Solving Trigonometric Equations

Learning Objectives

In this section, you will:

• Solve linear trigonometric equations in sine and cosine.
• Solve equations involving a single trigonometric function.
• Solve trigonometric equations using a calculator.
• Solve trigonometric equations that are quadratic in form.
• Solve trigonometric equations using fundamental identities.
• Solve trigonometric equations with multiple angles.
• Solve right triangle problems.

Figure 1 Egyptian pyramids standing near a modern city. (credit: Oisin Mulvihill)

Thales of Miletus (circa 625–547 BC) is known as the founder of geometry. The legend is that he calculated the height of the Great Pyramid of Giza in Egypt using the theory of similar triangles, which he developed by measuring the shadow of his staff. He reasoned that when the height of his staff’s shadow was exactly equal to the actual height of the staff, then the height of the nearby pyramid’s shadow must also be equal to the height of the actual pyramid. Since the structures and their shadows were creating a right triangle with two equal sides, they were similar triangles. By measuring the length of the pyramid’s shadow at that moment, he could obtain the height of the pyramid. Based on proportions, this theory has applications in a number of areas, including fractal geometry, engineering, and architecture. Often, the angle of elevation and the angle of depression are found using similar triangles.

In earlier sections of this chapter, we looked at trigonometric identities. Identities are true for all values in the domain of the variable. In this section, we begin our study of trigonometric equations to study real-world scenarios such as the finding the dimensions of the pyramids.

Solving Linear Trigonometric Equations in Sine and Cosine

Trigonometric equations are, as the name implies, equations that involve trigonometric functions. Similar in many ways to solving polynomial equations or rational equations, only specific values of the variable will be solutions, if there are solutions at all. Often we will solve a trigonometric equation over a specified interval. However, just as often, we will be asked to find all possible solutions, and as trigonometric functions are periodic, solutions are repeated within each period. In other words, trigonometric equations may have an infinite number of solutions. Additionally, like rational equations, the domain of the function must be considered before we assume that any solution is valid. The period of both the sine function and the cosine function is 2π.2𝜋. In other words, every 2π2𝜋 units, the y-values repeat. If we need to find all possible solutions, then we must add 2πk,2𝜋𝑘, where k𝑘 is an integer, to the initial solution. Recall the rule that gives the format for stating all possible solutions for a function where the period is 2π:2𝜋:

sinθ=sin(θ±2kπ)sin𝜃=sin(𝜃±2𝑘𝜋)

There are similar rules for indicating all possible solutions for the other trigonometric functions. Solving trigonometric equations requires the same techniques as solving algebraic equations. We read the equation from left to right, horizontally, like a sentence. We look for known patterns, factor, find common denominators, and substitute certain expressions with a variable to make solving a more straightforward process. However, with trigonometric equations, we also have the advantage of using the identities we developed in the previous sections.

EXAMPLE 1

Solving a Linear Trigonometric Equation Involving the Cosine Function

Find all possible exact solutions for the equation cosθ=12.cos𝜃=12.

Solution

From the unit circle, we know that

cosθθ==12π3,5π3cos𝜃=12𝜃=𝜋3,5𝜋3

These are the solutions in the interval [0,2π].[0,2𝜋]. All possible solutions are given by

θ=π3±2kπ  and  θ=5π3±2kπ𝜃=𝜋3±2𝑘𝜋  and  𝜃=5𝜋3±2𝑘𝜋

where k𝑘 is an integer.

EXAMPLE 2

Solving a Linear Equation Involving the Sine Function

Find all possible exact solutions for the equation sint=12.sin𝑡=12.

Solution

Solving for all possible values of t means that solutions include angles beyond the period of 2π.2𝜋. From Figure 2, we can see that the solutions are t=π6𝑡=𝜋6 and t=5π6.𝑡=5𝜋6. But the problem is asking for all possible values that solve the equation. Therefore, the answer is

t=π6±2πk  and  t=5π6±2πk𝑡=𝜋6±2𝜋𝑘  and  𝑡=5𝜋6±2𝜋𝑘

where k𝑘 is an integer.

HOW TO

Given a trigonometric equation, solve using algebra.

1. Look for a pattern that suggests an algebraic property, such as the difference of squares or a factoring opportunity.
2. Substitute the trigonometric expression with a single variable, such as x𝑥 or u.𝑢.
3. Solve the equation the same way an algebraic equation would be solved.
4. Substitute the trigonometric expression back in for the variable in the resulting expressions.
5. Solve for the angle.

EXAMPLE 3

Solve the Linear Trigonometric Equation

Solve the equation exactly: 2cosθ−3=−5,0≤θ<2π.2cos𝜃−3=−5,0≤𝜃<2𝜋.

Solution

Use algebraic techniques to solve the equation.

2cosθ−32cosθcosθθ====−5−2−1π2cos𝜃−3=−52cos𝜃=−2cos𝜃=−1𝜃=𝜋

TRY IT #1

Solve exactly the following linear equation on the interval [0,2π):2sinx+1=0.[0,2𝜋):2sin𝑥+1=0.

Solving Equations Involving a Single Trigonometric Function

When we are given equations that involve only one of the six trigonometric functions, their solutions involve using algebraic techniques and the unit circle (see Figure 2). We need to make several considerations when the equation involves trigonometric functions other than sine and cosine. Problems involving the reciprocals of the primary trigonometric functions need to be viewed from an algebraic perspective. In other words, we will write the reciprocal function, and solve for the angles using the function. Also, an equation involving the tangent function is slightly different from one containing a sine or cosine function. First, as we know, the period of tangent is π,𝜋, not 2π.2𝜋. Further, the domain of tangent is all real numbers with the exception of odd integer multiples of π2,𝜋2, unless, of course, a problem places its own restrictions on the domain.

EXAMPLE 4

Solving a Problem Involving a Single Trigonometric Function

Solve the problem exactly: 2sin2θ−1=0,0≤θ<2π.2sin2𝜃−1=0,0≤𝜃<2𝜋.

Solution

As this problem is not easily factored, we will solve using the square root property. First, we use algebra to isolate sinθ.sin𝜃. Then we will find the angles.

2sin2θ−1 2sin2θsin2θsin2θ−−−−√sinθθ======0112±12−−√±12√=±2√2π4,3π4,5π4,7π42sin2𝜃−1=0 2sin2𝜃=1sin2𝜃=12sin2𝜃=±12sin𝜃=±12=±22𝜃=𝜋4,3𝜋4,5𝜋4,7𝜋4

EXAMPLE 5

Solving a Trigonometric Equation Involving Cosecant

Solve the following equation exactly: cscθ=−2,0≤θ<4π.csc𝜃=−2,0≤𝜃<4𝜋.

Solution

We want all values of θ𝜃 for which cscθ=−2csc𝜃=−2 over the interval 0≤θ<4π.0≤𝜃<4𝜋.

cscθ1sinθsinθθ====−2−2−127π6,11π6,19π6,23π6csc𝜃=−21sin𝜃=−2sin𝜃=−12𝜃=7𝜋6,11𝜋6,19𝜋6,23𝜋6

Analysis

As sinθ=−12,sin𝜃=−12, notice that all four solutions are in the third and fourth quadrants.

EXAMPLE 6

Solving an Equation Involving Tangent

Solve the equation exactly: tan(θ−π2)=1,0≤θ<2π.tan(𝜃−𝜋2)=1,0≤𝜃<2𝜋.

Solution

Recall that the tangent function has a period of π.𝜋. On the interval [0,π),[0,𝜋), and at the angle of π4,𝜋4, the tangent has a value of 1. However, the angle we want is (θ−π2).(𝜃−𝜋2). Thus, if tan(π4)=1,tan(𝜋4)=1, then

θ−π2θ==π43π4±kπ𝜃−𝜋2=𝜋4𝜃=3𝜋4±𝑘𝜋

Over the interval [0,2π),[0,2𝜋), we have two solutions:

θ=3π4  and θ=3π4+π=7π4𝜃=3𝜋4  and 𝜃=3𝜋4+𝜋=7𝜋4

TRY IT #2

Find all solutions for tanx=3–√.tan𝑥=3.

EXAMPLE 7

Identify all Solutions to the Equation Involving Tangent

Identify all exact solutions to the equation 2(tanx+3)=5+tanx,0≤x<2π.2(tan𝑥+3)=5+tan𝑥,0≤𝑥<2𝜋.

Solution

We can solve this equation using only algebra. Isolate the expression tanxtan𝑥 on the left side of the equals sign.

2(tanx)+2(3)2tanx+6 2tanx−tanxtanx====5+tanx5+tanx5−6−12(tan𝑥)+2(3)=5+tan𝑥2tan𝑥+6=5+tan𝑥 2tan𝑥−tan𝑥=5−6tan𝑥=−1

There are two angles on the unit circle that have a tangent value of −1:θ=3π4−1:𝜃=3𝜋4 and θ=7π4.𝜃=7𝜋4.

Solve Trigonometric Equations Using a Calculator

Not all functions can be solved exactly using only the unit circle. When we must solve an equation involving an angle other than one of the special angles, we will need to use a calculator. Make sure it is set to the proper mode, either degrees or radians, depending on the criteria of the given problem.

EXAMPLE 8

Using a Calculator to Solve a Trigonometric Equation Involving Sine

Use a calculator to solve the equation sinθ=0.8,sin𝜃=0.8, where θ𝜃 is in radians.

Solution

Make sure mode is set to radians. To find θ,𝜃, use the inverse sine function. On most calculators, you will need to push the 2^ND button and then the SIN button to bring up the sin−1sin−1 function. What is shown on the screen is sin−1(.sin−1(. The calculator is ready for the input within the parentheses. For this problem, we enter sin−1(0.8),sin−1(0.8), and press ENTER. Thus, to four decimals places,

sin−1(0.8)≈0.9273sin−1(0.8)≈0.9273

The solution is

θ≈0.9273±2πk𝜃≈0.9273±2𝜋𝑘

The angle measurement in degrees is

θθ≈≈≈53.1°180°−53.1°126.9°𝜃≈53.1°𝜃≈180°−53.1°≈126.9°

Analysis

Note that a calculator will only return an angle in quadrants I or IV for the sine function, since that is the range of the inverse sine. The other angle is obtained by using π−θ.𝜋−𝜃. Thus, the additional solution is ≈2.2143±2πk≈2.2143±2π𝑘

EXAMPLE 9

Using a Calculator to Solve a Trigonometric Equation Involving Secant

Use a calculator to solve the equation secθ=−4,sec𝜃=−4, giving your answer in radians.

Solution

We can begin with some algebra.

secθ1cosθcosθ===−4−4−14sec𝜃=−41cos𝜃=−4cos𝜃=−14

Check that the MODE is in radians. Now use the inverse cosine function.

cos−1(−14)θ≈≈1.82351.8235+2πkcos−1(−14)≈1.8235𝜃≈1.8235+2𝜋𝑘

Since π2≈1.57𝜋2≈1.57 and π≈3.14,𝜋≈3.14, 1.8235 is between these two numbers, thus θ≈1.8235𝜃≈1.8235 is in quadrant II. Cosine is also negative in quadrant III. Note that a calculator will only return an angle in quadrants I or II for the cosine function, since that is the range of the inverse cosine. See Figure 2.

Figure 2

So, we also need to find the measure of the angle in quadrant III. In quadrant II, the reference angle is θ​​’≈π−1.8235≈1.3181.𝜃​​’≈𝜋−1.8235≈1.3181. The other solution in quadrant III is θ​​’≈π+1.3181≈4.4597.𝜃​​’≈𝜋+1.3181≈4.4597.

The solutions are θ≈1.8235±2πk𝜃≈1.8235±2𝜋𝑘 and θ≈4.4597±2πk.𝜃≈4.4597±2𝜋𝑘.

TRY IT #3

Solve cosθ=−0.2.cos𝜃=−0.2.

Solving Trigonometric Equations in Quadratic Form

Solving a quadratic equation may be more complicated, but once again, we can use algebra as we would for any quadratic equation. Look at the pattern of the equation. Is there more than one trigonometric function in the equation, or is there only one? Which trigonometric function is squared? If there is only one function represented and one of the terms is squared, think about the standard form of a quadratic. Replace the trigonometric function with a variable such as x𝑥 or u.𝑢. If substitution makes the equation look like a quadratic equation, then we can use the same methods for solving quadratics to solve the trigonometric equations.

EXAMPLE 10

Solving a Trigonometric Equation in Quadratic Form

Solve the equation exactly: cos2θ+3cosθ−1=0,0≤θ<2π.cos2𝜃+3cos𝜃−1=0,0≤𝜃<2𝜋.

Solution

We begin by using substitution and replacing cos θ𝜃 with x.𝑥. It is not necessary to use substitution, but it may make the problem easier to solve visually. Let cosθ=x.cos𝜃=𝑥. We have

x2+3x−1=0𝑥2+3𝑥−1=0

The equation cannot be factored, so we will use the quadratic formula x=−b±b2−4ac√2a.𝑥=−𝑏±𝑏2−4𝑎𝑐2𝑎.

x==−3±(−3)2−4(1)(−1)√2−3±13√2𝑥=−3±(−3)2−4(1)(−1)2=−3±132

Replace x𝑥 with cosθ,cos𝜃, and solve.

cosθθ==−3±13√2cos−1(−3+13√2)cos𝜃=−3±132𝜃=cos−1(−3+132)

Note that only the + sign is used. This is because we get an error when we solve θ=cos−1(−3−13√2)𝜃=cos−1(−3−132) on a calculator, since the domain of the inverse cosine function is [−1,1].[−1,1]. However, there is a second solution:

θ=≈cos−1(−3+13√2)1.26𝜃=cos−1(−3+132)≈1.26

This terminal side of the angle lies in quadrant I. Since cosine is also positive in quadrant IV, the second solution is

θ=≈2π−cos−1(−3+13√2) 5.02𝜃=2𝜋−cos−1(−3+132)≈ 5.02

EXAMPLE 11

Solving a Trigonometric Equation in Quadratic Form by Factoring

Solve the equation exactly: 2sin2θ−5sinθ+3=0,0≤θ≤2π.2sin2𝜃−5sin𝜃+3=0,0≤𝜃≤2𝜋.

Solution

Using grouping, this quadratic can be factored. Either make the real substitution, sinθ=u,sin𝜃=𝑢, or imagine it, as we factor:

 2sin2θ−5sinθ+3(2sinθ−3)(sinθ−1)==00 2sin2𝜃−5sin𝜃+3=0(2sin𝜃−3)(sin𝜃−1)=0

Now set each factor equal to zero.

2sinθ−32sinθsinθsinθ−1sinθ=====0332012sin𝜃−3=02sin𝜃=3sin𝜃=32sin𝜃−1=0sin𝜃=1

Next solve for θ:sinθ≠32,𝜃:sin𝜃≠32, as the range of the sine function is [−1,1].[−1,1]. However, sinθ=1,sin𝜃=1, giving the solution θ=π2.𝜃=𝜋2.

Analysis

Make sure to check all solutions on the given domain as some factors have no solution.

TRY IT #4

Solve sin2θ=2cosθ+2,0≤θ≤2π.sin2𝜃=2cos𝜃+2,0≤𝜃≤2𝜋. [Hint: Make a substitution to express the equation only in terms of cosine.]

EXAMPLE 12

Solving a Trigonometric Equation Using Algebra

Solve exactly:

2sin2θ+sinθ=0;0≤θ<2π2sin2𝜃+sin𝜃=0;0≤𝜃<2𝜋

Solution

This problem should appear familiar as it is similar to a quadratic. Let sinθ=x.sin𝜃=𝑥. The equation becomes 2×2+x=0.2𝑥2+𝑥=0. We begin by factoring:

2×2+xx(2x+1)==002𝑥2+𝑥=0𝑥(2𝑥+1)=0

Set each factor equal to zero.

x(2x+1)x===00−12𝑥=0(2𝑥+1)=0𝑥=−12

Then, substitute back into the equation the original expression sinθsin𝜃 for x.𝑥. Thus,

sinθθsinθθ====00,π−127π6,11π6sin𝜃=0𝜃=0,𝜋sin𝜃=−12𝜃=7𝜋6,11𝜋6

The solutions within the domain 0≤θ<2π0≤𝜃<2𝜋 are θ=0,π,7π6,11π6.𝜃=0,𝜋,7𝜋6,11𝜋6.

If we prefer not to substitute, we can solve the equation by following the same pattern of factoring and setting each factor equal to zero.

2sin2θ+sinθsinθ(2sinθ+1)sinθθ2sinθ+12sinθsinθθ========0000,π0−1−127π6,11π62sin2𝜃+sin𝜃=0sin𝜃(2sin𝜃+1)=0sin𝜃=0𝜃=0,𝜋2sin𝜃+1=02sin𝜃=−1sin𝜃=−12𝜃=7𝜋6,11𝜋6

Analysis

We can see the solutions on the graph in Figure 3. On the interval 0≤θ<2π,0≤𝜃<2𝜋, the graph crosses the x-axis four times, at the solutions noted. Notice that trigonometric equations that are in quadratic form can yield up to four solutions instead of the expected two that are found with quadratic equations. In this example, each solution (angle) corresponding to a positive sine value will yield two angles that would result in that value.

Figure 3

We can verify the solutions on the unit circle in Figure 2 as well.

EXAMPLE 13

Solving a Trigonometric Equation Quadratic in Form

Solve the equation quadratic in form exactly: 2sin2θ−3sinθ+1=0,0≤θ<2π.2sin2𝜃−3sin𝜃+1=0,0≤𝜃<2𝜋.

Solution

We can factor using grouping. Solution values of θ𝜃 can be found on the unit circle.

(2sinθ−1)(sinθ−1) 2sinθ−1sinθθsinθθ======0012π6,5π61π2(2sin𝜃−1)(sin𝜃−1)=0 2sin𝜃−1=0sin𝜃=12𝜃=𝜋6,5𝜋6sin𝜃=1𝜃=𝜋2

TRY IT #5

Solve the quadratic equation 2cos2θ+cosθ=0.2cos2𝜃+cos𝜃=0.

Solving Trigonometric Equations Using Fundamental Identities

While algebra can be used to solve a number of trigonometric equations, we can also use the fundamental identities because they make solving equations simpler. Remember that the techniques we use for solving are not the same as those for verifying identities. The basic rules of algebra apply here, as opposed to rewriting one side of the identity to match the other side. In the next example, we use two identities to simplify the equation.

EXAMPLE 14

Use Identities to Solve an Equation

Use identities to solve exactly the trigonometric equation over the interval 0≤x<2π.0≤𝑥<2𝜋.

cosxcos(2x)+sinxsin(2x)=3–√2cos𝑥cos(2𝑥)+sin𝑥sin(2𝑥)=32

Solution

Notice that the left side of the equation is the difference formula for cosine.

cosxcos(2x)+sinxsin(2x)cos(x−2x)cos(−x)cosx====3√23√23√23√2Difference formula for cosineUse the negative angle identity.cos𝑥cos(2𝑥)+sin𝑥sin(2𝑥)=32cos(𝑥−2𝑥)=32Difference formula for cosinecos(−𝑥)=32Use the negative angle identity.cos𝑥=32

From the unit circle in Figure 2, we see that cosx=3√2cos𝑥=32 when x=π6,11π6.𝑥=𝜋6,11𝜋6.

EXAMPLE 15

Solving the Equation Using a Double-Angle Formula

Solve the equation exactly using a double-angle formula: cos(2θ)=cosθ.cos(2𝜃)=cos𝜃.

Solution

We have three choices of expressions to substitute for the double-angle of cosine. As it is simpler to solve for one trigonometric function at a time, we will choose the double-angle identity involving only cosine:

cos(2θ)2cos2θ−12cos2θ−cosθ−1(2cosθ+1)(cosθ−1)2cosθ+1cosθcosθ−1cosθ========cosθcosθ000−1201cos(2𝜃)=cos𝜃2cos2𝜃−1=cos𝜃2cos2𝜃−cos𝜃−1=0(2cos𝜃+1)(cos𝜃−1)=02cos𝜃+1=0cos𝜃=−12cos𝜃−1=0cos𝜃=1

So, if cosθ=−12,cos𝜃=−12, then θ=2π3±2πk𝜃=2𝜋3±2𝜋𝑘 and θ=4π3±2πk;𝜃=4𝜋3±2𝜋𝑘; if cosθ=1,cos𝜃=1, then θ=0±2πk.𝜃=0±2𝜋𝑘.

EXAMPLE 16

Solving an Equation Using an Identity

Solve the equation exactly using an identity: 3cosθ+3=2sin2θ,0≤θ<2π.3cos𝜃+3=2sin2𝜃,0≤𝜃<2𝜋.

Solution

If we rewrite the right side, we can write the equation in terms of cosine:

3cosθ+33cosθ+33cosθ+32cos2θ+3cosθ+1(2cosθ+1)(cosθ+1)2cosθ+1cosθθcosθ+1cosθθ===========2sin2θ2(1−cos2θ)2−2cos2θ000−122π3,4π30−1π3cos𝜃+3=2sin2𝜃3cos𝜃+3=2(1−cos2𝜃)3cos𝜃+3=2−2cos2𝜃2cos2𝜃+3cos𝜃+1=0(2cos𝜃+1)(cos𝜃+1)=02cos𝜃+1=0cos𝜃=−12𝜃=2𝜋3,4𝜋3cos𝜃+1=0cos𝜃=−1𝜃=𝜋

Our solutions are θ=2π3,4π3,π.𝜃=2𝜋3,4𝜋3,𝜋.

Solving Trigonometric Equations with Multiple Angles

Sometimes it is not possible to solve a trigonometric equation with identities that have a multiple angle, such as sin(2x)sin(2𝑥) or cos(3x).cos(3𝑥). When confronted with these equations, recall that y=sin(2x)𝑦=sin(2𝑥) is a horizontal compression by a factor of 2 of the function y=sinx.𝑦=sin𝑥. On an interval of 2π,2𝜋, we can graph two periods of y=sin(2x),𝑦=sin(2𝑥), as opposed to one cycle of y=sinx.𝑦=sin𝑥. This compression of the graph leads us to believe there may be twice as many x-intercepts or solutions to sin(2x)=0sin(2𝑥)=0 compared to sinx=0.sin𝑥=0. This information will help us solve the equation.

EXAMPLE 17

Solving a Multiple Angle Trigonometric Equation

Solve exactly: cos(2x)=12cos(2𝑥)=12 on [0,2π).[0,2𝜋).

Solution

We can see that this equation is the standard equation with a multiple of an angle. If cos(α)=12,cos(𝛼)=12, we know α𝛼 is in quadrants I and IV. While θ=cos−112𝜃=cos−112 will only yield solutions in quadrants I and II, we recognize that the solutions to the equation cosθ=12cos𝜃=12 will be in quadrants I and IV.

Therefore, the possible angles are θ=π3𝜃=𝜋3 and θ=5π3.𝜃=5𝜋3. So, 2x=π32𝑥=𝜋3 or 2x=5π3,2𝑥=5𝜋3, which means that x=π6𝑥=𝜋6 or x=5π6.𝑥=5𝜋6. Does this make sense? Yes, because cos(2(π6))=cos(π3)=12.cos(2(𝜋6))=cos(𝜋3)=12.

Are there any other possible answers? Let us return to our first step.

In quadrant I, 2x=π3,2𝑥=𝜋3, so x=π6𝑥=𝜋6 as noted. Let us revolve around the circle again:

2x===π3+2ππ3+6π37π32𝑥=𝜋3+2𝜋=𝜋3+6𝜋3=7𝜋3

so x=7π6.𝑥=7𝜋6.

One more rotation yields

2x===π3+4ππ3+12π313π32𝑥=𝜋3+4𝜋=𝜋3+12𝜋3=13𝜋3

x=13π6>2π,𝑥=13𝜋6>2𝜋, so this value for x𝑥 is larger than 2π,2𝜋, so it is not a solution on [0,2π).[0,2𝜋).

In quadrant IV, 2x=5π3,2𝑥=5𝜋3, so x=5π6𝑥=5𝜋6 as noted. Let us revolve around the circle again:

2x===5π3+2π5π3+6π311π32𝑥=5𝜋3+2𝜋=5𝜋3+6𝜋3=11𝜋3

so x=11π6.𝑥=11𝜋6.

One more rotation yields

2x===5π3+4π5π3+12π317π32𝑥=5𝜋3+4𝜋=5𝜋3+12𝜋3=17𝜋3

x=17π6>2π,𝑥=17𝜋6>2𝜋, so this value for x𝑥 is larger than 2π,2𝜋, so it is not a solution on [0,2π).[0,2𝜋).

Our solutions are x=π6,5π6,7π6,and 11π6𝑥=𝜋6,5𝜋6,7𝜋6,and 11𝜋6. Note that whenever we solve a problem in the form of sin(nx)=c,sin(𝑛𝑥)=𝑐, we must go around the unit circle n𝑛 times.

Solving Right Triangle Problems

We can now use all of the methods we have learned to solve problems that involve applying the properties of right triangles and the Pythagorean Theorem. We begin with the familiar Pythagorean Theorem, a2+b2=c2,𝑎2+𝑏2=𝑐2, and model an equation to fit a situation.

EXAMPLE 18

Using the Pythagorean Theorem to Model an Equation

Use the Pythagorean Theorem, and the properties of right triangles to model an equation that fits the problem.

One of the cables that anchors the center of the London Eye Ferris wheel to the ground must be replaced. The center of the Ferris wheel is 69.5 meters above the ground, and the second anchor on the ground is 23 meters from the base of the Ferris wheel. Approximately how long is the cable, and what is the angle of elevation (from ground up to the center of the Ferris wheel)? See Figure 4.

Figure 4

Solution

Using the information given, we can draw a right triangle. We can find the length of the cable with the Pythagorean Theorem.

a2+b2(23)2+(69.5)25359−−−−√=≈≈c2535973.2 m𝑎2+𝑏2=𝑐2(23)2+(69.5)2≈53595359≈73.2 m

The angle of elevation is θ,𝜃, formed by the second anchor on the ground and the cable reaching to the center of the wheel. We can use the tangent function to find its measure. Round to two decimal places.

tanθtan−1(69.523)=≈≈69.5231.252271.69°tan𝜃=69.523tan−1(69.523)≈1.2522≈71.69°

The angle of elevation is approximately 71.7°,71.7°, and the length of the cable is 73.2 meters.

EXAMPLE 19

Using the Pythagorean Theorem to Model an Abstract Problem

OSHA safety regulations require that the base of a ladder be placed 1 foot from the wall for every 4 feet of ladder length. Find the angle that a ladder of any length forms with the ground and the height at which the ladder touches the wall.

Solution

For any length of ladder, the base needs to be a distance from the wall equal to one fourth of the ladder’s length. Equivalently, if the base of the ladder is “a” feet from the wall, the length of the ladder will be 4a feet. See Figure 5.

Figure 5

The side adjacent to θ𝜃 is a and the hypotenuse is 4a.4𝑎. Thus,

cosθcos−1(14)=≈a4a=1475.5°cos𝜃=𝑎4𝑎=14cos−1(14)≈75.5°

The elevation of the ladder forms an angle of 75.5°75.5° with the ground. The height at which the ladder touches the wall can be found using the Pythagorean Theorem:

a2+b2b2b2b2b=====(4a)2(4a)2−a216a2−a215a2a15−−√𝑎2+𝑏2=(4𝑎)2𝑏2=(4𝑎)2−𝑎2𝑏2=16𝑎2−𝑎2𝑏2=15𝑎2𝑏=𝑎15

Thus, the ladder touches the wall at a15−−√𝑎15 feet from the ground.