# High School Trigonometry/Applications of Radian Measure

In this lesson you will apply radian measure to various problem-solving contexts involving rotations.

## Lesson Objectives

• Solve problems involving angles of rotation using radian measure.
• Solve problems by calculating the length of an arc.
• Solve problems by calculating the area of a sector.
• Approximate the length of a chord given the central angle and radius.

## Rotations

 Example 1 The hands of a clock show 11:20. Express the obtuse angle formed by the hour and minute hands in radian measure to the nearest tenth of a radian. The following diagram shows the location of the hands at the specified time. Solution: Because there are 12 increments on a clock, the angle between each hour marking on the clock is ${\tfrac {2\pi }{12}}$ = ${\tfrac {\pi }{6}}$ (or 30°). So, the angle between the 12 and the 4 is 4 · ${\tfrac {\pi }{6}}$ = ${\tfrac {2\pi }{3}}$ (or 120°). Because the rotation from 12 to 4 is one-third of a complete rotation, it seems reasonable to assume that the hour hand is moving continuously and has therefore moved one-third of the distance between the 11 and the 12. So, ${\tfrac {2}{3}}$ · ${\tfrac {\pi }{6}}$ = ${\tfrac {2\pi }{18}}$ , and the total measure of the angle is therefore ${\tfrac {2\pi }{18}}$ + ${\tfrac {2\pi }{3}}$ = ${\tfrac {2\pi }{18}}$ + ${\tfrac {12\pi }{18}}$ = ${\tfrac {14\pi }{18}}$ . Using a calculator to approximate the angle would give: 14π/182.44346 To the nearest tenth of a radian it is 2.4 radians.

## Length of an Arc

The length of an arc on a circle depends on both the angle of rotation and the radius length of the circle. If you recall from the last lesson, we defined a radian as the length of the arc the measure of an angle θ in radians is defined as the length of the arc cut off by one radius length, so that a half-rotation is π radians, or a little more than 3 radius lengths around the circle. What if the radius is 4 cm? The length of the half-circle arc would be π radius lengths, or 4π cm in length. This results in a formula that can be used to calculate the length of any arc.

$s=r\theta \,\!$ where s is the length of the arc, r is the radius, and θ is the measure of the angle in radians.

Solving this equation for θ will give us a formula for finding the radian measure given the arc length and the radius length:

$\theta ={\frac {s}{r}}$ Example 2

The free-throw line on an NCAA basketball court is 12 ft wide. In international competition, it is only about 11.81 ft. How much longer is the half circle above the free-throw line on the NCAA court? Solution:

Arc length calculations:

NCAA International
s1 = s2 =
s1 = 6(π) s2 ≈ 5.905(π)
s1 = 6π s2 ≈ 5.905π

So the answer is approximately 6π − 5.905π ≈ 0.095π.

.095π
.0.29845130209103

This is approximately 0.3 ft, or about 3.6 inches longer.

 Example 3 Two connected gears are rotating. The smaller gear has a radius of 4 inches and the larger gear's radius is 7 inches. What is the angle through which the larger gear has rotated when the smaller gear has made one complete rotation? Solution: Because the blue gear performs one complete rotation, the length of the arc traveled is: s = rθ s = 4 · 2π So, an 8π arc length on the larger circle would form an angle as follows: θ = ${\tfrac {s}{r}}$ θ = ${\tfrac {8\pi }{7}}$ θ ≈ 3.6 So the angle is approximately 3.6 radians. 3.6 · ${\tfrac {180}{\pi }}$ ≈ 206°

## Area of a Sector

One of the most common geometric formulas is the area of a circle:

$A=\pi r^{2}\,\!$ In terms of angle rotation, this is the area created by 2π radians.

$2\pi {\text{ radian angle}}=\pi r^{2}{\text{ area}}\,\!$ A half-circle, or π radian rotation would create a section, or sector of the circle equal to half the area or:

${\frac {1}{2}}\pi r^{2}$ So an angle of 1 radian would define an area of a sector equal to:

 ${\frac {2\pi }{2\pi }}=$ ${\frac {\pi r^{2}}{2\pi }}$ $1=\,\!$ ${\frac {1}{2}}r^{2}$ From this we can determine the area of the sector created by any angle, θ radians, to be:

$A={\frac {1}{2}}r^{2}\theta$ Example 4 Crops are often grown using a technique called center pivot irrigation that results in circular shaped fields. Here is a satellite image taken over fields in Kansas that use this type of irrigation system. If the irrigation pipe is 450 m in length, what is the area that can be irrigated after a rotation of ${\tfrac {2\pi }{3}}$ radians? Solution: Using the formula: $A={\frac {1}{2}}r^{2}\theta$ $A={\frac {1}{2}}(450)^{2}\left({\frac {2\pi }{3}}\right)$ .5*4502*2π/3212057.5041 The area is approximately 212,058 square meters.

## Length of a Chord

You may recall from your Geometry studies that a chord is a segment that begins and ends on a circle. ${\overline {AB}}$ is a chord in the circle.

We can calculate the length of any chord if we know the angle measure and the length of the radius. Because each endpoint of the chord is on the circle, the distance from the center to A and B is the same as the radius length. Next, if we bisect the angle, the angle bisector must be perpendicular to the chord (we will leave the proof of this to your Geometry class). This forms a right triangle. We can now use a simple sine ratio to find half the chord, called c here, and double the result to find the length of the chord. $\sin \left({\frac {\theta }{2}}\right)=$ ${\frac {c}{r}}$ $c=\,\!$ $r\cdot \sin \left({\frac {\theta }{2}}\right)$ So the length of the chord is:

$2c=2r\sin \left({\frac {\theta }{2}}\right)$ Example 5 Find the length of the chord of a circle with radius 8 cm and a central angle of 110°. Approximate your answer to the nearest mm. Solution: It's always a good problem solving technique to estimate the answer first. A thought process for estimating the measure might look something like this: The angle is slightly more than 90°, or ${\tfrac {\pi }{2}}$ radians. ${\tfrac {\pi }{2}}$ radians is slightly more than 1.5 radius lengths. One and a half radii would be 12, so we might expect the answer to be a little more than 12 cm. Let's see how the actual answer compares. We must first convert the angle measure to radians: $100\cdot {\frac {\pi }{180}}={\frac {11\pi }{18}}$ Using the formula, half of the chord length should be the radius of the circle times the sine of half the angle. ${\frac {11\pi }{18}}\cdot {\frac {1}{2}}={\frac {11\pi }{36}}$ 8 · sin $\left({\tfrac {11\pi }{36}}\right)$ (Make sure your calculator is in radians!) Multiply this result by 2. 8sin(11π/36)6.553216354 Ans*213.10643271 So, the length of the arc is approximately 13.1 cm. This seems very reasonable based on our estimate.

## Review Questions

1. The image at right shows a 24−hour clock in Curitiba, Paraná, Brasil.
(a) What is the angle between each number of the clock expressed in:
i. exact radian measure in terms of π?
ii. to the nearest tenth of a radian?
iii. in degree measure?
(b) Estimate the measure of the angle between the hands at the time shown in:
i. to the nearest whole degree
ii. in radian measure in terms of π
2. The picture at right is a window of a building on the campus of Princeton University in Princeton, New Jersey.
(a) What is the exact radian measure in terms of π between two consecutive circular dots on the small circle in the center of the window? (b) If the radius of this circle is about 0.5 m, what is the length of the arc between the centers of each consecutive dot? Round your answer to the nearest cm. 3. Now look at the next larger circle in the window. (a) Find the exact radian measure in terms of π between two consecutive dots in this window.
(b) The radius of the glass portion of this window is approximately 1.20 m. Calculate an estimate of the length of the highlighted chord to the nearest cm. Explain the reasoning behind your solution. 4. The state championship game is to be held at Ray Diaz Memorial Arena. The seating forms a perfect circle around the court. The principal of Archimedes High School is sent the following diagram showing the seating allotted to the students at her school. It is 55 ft from the center of the court to the beginning of the stands and 110 ft from the center to the end. Calculate the approximate number of square feet each of the following groups has been granted:
(a) the students from Archimedes
(c) the press and officials
5. To the right is an image of the state flag of Colorado. It turns out that the diameter of the gold circle is ${\tfrac {1}{3}}$ the total height of the flag (the same width as the yellow stripe) and the outer diameter of the red circle is ${\tfrac {2}{3}}$ of the total height of the flag. The angle formed by the missing portion of the red band is ${\tfrac {\pi }{4}}$ radians. In a flag that is 33 inches tall, what is the area of the red portion of the flag to the nearest square inch?

1.
(a) i. ${\tfrac {\pi }{12}}$ iii. 15°
(b) i. 20°. Answers may vary, anything about 15° and less than 25° is reasonable.
ii. ${\tfrac {\pi }{9}}$ . Again, answers may vary.
2.
(a) ${\tfrac {\pi }{6}}$ (b) ≈ 26 cm
3.
(a) ${\tfrac {\pi }{16}}$ (b) Let's assume, to simplify, that the chord stretches to the center of each of the dots. We need to find the measure of the central angle of the circle that connects those two dots. Since there are 13 dots, this angle is ${\tfrac {13\pi }{16}}$ . The length of the chord then is:
$=2r\sin \left({\frac {\theta }{2}}\right)$ $=2\cdot 1.2\cdot \sin \left({\frac {1}{2}}\cdot {\frac {13\pi }{16}}\right)$ 2*1.2sin(13π/32)
2.296656806
The chord is approximately 2.30 cm.
4. Each section is ${\tfrac {\pi }{6}}$ radians. The area of one section of the stands is therefore the area of the outer sector minus the area of the inner sector:
(a) The students have four sections or ≈ 9,503 ft2
(b) There are three general admission sections or ≈ 7,127 ft2
(c) There is only one press and officials section or ≈ 2,376 ft2
$A=A_{\text{outer}}-A_{\text{inner}}\,\!$ $A={\frac {1}{2}}(r_{\text{outer}})^{2}\cdot {\frac {\pi }{6}}-{\frac {1}{2}}(r_{\text{inner}})^{2}\cdot {\frac {\pi }{6}}$ $A={\frac {1}{2}}(110)^{2}\cdot {\frac {\pi }{6}}-{\frac {1}{2}}(55)^{2}\cdot {\frac {\pi }{6}}$ .5*1102*π/6-.5*552
*π/6
2375.829444
The area of each section is approximately 2376 ft2.
5. There are many different approaches to the problem. Here is one possibility:
First, calculate the area of the red ring as if it went completely around the circle:
$A=A_{\text{total}}-A_{\text{gold}}\,\!$ $A=\pi \left({\frac {2}{3}}\cdot 33\right)^{2}-\pi \left({\frac {1}{3}}\cdot 33\right)^{2}$ $A=\pi \cdot 22^{2}-\pi \cdot 11^{2}\,\!$ $A=484\pi -121\pi =363\pi \,\!$ $A\approx 1140.4{\text{ in}}^{2}\,\!$  Next, calculate the area of the total sector that would form the opening of the "c":
$A={\frac {1}{2}}r^{2}\theta$ $A={\frac {1}{2}}(22)^{2}\left({\frac {\pi }{4}}\right)$ $A\approx 190.1{\text{ in}}^{2}\,\!$  Then, calculate the area of the yellow sector and subtract it from the previous answer.
$A={\frac {1}{2}}r^{2}\theta$ $A={\frac {1}{2}}(11)^{2}\left({\frac {\pi }{4}}\right)$ $A\approx 47.5{\text{ in}}^{2}\,\!$ $190.1-47.5=142.6{\text{ in}}^{2}\,\!$ Finally, subtract this answer from the first area calculated. The area is approximately 998 in2.