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Trigonometry

Module by: Scott Starks. E-mail the author

Trigonometry

Introduction

Trigonometry is a very important tool for engineers. This reading mentions just a few of the many applications that involve trigonometry to solve engineering problems.

Trigonometry means the study of the triangle. Most often, it refers to finding angles of a triangle when the lengths of the sides are known, or finding the lengths of two sides when the angles and one of the side lengths are known.

As you complete this reading, be sure to pay special attention to the variety of areas in which engineers utilize trigonometry to develop the solutions to problems. Rest assured that there are an untold number of applications of trigonometry in engineering. This reading only introduces you to a few. You will learn many more as you progress through your engineering studies.

Virtually all engineers use trigonometry in their work on a regular basis. Things like the generation of electrical current or a computer use angles in ways that are difficult to see directly, but that rely on the fundamental rules of trigonometry to work properly. Any time angles appear in a problem, the use of trigonometry usually will not be far behind.

An excellent working knowledge of trigonometry is essential for practicing engineers. Many problems can be easily solved by applying the fundamental definitions of trigonometric functions. Figure 1 depicts a right triangle for which we will express the various trigonometric functions and relationships that are important to engineers.

Figure 1: Triangle for definition of trigonometric functions.
Figure 1 (graphics1.jpg)

Some of the most widely used trigonometric functions follow

sin ( θ ) = opposite side hypotenuse = B C sin ( θ ) = opposite side hypotenuse = B C size 12{"sin" \( θ \) = { { ital "opposite"` ital "side"} over { ital "hypotenuse"} } = { {B} over {C} } } {}
(1)
cos ( θ ) = adjacent side hypotenuse = A C cos ( θ ) = adjacent side hypotenuse = A C size 12{"cos" \( θ \) = { { ital "adjacent"` ital "side"} over { ital "hypotenuse"} } = { {A} over {C} } } {}
(2)
tan ( θ ) = opposite side adjacent side = B A tan ( θ ) = opposite side adjacent side = B A size 12{"tan" \( θ \) = { { ital "opposite"` ital "side"} over { ital "adjacent"` ital "side"} } = { {B} over {A} } } {}
(3)

The Pythagorean Theorem often plays a key role in applications involving trigonometry in engineering. It states that the square of the hypotenuse is equal to the sum of the squares of the adjacent side and of the opposite side. This theorem can be stated mathematically by means of the equation that follows. Here we make use of the symbols (A, B, and C) that designate associated lengths.

A 2 + B 2 = C 2 A 2 + B 2 = C 2 size 12{A rSup { size 8{2} } +B rSup { size 8{2} } =C rSup { size 8{2} } } {}
(4)

In the remainder of this module, we will make use of these formulas in addressing various applications.

Flight Path of an Aircraft

The following is representative of the type of problem one is likely to encounter in the introductory Physics course as well as in courses in Mechanical Engineering. This represents an example of how trigonometry can be applied to determine the motion of an aircraft.

Assume that an airplane climbs at a constant angle of 30 from its departure point situated at sea level. It continues to climb at this angle until it reaches its cruise altitude. Suppose that its cruise altitude is 31,680 ft above sea level.

With the information stated above, we may sketch an illustration of the situation

Figure 2: Depiction of aircraft motion.
Figure 2 (graphics2.jpg)

Question 1: What is the distance traveled by the plane from its departure point to its cruise altitude?

The distance traveled by the plane is equal to the length of the hypotenuse of the right triangle depicted in Figure 2. Let us denote the distance measured in feet that the plane travels by the symbol C. We may apply the definition of the sine function to enable us to solve for the symbol C as follows.

sin ( 3 0 ) = 31 , 680 C sin ( 3 0 ) = 31 , 680 C size 12{"sin" \( 3 rSup { size 8{0} } \) = { {"31","680"} over {C} } } {}
(5)

Rearranging terms we find

C = 31 , 680 sin ( 3 o ) C = 31 , 680 sin ( 3 o ) size 12{C= { {"31","680"} over {"sin" \( 3 rSup { size 8{o} } \) } } } {}
(6)

Calculation and rounding to 3 significant digits yields the result

C = 605 , 320 ft C = 605 , 320 ft size 12{C="605","320"` ital "ft"} {}
(7)

Let us ponder a second question based upon the data presented in the original problem.

Question 2: What is the ground distance traveled by the airplane as it moves from its departure point to its cruise altitude?

Solution: Referring to Figure 2, we observe that we must find the length of the adjacent side in order to answer the question. We can use the definition of the tangent to guide our solution.

tan ( 3 0 ) = Opposite side Adjacent side tan ( 3 0 ) = Opposite side Adjacent side size 12{"tan" \( 3 rSup { size 8{0} } \) = { { ital "Opposite"` ital "side"} over { ital "Adjacent"` ital "side"} } } {}
(8)

Denoting the adjacent side by the symbol A, we obtain

A = Opposite side tan ( 3 0 ) A = Opposite side tan ( 3 0 ) size 12{A= { { ital "Opposite"` ital "side"} over {"tan" \( 3 rSup { size 8{0} } \) } } } {}
(9)
A = 31 , 680 0 . 0524 A = 31 , 680 0 . 0524 size 12{A= { {"31","680"} over {0 "." "0524"} } } {}
(10)

After rounding to 3 significant digits, we obtain the solution

A = 604 , 580 ft A = 604 , 580 ft size 12{A="604","580"` ital "ft"} {}
(11)

Inclined Plane

Work is an important concept in virtually every field of science and engineering. It takes work to move an object; it takes work to move an electron through an electric field; it takes work to overcome the force of gravity; etc.

Let’s consider the case where we use an inclined plane to assist in the raising of a 300 pound weight. The inclined plane situated such that one end rests on the ground and the other end rests upon a surface 4 feet aove the ground. This situation is depicted in Figure 3.

Figure 3: Object on an inclined plane.
Figure 3 (graphics3.jpg)

Question 3: Suppose that the length of the inclined plane is 12 feet. What is the angle that the plane makes with the ground?

Clearly, the length of the inclined plane is same as that of the hypotenuse shown in the figure. Thus, we may use the sine function to solve for the angle

sin ( θ ) = 4 12 = 0 . 333 sin ( θ ) = 4 12 = 0 . 333 size 12{"sin" \( θ \) = { {4} over {"12"} } =0 "." "333"} {}
(12)

In order to solve for the angle, we must make use of the inverse sine function as shown below

sin − 1 ( sin ( θ ) ) = sin − 1 ( 0 . 333 ) sin − 1 ( sin ( θ ) ) = sin − 1 ( 0 . 333 ) size 12{"sin" rSup { size 8{ - 1} } \( "sin" \( θ \) \) ="sin" rSup { size 8{ - 1} } \( 0 "." "333" \) } {}
(13)
θ = 19 . 45 0 θ = 19 . 45 0 size 12{θ="19" "." "45" rSup { size 8{0} } } {}
(14)

So we conclude that the inclined plane makes a 19.850 angle with the ground.

Neglecting any effects of friction, we wish to determine the amount of work that is expended in moving the block a distance (L) along the surface of the inclined plane.

Surveying

Let us now turn our attention to an example in the field of surveying. In particular, we will investigate how trigonometry can be used to help forest rangers combat fires. Let us suppose that a fire guard observes a fire due south of her Hilltop Lookout location. A second fire guard is on duty at a Watch Tower that is located 11 miles due east of the Hilltop Lookout location. This second guard spots the same fire and measures the bearing (angle) at 2150 from North. The figure below illustrates the geometry of the situation.

Figure 4: Depiction of a scenario associated with a forest fire.
Figure 4 (graphics4.jpg)

Question: How far away is the fire from the Hilltop Lookout location?

We begin by identifying the angle θ in the figure below.

Figure 5: Refined depiction of scenario.
Figure 5 (graphics5.jpg)

The value of θ can be found via the equation

θ = 90 0 − 35 0 θ = 90 0 − 35 0 size 12{θ="90" rSup { size 8{0} } - "35" rSup { size 8{0} } } {}
(15)
θ = 55 0 θ = 55 0 size 12{θ="55" rSup { size 8{0} } } {}
(16)

So we can simplify the drawing as shown below.

Figure 6: Trigonometric representation of scenario.
Figure 6 (graphics6.jpg)

Our problem reduces to solving for the value of b.

tan ( 55 0 ) = b 11 miles tan ( 55 0 ) = b 11 miles size 12{"tan" \( "55" rSup { size 8{0} } \) = { {b} over {"11"` ital "miles"} } } {}
(17)
b = ( 1 . 43 ) ( 11 miles ) = 15 . 7 miles b = ( 1 . 43 ) ( 11 miles ) = 15 . 7 miles size 12{b= \( 1 "." "43" \) ` \( "11"` ital "miles" \) ="15" "." 7` ital "miles"} {}
(18)

We conclude that the fire is located 15.7 miles south of the Hilltop Lookout location.

Comment: Triangulation is a process that can be applied to solve problems in a number of areas of engineering including surveying, construction management, radar, sonar, lidar, etc.

Refraction

Refraction is a physical phenomenon that occurs when light passes from one transparent medium (such as air) through another (for example, glass.) It is known that light travels at different speeds through different transparent media. The index of refraction of a medium is a measure of how much the speed of light is reduced as it passes through the medium. In the case of glass, the index of refraction is approximately 1.5. This means that light travels as a speed of 11.5=2311.5=23 size 12{ { {1} over {1 "." 5} } = { {2} over {3} } } {}times the speed of light in a vacuum.

Two common properties of transparent materials can be attributed to the index of refraction. One is that light rays change direction as they pass from one medium through another. Secondly, light is partially reflected when it passes from one medium to another medium with a different index of refraction. We will focus on the first of these properties in this reading.

In optics, which is a field of physics, you will learn about Snell's law, which is also known as Descartes' law after the scientist, Rene Descartes. Snell’s law takes the form of an equation that states the relationship between the angle of incidence and the angle of refraction for light passing from one medium to another. Stated mathematically, Snell’s law is

sin ( θ incidence ) sin ( θ refraction ) = c incidence c refraction sin ( θ incidence ) sin ( θ refraction ) = c incidence c refraction size 12{ { {"sin" \( θ rSub { size 8{ ital "incidence"} } \) } over {"sin" \( θ rSub { size 8{ ital "refraction"} } \) } } = { {c rSub { size 8{ ital "incidence"} } } over {c rSub { size 8{ ital "refraction"} } } } } {}
(19)

It follows that

sin ( θ incidence ) sin ( θ refraction ) = I 2 I 1 sin ( θ incidence ) sin ( θ refraction ) = I 2 I 1 size 12{ { {"sin" \( θ rSub { size 8{ ital "incidence"} } \) } over {"sin" \( θ rSub { size 8{ ital "refraction"} } \) } } = { {I rSub { size 8{2} } } over {I rSub { size 8{1} } } } } {}
(20)

where I1andI2I1andI2 size 12{I rSub { size 8{1} } ` ital "and"`I rSub { size 8{2} } } {}are the Index of Refraction of medium 1 and medium 2 respectively.

Consider a situation where light rays pass are shined from air through a tank of water. This situation is illustrated below.

Figure 7: Depiction of light refraction.
Figure 7 (Object 5.png)

The Index of refraction for air is 1.0003 and that of water is 1.3000. Let us assume that the angle that light enters the water is 210 40’, what is the angle of refraction, w?

From Snell’s law, we know

I W I A = sin ( a ) sin ( w ) I W I A = sin ( a ) sin ( w ) size 12{ { {I rSub { size 8{W} } } over {I rSub { size 8{A} } } } = { {"sin" \( a \) } over {"sin" \( w \) } } } {}
(21)
sin ( w ) = I A I W sin ( a ) sin ( w ) = I A I W sin ( a ) size 12{"sin" \( w \) = { {I rSub { size 8{A} } } over {I rSub { size 8{W} } } } `"sin" \( a \) } {}
(22)

sin ( w ) = I A I W sin ( a ) sin ( w ) = I A I W sin ( a ) size 12{"sin" \( w \) = { {I rSub { size 8{A} } } over {I rSub { size 8{W} } } } `"sin" \( a \) } {}

Substituting in the numerical values for IA, IW and a yield

sin ( w ) = 0 . 2841 sin ( w ) = 0 . 2841 size 12{"sin" \( w \) =0 "." "2841"} {}
(23)

We now make use of the inverse sine function

w = sin − 1 ( 0 . 2841 ) w = sin − 1 ( 0 . 2841 ) size 12{w="sin" rSup { size 8{ - 1} } \( 0 "." "2841" \) } {}
(24)

This leads to the result

w = 16 0 30 ' w = 16 0 30 ' size 12{w="16" rSup { size 8{0} } `"30" rSup { size 8{'} } } {}
(25)

We conclude that the refracted ray will travel through the water at an angle of refraction of 160 30’.

Exercises

  1. A 50 ft ladder leans against the top of a building which is 30 ft tall. Determine the angle the ladder makes with the horizontal. Also determine the distance from the base of the ladder to the building.
  2. A straight trail leads from the Alpine Hotel at elevation 8,000 feet to a scenic overlook at elevation 11,100 feet. The length of the trail is 14,100 feet. What is the inclination angle β in degrees? What is the value of β in radians?
  3. A ray of light moves from a media whose index of refraction is 1.200 to another whose index of refraction is 1.450. The angle of incidence of the ray as it intersects the interface of the two media is 150. Sketch the geometry of the situation and determine the value of the angle of refraction.
  4. One-link planar robots can be used to place pick up and place parts on work table. A one-link planar robot consists of an arm that is attached to a work table at one end. The other end is left free to rotate about the work space. If l = 5 cm, sketch the position of the robot and determine the (x, y) coordinates of point p(x,y) for the following values for θ: (50˚, 2π/3 rad, -20˚, and -5 π/4 rad).

Figure 8: One-link planar robot.
Figure 8 (Picture 3.png)

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