Javascript required
Skip to content Skip to sidebar Skip to footer

Precalculus 9 2 Review Power and Radical Functions Answers

Learning Objectives

In this section, yous will:

  • Solve linear trigonometric equations in sine and cosine.
  • Solve equations involving a single trigonometric part.
  • Solve trigonometric equations using a calculator.
  • Solve trigonometric equations that are quadratic in form.
  • Solve trigonometric equations using primal identities.
  • Solve trigonometric equations with multiple angles.
  • Solve right triangle problems.
Photo of the Egyptian pyramids near a modern city.

Figure one Egyptian pyramids continuing near a modern urban center. (credit: Oisin Mulvihill)

Thales of Miletus (circa 625–547 BC) is known every bit the founder of geometry. The legend is that he calculated the peak of the Great Pyramid of Giza in Arab republic of egypt using the theory of similar triangles, which he developed by measuring the shadow of his staff. Based on proportions, this theory has applications in a number of areas, including fractal geometry, engineering, and architecture. Often, the bending of elevation and the bending of depression are constitute using similar triangles.

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

Solving Linear Trigonometric Equations in Sine and Cosine

Trigonometric equations are, as the proper noun implies, equations that involve trigonometric functions. Similar in many means to solving polynomial equations or rational equations, only specific values of the variable will be solutions, if there are solutions at all. Oftentimes we volition solve a trigonometric equation over a specified interval. However, just equally frequently, we will be asked to observe 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 exist considered before we assume that any solution is valid. The flow of both the sine function and the cosine role is ii π . 2 π . In other words, every 2 π ii π units, the y-values repeat. If we demand to detect all possible solutions, then we must add 2 π chiliad , ii π k , where k k 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 2 π : 2 π :

sin θ = sin ( θ ± 2 yard Ï€ ) sin θ = sin ( θ ± ii k Ï€ )

In that location are similar rules for indicating all possible solutions for the other trigonometric functions. Solving trigonometric equations requires the same techniques every bit solving algebraic equations. Nosotros read the equation from left to right, horizontally, similar a judgement. We expect for known patterns, factor, find common denominators, and substitute certain expressions with a variable to make solving a more straightforward process. Nonetheless, with trigonometric equations, nosotros besides have the advantage of using the identities we developed in the previous sections.

Example one

Solving a Linear Trigonometric Equation Involving the Cosine Function

Detect all possible exact solutions for the equation cos θ = ane 2 . cos θ = 1 2 .

Example 2

Solving a Linear Equation Involving the Sine Function

Find all possible exact solutions for the equation sin t = one two . sin t = 1 two .

How To

Given a trigonometric equation, solve using algebra.

  1. Look for a pattern that suggests an algebraic property, such as the departure of squares or a factoring opportunity.
  2. Substitute the trigonometric expression with a single variable, such as 10 x or u . u .
  3. Solve the equation the same way an algebraic equation would exist solved.
  4. Substitute the trigonometric expression dorsum in for the variable in the resulting expressions.
  5. Solve for the angle.

Example 3

Solve the Trigonometric Equation in Linear Form

Solve the equation exactly: 2 cos θ 3 = 5 , 0 θ < 2 π . 2 cos θ three = five , 0 θ < 2 π .

Try It #1

Solve exactly the following linear equation on the interval [ 0 , two π ) : ii sin 10 + 1 = 0. [ 0 , 2 π ) : 2 sin x + 1 = 0.

Solving Equations Involving a Single Trigonometric Function

When we are given equations that involve only one of the half dozen trigonometric functions, their solutions involve using algebraic techniques and the unit circle (run into 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 main trigonometric functions need to be viewed from an algebraic perspective. In other words, nosotros will write the reciprocal function, and solve for the angles using the function. Likewise, an equation involving the tangent office is slightly different from ane containing a sine or cosine function. Kickoff, as we know, the period of tangent is π , π , non 2 π . ii π . Further, the domain of tangent is all real numbers with the exception of odd integer multiples of π 2 , π 2 , unless, of course, a problem places its own restrictions on the domain.

Instance iv

Solving a Trouble Involving a Single Trigonometric Part

Solve the problem exactly: 2 sin two θ 1 = 0 , 0 θ < 2 π . two sin 2 θ 1 = 0 , 0 θ < 2 π .

Case 5

Solving a Trigonometric Equation Involving Cosecant

Solve the following equation exactly: csc θ = two , 0 θ < 4 π . csc θ = 2 , 0 θ < 4 π .

Analysis

Equally sin θ = 1 2 , sin θ = 1 two , observe that all four solutions are in the third and quaternary quadrants.

Instance six

Solving an Equation Involving Tangent

Solve the equation exactly: tan ( θ π ii ) = 1 , 0 θ < 2 π . tan ( θ π ii ) = ane , 0 θ < 2 π .

Attempt Information technology #2

Find all solutions for tan x = 3 . tan ten = 3 .

Case 7

Identify all Solutions to the Equation Involving Tangent

Identify all exact solutions to the equation 2 ( tan 10 + 3 ) = 5 + tan x , 0 x < 2 π . two ( tan x + 3 ) = 5 + tan x , 0 x < ii π .

Solve Trigonometric Equations Using a Calculator

Non 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, nosotros volition demand to use a figurer. Make certain it is set up to the proper fashion, either degrees or radians, depending on the criteria of the given problem.

Instance 8

Using a Calculator to Solve a Trigonometric Equation Involving Sine

Utilize a estimator to solve the equation sin θ = 0.8 , sin θ = 0.8 , where θ θ is in radians.

Analysis

Annotation that a calculator will only return an bending in quadrants I or 4 for the sine function, since that is the range of the inverse sine. The other angle is obtained past using π θ . π θ .

Example 9

Using a Calculator to Solve a Trigonometric Equation Involving Secant

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

Effort Information technology #3

Solve cos θ = 0.2. cos θ = 0.ii.

Solving Trigonometric Equations in Quadratic Form

Solving a quadratic equation may be more complicated, but once again, nosotros tin utilise algebra every bit we would for whatever quadratic equation. Look at the pattern of the equation. Is there more than one trigonometric office in the equation, or is at that place merely one? Which trigonometric function is squared? If in that location is simply one function represented and one of the terms is squared, recollect about the standard form of a quadratic. Supplant the trigonometric role with a variable such as x x or u . u . If substitution makes the equation look like a quadratic equation, and then we can utilize the same methods for solving quadratics to solve the trigonometric equations.

Example 10

Solving a Trigonometric Equation in Quadratic Form

Solve the equation exactly: cos 2 θ + three cos θ 1 = 0 , 0 θ < ii π . cos 2 θ + iii cos θ 1 = 0 , 0 θ < 2 π .

Instance eleven

Solving a Trigonometric Equation in Quadratic Form past Factoring

Solve the equation exactly: 2 sin 2 θ 5 sin θ + 3 = 0 , 0 θ 2 π . 2 sin 2 θ 5 sin θ + iii = 0 , 0 θ 2 π .

Analysis

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

Try Information technology #four

Solve sin 2 θ = 2 cos θ + 2 , 0 θ 2 π . sin ii θ = 2 cos θ + 2 , 0 θ 2 π . [Hint: Make a substitution to express the equation simply in terms of cosine.]

Case 12

Solving a Trigonometric Equation Using Algebra

Solve exactly:

2 sin 2 θ + sin θ = 0 ; 0 θ < two π 2 sin 2 θ + sin θ = 0 ; 0 θ < ii π

Assay

Nosotros tin see the solutions on the graph in Figure 3. On the interval 0 θ < ii π , 0 θ < 2 π , the graph crosses the x-centrality 4 times, at the solutions noted. Find that trigonometric equations that are in quadratic form can yield up to four solutions instead of the expected ii that are found with quadratic equations. In this case, each solution (angle) respective to a positive sine value will yield two angles that would result in that value.

Graph of 2*(sin(theta))^2 + sin(theta) from 0 to 2pi. Zeros are at 0, pi, 7pi/6, and 11pi/6.

Figure iii

Nosotros can verify the solutions on the unit circumvolve in Figure ii as well.

Example xiii

Solving a Trigonometric Equation Quadratic in Form

Solve the equation quadratic in grade exactly: 2 sin two θ 3 sin θ + 1 = 0 , 0 θ < 2 π . 2 sin 2 θ 3 sin θ + ane = 0 , 0 θ < 2 π .

Endeavor It #v

Solve the quadratic equation ii cos 2 θ + cos θ = 0. two cos 2 θ + cos θ = 0.

Solving Trigonometric Equations Using Cardinal Identities

While algebra tin can be used to solve a number of trigonometric equations, we can as well utilize the cardinal identities considering they make solving equations simpler. Retrieve that the techniques we use for solving are not the same as those for verifying identities. The basic rules of algebra utilize 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.

Case 14

Use Identities to Solve an Equation

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

cos x cos ( two x ) + sin x sin ( ii x ) = iii ii cos ten cos ( 2 x ) + sin x sin ( ii x ) = 3 two

Example 15

Solving the Equation Using a Double-Angle Formula

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

Example xvi

Solving an Equation Using an Identity

Solve the equation exactly using an identity: three cos θ + 3 = 2 sin 2 θ , 0 θ < 2 π . three cos θ + 3 = 2 sin 2 θ , 0 θ < 2 π .

Solving Trigonometric Equations with Multiple Angles

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

Case 17

Solving a Multiple Angle Trigonometric Equation

Solve exactly: cos ( ii ten ) = ane 2 cos ( two x ) = 1 2 on [ 0 , 2 π ) . [ 0 , ii π ) .

Solving Correct Triangle Problems

We can now use all of the methods nosotros have learned to solve issues that involve applying the properties of right triangles and the Pythagorean Theorem. We brainstorm with the familiar Pythagorean Theorem, a 2 + b 2 = c 2 , a ii + b two = c two , and model an equation to fit a situation.

Case 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 eye of the London Eye Ferris bicycle 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 footing is 23 meters from the base of the Ferris cycle. Approximately how long is the cable, and what is the angle of pinnacle (from ground upwards to the heart of the Ferris bike)? See Figure 4.

Basic diagram of a ferris wheel (circle) and its support cables (form a right triangle). One cable runs from the center of the circle to the ground (outside the circle), is perpendicular to the ground, and has length 69.5. Another cable of unknown length (the hypotenuse) runs from the center of the circle to the ground 23 feet away from the other cable at an angle of theta degrees with the ground. So, in closing, there is a right triangle with base 23, height 69.5, hypotenuse unknown, and angle between base and hypotenuse of theta degrees.

Figure iv

Example 19

Using the Pythagorean Theorem to Model an Abstract Problem

OSHA safety regulations crave that the base of operations of a ladder exist placed one 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.

7.v Section Exercises

Verbal

1 .

Will there always be solutions to trigonometric part equations? If non, depict an equation that would not take a solution. Explain why or why not.

2 .

When solving a trigonometric equation involving more than one trig role, do we ever want to attempt to rewrite the equation so it is expressed in terms of one trigonometric function? Why or why not?

iii .

When solving linear trig equations in terms of but sine or cosine, how practise we know whether there will be solutions?

Algebraic

For the following exercises, find all solutions exactly on the interval 0 θ < 2 π . 0 θ < 2 π .

eleven .

4 sin 2 x 2 = 0 four sin two x two = 0

12 .

csc 2 x 4 = 0 csc ii ten 4 = 0

For the following exercises, solve exactly on [ 0 , ii π ) . [ 0 , 2 π ) .

17 .

2 sin ( 3 θ ) = 1 2 sin ( three θ ) = one

18 .

2 sin ( 2 θ ) = 3 2 sin ( 2 θ ) = 3

19 .

2 cos ( 3 θ ) = two ii cos ( 3 θ ) = 2

xx .

cos ( 2 θ ) = 3 2 cos ( 2 θ ) = three 2

21 .

ii sin ( π θ ) = ane two sin ( π θ ) = 1

22 .

2 cos ( π 5 θ ) = 3 2 cos ( π five θ ) = 3

For the post-obit exercises, discover all exact solutions on [ 0 , two π ) . [ 0 , 2 π ) .

23 .

sec ( x ) sin ( ten ) 2 sin ( x ) = 0 sec ( ten ) sin ( ten ) 2 sin ( 10 ) = 0

24 .

tan ( x ) 2 sin ( x ) tan ( 10 ) = 0 tan ( ten ) 2 sin ( x ) tan ( ten ) = 0

25 .

two cos 2 t + cos ( t ) = one 2 cos 2 t + cos ( t ) = ane

26 .

2 tan 2 ( t ) = 3 sec ( t ) 2 tan 2 ( t ) = 3 sec ( t )

27 .

2 sin ( x ) cos ( 10 ) sin ( x ) + two cos ( 10 ) 1 = 0 ii sin ( x ) cos ( x ) sin ( x ) + two cos ( 10 ) 1 = 0

28 .

cos ii θ = 1 2 cos ii θ = i 2

30 .

tan 2 ( 10 ) = 1 + two tan ( x ) tan 2 ( x ) = 1 + 2 tan ( x )

31 .

eight sin 2 ( x ) + 6 sin ( x ) + 1 = 0 8 sin 2 ( x ) + six sin ( 10 ) + 1 = 0

32 .

tan five ( x ) = tan ( x ) tan 5 ( x ) = tan ( 10 )

For the following exercises, solve with the methods shown in this section exactly on the interval [ 0 , ii π ) . [ 0 , 2 π ) .

33 .

sin ( iii x ) cos ( half dozen x ) cos ( 3 x ) sin ( 6 x ) = −0.ix sin ( three 10 ) cos ( half-dozen x ) cos ( 3 x ) sin ( half dozen x ) = −0.9

34 .

sin ( six x ) cos ( 11 ten ) cos ( 6 10 ) sin ( 11 x ) = −0.1 sin ( 6 ten ) cos ( 11 x ) cos ( half dozen ten ) sin ( 11 10 ) = −0.1

35 .

cos ( ii 10 ) cos x + sin ( two x ) sin x = i cos ( two ten ) cos x + sin ( ii x ) sin 10 = i

36 .

6 sin ( 2 t ) + 9 sin t = 0 vi sin ( 2 t ) + 9 sin t = 0

37 .

ix cos ( ii θ ) = ix cos 2 θ 4 nine cos ( 2 θ ) = 9 cos 2 θ 4

38 .

sin ( ii t ) = cos t sin ( 2 t ) = cos t

39 .

cos ( 2 t ) = sin t cos ( 2 t ) = sin t

40 .

cos ( six x ) cos ( 3 x ) = 0 cos ( 6 x ) cos ( 3 ten ) = 0

For the following exercises, solve exactly on the interval [ 0 , two π ) . [ 0 , 2 π ) . Use the quadratic formula if the equations do non factor.

41 .

tan 2 x iii tan x = 0 tan two 10 3 tan x = 0

42 .

sin 2 x + sin x 2 = 0 sin 2 10 + sin x 2 = 0

43 .

sin two x 2 sin x 4 = 0 sin 2 x two sin ten 4 = 0

44 .

five cos 2 10 + 3 cos x one = 0 5 cos two x + 3 cos 10 1 = 0

45 .

iii cos 2 x 2 cos ten 2 = 0 3 cos 2 10 2 cos x ii = 0

46 .

five sin ii x + 2 sin x 1 = 0 5 sin two ten + 2 sin x 1 = 0

47 .

tan 2 x + 5 tan ten ane = 0 tan 2 x + 5 tan 10 1 = 0

48 .

cot 2 x = cot x cot 2 10 = cot x

49 .

tan 2 x tan 10 2 = 0 tan 2 x tan x two = 0

For the following exercises, find exact solutions on the interval [ 0 , 2 π ) . [ 0 , 2 π ) . Look for opportunities to apply trigonometric identities.

50 .

sin 2 ten cos 2 x sin x = 0 sin two x cos 2 x sin x = 0

51 .

sin ii x + cos two x = 0 sin 2 x + cos two x = 0

52 .

sin ( two ten ) sin x = 0 sin ( ii x ) sin 10 = 0

53 .

cos ( two ten ) cos x = 0 cos ( ii x ) cos x = 0

54 .

2 tan 10 2 sec 2 x sin 2 ten = cos 2 x 2 tan x 2 sec 2 ten sin 2 x = cos two 10

55 .

1 cos ( two x ) = 1 + cos ( 2 ten ) 1 cos ( two x ) = 1 + cos ( ii x )

57 .

x sin 10 cos x = vi cos 10 10 sin ten cos x = vi cos ten

58 .

−3 sin t = fifteen cos t sin t −3 sin t = 15 cos t sin t

59 .

4 cos ii x 4 = 15 cos x 4 cos two 10 4 = 15 cos x

60 .

8 sin 2 x + 6 sin x + ane = 0 8 sin 2 x + 6 sin 10 + 1 = 0

61 .

8 cos 2 θ = three two cos θ 8 cos 2 θ = 3 2 cos θ

62 .

6 cos ii ten + 7 sin x 8 = 0 6 cos 2 x + 7 sin x 8 = 0

63 .

12 sin 2 t + cos t six = 0 12 sin 2 t + cos t 6 = 0

64 .

tan 10 = 3 sin x tan x = 3 sin 10

65 .

cos iii t = cos t cos 3 t = cos t

Graphical

For the following exercises, algebraically determine all solutions of the trigonometric equation exactly, then verify the results by graphing the equation and finding the zeros.

66 .

six sin two x five sin 10 + 1 = 0 vi sin 2 x 5 sin x + one = 0

67 .

8 cos 2 x 2 cos x 1 = 0 8 cos two 10 2 cos x 1 = 0

68 .

100 tan 2 x + twenty tan ten three = 0 100 tan 2 x + 20 tan x 3 = 0

69 .

2 cos 2 x cos 10 + 15 = 0 2 cos 2 x cos x + 15 = 0

lxx .

20 sin two ten 27 sin x + 7 = 0 20 sin 2 ten 27 sin x + vii = 0

71 .

2 tan two x + 7 tan x + vi = 0 2 tan 2 10 + 7 tan x + 6 = 0

72 .

130 tan ii x + 69 tan ten 130 = 0 130 tan 2 x + 69 tan x 130 = 0

Engineering science

For the following exercises, utilise a estimator to notice all solutions to four decimal places.

74 .

sin 10 = −0.55 sin 10 = −0.55

75 .

tan x = −0.34 tan x = −0.34

For the post-obit exercises, solve the equations algebraically, and so utilise a calculator to find the values on the interval [ 0 , 2 π ) . [ 0 , ii π ) . Round to four decimal places.

77 .

tan 2 ten + iii tan x iii = 0 tan 2 x + 3 tan x 3 = 0

78 .

6 tan 2 ten + 13 tan x = −6 6 tan 2 x + 13 tan x = −half dozen

79 .

tan 2 x sec x = i tan ii x sec x = 1

80 .

sin 2 x two cos ii x = 0 sin 2 ten 2 cos 2 x = 0

81 .

2 tan 2 x + nine tan x half-dozen = 0 ii tan 2 x + ix tan ten 6 = 0

82 .

4 sin 2 ten + sin ( 2 10 ) sec 10 three = 0 4 sin two ten + sin ( 2 x ) sec x 3 = 0

Extensions

For the post-obit exercises, find all solutions exactly to the equations on the interval [ 0 , two π ) . [ 0 , 2 π ) .

83 .

csc ii x iii csc 10 4 = 0 csc 2 10 iii csc 10 4 = 0

84 .

sin ii x cos 2 x 1 = 0 sin 2 10 cos 2 ten i = 0

85 .

sin 2 x ( 1 sin ii x ) + cos 2 x ( 1 sin 2 x ) = 0 sin two x ( 1 sin two x ) + cos 2 10 ( one sin 2 x ) = 0

86 .

iii sec 2 10 + 2 + sin 2 x tan 2 x + cos two x = 0 3 sec two ten + 2 + sin 2 ten tan 2 ten + cos 2 x = 0

87 .

sin two x 1 + 2 cos ( 2 x ) cos 2 x = 1 sin 2 10 1 + 2 cos ( 2 x ) cos 2 x = 1

88 .

tan 2 x 1 sec 3 x cos x = 0 tan two x 1 sec three 10 cos x = 0

89 .

sin ( 2 x ) sec 2 10 = 0 sin ( 2 x ) sec 2 x = 0

xc .

sin ( 2 x ) 2 csc 2 ten = 0 sin ( 2 ten ) two csc 2 x = 0

91 .

2 cos 2 x sin two 10 cos x five = 0 2 cos 2 x sin 2 x cos ten 5 = 0

92 .

1 sec 2 10 + 2 + sin two x + 4 cos 2 x = 4 1 sec two x + 2 + sin ii 10 + 4 cos ii x = 4

Real-Earth Applications

93 .

An airplane has but plenty gas to fly to a metropolis 200 miles northeast of its current location. If the pilot knows that the city is 25 miles n, how many degrees north of due east should the aeroplane fly?

94 .

If a loading ramp is placed next to a truck, at a superlative of 4 feet, and the ramp is 15 anxiety long, what bending does the ramp brand with the ground?

95 .

If a loading ramp is placed next to a truck, at a pinnacle of 2 anxiety, and the ramp is 20 feet long, what angle does the ramp make with the ground?

96 .

A adult female is watching a launched rocket currently 11 miles in distance. If she is standing 4 miles from the launch pad, at what angle is she looking up from horizontal?

97 .

An astronaut is in a launched rocket currently 15 miles in altitude. If a human is standing 2 miles from the launch pad, at what angle is she looking down at him from horizontal? (Hint: this is called the angle of depression.)

98 .

A woman is continuing 8 meters away from a x-meter tall building. At what angle is she looking to the summit of the building?

99 .

A man is standing 10 meters away from a 6-meter tall building. Someone at the top of the edifice is looking down at him. At what angle is the person looking at him?

100 .

A twenty-foot tall building has a shadow that is 55 feet long. What is the angle of elevation of the dominicus?

101 .

A ninety-foot tall building has a shadow that is 2 feet long. What is the angle of elevation of the sun?

102 .

A spotlight on the ground three meters from a 2-meter tall homo casts a 6 meter shadow on a wall 6 meters from the homo. At what bending is the lite?

103 .

A spotlight on the basis 3 feet from a 5-foot tall adult female casts a 15-pes tall shadow on a wall six feet from the woman. At what bending is the light?

For the post-obit exercises, discover a solution to the word problem algebraically. So use a calculator to verify the result. Round the reply to the nearest tenth of a degree.

104 .

A person does a handstand with his feet touching a wall and his easily 1.5 feet away from the wall. If the person is 6 anxiety alpine, what angle practice his anxiety make with the wall?

105 .

A person does a handstand with her feet touching a wall and her hands three anxiety abroad from the wall. If the person is 5 feet tall, what angle do her feet make with the wall?

106 .

A 23-pes ladder is positioned next to a house. If the ladder slips at 7 feet from the business firm when in that location is not plenty traction, what angle should the ladder brand with the ground to avoid slipping?

macdonnellfruck1991.blogspot.com

Source: https://openstax.org/books/precalculus-2e/pages/7-5-solving-trigonometric-equations