ﬁnal ques- tion, we need to solve the equation A(t) = 1000; i.e., 120,000πt = 1000, so 1 3600 seconds t = hours = 9.55 seconds. 120π × hour 15.5. AREAS OF WEDGES 201

15.5.1 Chord Approximation Our ability to compute arc lengths can be used as an estimating tool for distances between two points. Let’s return to the situation posed at the beginning of this section: Cosmo the dog, tied by a 20 foot long tether to a post in the ground, begins at location R and walks counterclockwise to location S. Furthermore, let’s suppose you are standing at the center of the circle determined by the tether and you measure the angle from R to S to be 5◦; see the left-hand ﬁgure. Because the angle is small, notice that the straight line distance d from R to S is approximately the same as the arc length s subtended by the angle ∠RPS; the right-hand picture in Figure 15.13 is a blow-up:

5◦ s R d P 20 feet 5◦ P 20 feet R

Figure 15.13: Using the arc length s to approximate the chord d.

Example 15.5.4. Estimate the distance from R to S.

Solution. We ﬁrst convert the angle into radian measure via (15.1): 5◦ = 0.0873 radians. By Fact 15.3.2, the arc s has length 1.745 feet = 20.94 inches. This is approximately equal to the distance from R to S, since the angle is small. 202 CHAPTER 15. MEASURING AN ANGLE

S We call a line segment connecting two points on a circle a chord of the circle. The above example illustrates s a general principal for approximating the length of any chord chord. A smaller angle will improve the accuracy of the R O arc length approximation. Important Fact 15.5.5 (Chord Approximation). In Fig- ure 15.14, if the central angle is small, then s |RS|. ≈

Figure 15.14: Chord approximation. 15.6 Great Circle Navigation A basic problem is to ﬁnd the shortest route between any two locations on the earth. We will review how to coordinatize the surface of the earth and recall the fact that the shortest path between two points is measured along a great circle. View the earth as a sphere of radius r = 3,960 miles. We could slice the earth with a two-dimensional plane 0 which is both perpendicular to a line connecting the North and SouthP poles and passes through the center of the earth. Of course, the resulting intersection will trace out a circle of radius r = 3,960 miles on the surface of the earth, which we call the equator. We call the plane the equatorial plane. Slicing the earth P0 with any other plane parallel to 0, we can consider the right triangle pictured below and theP angle θ: P

North Pole line of latitude North Pole b b θ r r θ equatorial plane P r 0 r equator 90◦ − θ◦ equator

center of earth center of earth South Pole South Pole

Figure 15.15: Measuring latitude.

Essentially two cases arise, depending on whether or not the plane is above or below the equatorial plane. The plane slices the surface ofP the earth in a circle, which we call a line of latitudeP . This terminology is somewhat incorrect, since these lines of latitude are actually circles on the surface of the earth, but the terminology is by now standard. De- pending on whether this line of latitude lies above or below the equatorial plane, we refer to it as the θ◦ North line of latitude (denoted θ◦ N) or the θ◦ South line of latitude (denoted θ◦ S). Notice, the radius b of a line of latitude can vary from a maximum of 3,960 miles (in the case of θ = 0◦), 15.6. GREAT CIRCLE NAVIGATION 203

to a minimum of 0 miles, (when θ = 90◦). When b = 0, we are at the North or South poles on the earth. In a similar spirit, we could imagine slicing the earth with a plane which is perpendicular to the equatorial plane and passes through theQ center of the earth. The resulting intersection will trace out a circle of radius 3,960 miles on the surface of the earth, which is called a line of longitude. Half of a line of longitude from the North Pole to the South Pole is called a meridian. We distinguish one such meridian; the one which passes through Greenwich, England as the Greenwich meridian. Longitudes are measured using angles East or West of Greenwich. Pic- tured below, the longitude of A is θ. Because θ is east of Greenwich, θ measures longitude East, typically written θ◦ E; west longitudes would be denoted as θ◦ W. All longitudes are between 0◦ and 180◦. The meridian which is 180◦ West (and 180◦ E) is called the International Date Line. Introducing the grid of latitude and longitude lines on Greenwich, England the earth amounts to imposing a coordinate system. In North Pole line of longitude other words, any position on the earth can be determined A Greenwich International meridian by providing the longitude and latitude of the point. The r Date Line θ usual convention is to list longitude ﬁrst. For example, r Equator Seattle has coordinates 122.0333◦ W, 47.6◦ N. Since the labels “N and S” are attached to latitudes and the labels “E center of earth South Pole and W” are attached to longitudes, there is no ambiguity Figure 15.16: The Interna- here. This means that Seattle is on the line of longitude tional Date Line. 122.0333◦ West of the Greenwich meridian and on the line of latitude 47.6◦ North of the equator. In the ﬁgure below, we indicate the key angles ψ = 47.6◦ and θ = 122.0333◦ by inserting the three indicated radial line segments.

not great circle

N Seattle, WA great circles

Ψ r θ

Greenwich Meridian

S not great circle

Figure 15.17: Distances along great circles. 204 CHAPTER 15. MEASURING AN ANGLE

Now that we have imposed a coordinate system on the earth, it is natural to study the distance between two locations. A great circle of a sphere is deﬁned to be a circle lying on the sphere with the same center as the sphere. For example, the equator and any line of longitude are great circles. However, lines of latitude are not great circles (except the special case of the equator). Great circles are very important because they are used to ﬁnd the shortest distance between two points on the earth. The important fact from geometry is summarized below.

Important Fact 15.6.1 (Great Circles). The shortest distance between two points on the earth is measured along a great circle connecting them.

Example 15.6.2. What is the shortest distance from the North Pole to Seat- tle, WA ?

Solution. 122.0333 N The line of longitude ◦ W is a great circle connecting the North Pole and Seattle. So, the shortest W distance will be the arc length s subtended by the angle ∠NOW pictured in Figure 15.18. Since the latitude of

O Seattle is 47.6◦, the angle ∠EOW has measure 47.6◦. Since E equator ∠EON is a right angle (i.e., 90◦), ∠NOW has measure 42.4◦. By Fact 15.4.1 and Equation 15.1,

Greenwich S Meridian s = (3960 miles)(42.4◦)(0.01745 radians/degree)

Figure 15.18: Distance be- = 2943.7 miles, tween the North Pole and which is the shortest distance from the pole to Seattle. Seattle, Washington.

15.7 Summary

360◦ = 2π radians • A circular arc with radius r and angle θ has length s, with • s = rθ

when θ is measured in radians.

A circular wedge with radius r and angle θ has area A, with • 1 A = r2θ 2 when θ is measured in radians. 15.8. EXERCISES 205 15.8 Exercises

Problem 15.1. Let ∠AOB be an angle of mea- (c) Suppose bug A lands on the end of the sure θ. blade farthest from the pivot. Assume the wiper turns through an angle of 110 . (a) Convert θ = 13.4o into degrees/ min- ◦ In one cycle (back and forth) of the wiper utes/ seconds and into radians. blade, how far has the bug traveled? o (b) Convert θ = 1 4 ′44 ′′ into degrees and radians. (d) Suppose bug B lands on the end of the (c) Convert θ = 0.1 radian into degrees and wiper blade closest to the pivot. Assume degrees/ minutes/ seconds. the wiper turns through an angle of 110◦. In one cycle of the wiper blade, how far has the bug traveled? Problem 15.2. A nautical mile is a unit of distance frequently used in ocean navigation. It is deﬁned as the length of an arc s along a great (e) Suppose bug C lands on an intermediate circle on the earth when the subtending angle location of the wiper blade. Assume the wiper turns through an angle of 110◦. If has measure 1 ′ = “one minute” = 1/60 of one degree. Assume the radius of the earth is bug C travels 28 inches after one cycle 3,960 miles. of the wiper blade, determine the location of bug C on the wiper blade. (a) Find the length of one nautical mile to the nearest 10 feet. (b) A vessel which travels one nautical mile in one hours time is said to have the speed of one knot; this is the usual nav- Problem 15.4. A water treatment facility oper- igational measure of speed. If a vessel is ates by dripping water from a 60 foot long arm traveling 26 knots, what is the speed in whose end is mounted to a central pivot. The mph (miles per hour)? water then ﬁlters through a layer of charcoal. The arm rotates once every 8 minutes. (c) If a vessel is traveling 18 mph, what is the speed in knots? (a) Find the area of charcoal covered with water after 1 minute. Problem 15.3. The rear window wiper blade on a station wagon has a length of 16 inches. (b) Find the area of charcoal covered with The wiper blade is mounted on a 22 inch arm, water after 1 second. 6 inches from the pivot point. (c) How long would it take to cover 100 square feet of charcoal with water? 6"

16" (d) How long would it take to cover 3245 square feet of charcoal with water?

Problem 15.5. Astronomical measurements are often made by computing the small angle formed by the extremities of a distant object (a) If the wiper turns through an angle of and using the estimating technique in 15.5.1. 110 , how much area is swept clean? ◦ In the picture below, the full moon is shown 1 o (b) Through how much of an angle would to form an angle of 2 when the distance indi- the wiper sweep if the area cleaned was cated is 248,000 miles. Estimate the diameter 10 square inches? of the moon. 206 CHAPTER 15. MEASURING AN ANGLE

moon (a) θ = 45◦ (b) θ = 80o 248,000 miles (c) θ = 3 radians (d) θ = 2.46 radians (e) θ = 97o23 3 o ′ ′′ 1/2 o (f) θ = 35 24 ′2 ′′

earth Problem 15.8. Matilda is planning a walk around the perimeter of Wedge Park, which is shaped like a circular wedge, as shown below. Problem 15.6. An aircraft is ﬂying at the The walk around the park is 2.1 miles, and the speed of 500 mph at an elevation of 10 park has an area of 0.25 square miles. miles above the earth, beginning at the North If θ is less than 90 degrees, what is the pole and heading South along the Greenwich value of the radius, r? meridian. A spy satellite is orbiting the earth at an elevation of 4800 miles above the earth r in a circular orbit in the same plane as the Greenwich meridian. Miraculously, the plane and satellite always lie on the same radial line θ from the center of the earth. Assume the radius of the earth is 3960 miles. r satellite

Problem 15.9. Let C be the circle of radius plane 6 6 inches centered at the origin in the xy- coordinate system. Compute the areas of the shaded regions in the picture below; the inner earth circle in the rightmost picture is the unit circle: y=x y=−x

(a) When is the plane directly over a loca- C tion with latitude 74◦30 ′18 ′′ N for the ﬁrst C 6 time? 6 (b) How fast is the satellite moving? y=x (c) When is the plane directly over the equator and how far has it traveled? (d) How far has the satellite traveled when the plane is directly over the equator? y=−(1/4)x + 2

Problem 15.7. Find the area of the sector of a C circle of radius 11 inches if the measure θ of a 6 central angle of this sector is: