problem solving using angle properties and algebraic equations

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CHAPTER 9 Trigonometry

9.1 Use Properties of Angles, Triangles, and the Pythagorean Theorem

Learning Objectives

By the end of this section, you will be able to:

• Use the properties of angles
• Use the properties of triangles

Use the Pythagorean Theorem

Use the properties of angles.

Supplementary and Complementary Angles

In this section and the next, you will be introduced to some common geometry formulas. We will adapt our Problem Solving Strategy for Geometry Applications. The geometry formula will name the variables and give us the equation to solve.

In addition, since these applications will all involve geometric shapes, it will be helpful to draw a figure and then label it with the information from the problem. We will include this step in the Problem Solving Strategy for Geometry Applications.

HOW TO: Use a Problem Solving Strategy for Geometry Applications

• Read the problem and make sure you understand all the words and ideas. Draw a figure and label it with the given information.
• Identify what you are looking for.
• Name what you are looking for and choose a variable to represent it.
• Translate into an equation by writing the appropriate formula or model for the situation. Substitute in the given information.
• Solve the equation using good algebra techniques.
• Check the answer in the problem and make sure it makes sense.
• Answer the question with a complete sentence.

The next example will show how you can use the Problem Solving Strategy for Geometry Applications to answer questions about supplementary and complementary angles.

Use the Properties of Triangles

Sum of the Measures of the Angles of a Triangle

Right Triangles

In the examples so far, we could draw a figure and label it directly after reading the problem. In the next example, we will have to define one angle in terms of another. So we will wait to draw the figure until we write expressions for all the angles we are looking for.

20°, 70°, 90°

30°, 60°, 90°

Similar Triangles

When we use a map to plan a trip, a sketch to build a bookcase, or a pattern to sew a dress, we are working with similar figures. In geometry, if two figures have exactly the same shape but different sizes, we say they are similar figures. One is a scale model of the other. The corresponding sides of the two figures have the same ratio, and all their corresponding angles are have the same measures.

Properties of Similar Triangles

If two triangles are similar, then their corresponding angle measures are equal and their corresponding side lengths are in the same ratio.

We will often use this notation when we solve similar triangles because it will help us match up the corresponding side lengths.

The Pythagorean Theorem tells how the lengths of the three sides of a right triangle relate to each other. It states that in any right triangle, the sum of the squares of the two legs equals the square of the hypotenuse.

The Pythagorean Theorem

We will use this definition of square roots to solve for the length of a side in a right triangle.

Use the Pythagorean Theorem to find the length of the hypotenuse.

Use the Pythagorean Theorem to find the length of the longer leg.

Use the Pythagorean Theorem to find the length of the leg.

Key Concepts

• If the sum of the measures of two angles is 180°, then the angles are supplementary.

• If the sum of the measures of two angles is 90°, then the angles are complementary.

• A right triangle is a triangle that has one 90° angle, which is often marked with a ⦜ symbol.
• If two triangles are similar, then their corresponding angle measures are equal and their corresponding side lengths have the same ratio.

Practice Makes Perfect

In the following exercises, find a) the supplement and b) the complement of the given angle.

In the following exercises, use the properties of angles to solve.

In the following exercises, solve using properties of triangles.

Find the Length of the Missing Side

In the following exercises, use the Pythagorean Theorem to find the length of the hypotenuse.

In the following exercises, use the Pythagorean Theorem to find the length of the missing side. Round to the nearest tenth, if necessary.

In the following exercises, solve. Approximate to the nearest tenth, if necessary.

Everyday Math

Writing Exercises

Attributions

This chapter has been adapted from “Use Properties of Angles, Triangles, and the Pythagorean Theorem” in Prealgebra (OpenStax) by Lynn Marecek, MaryAnne Anthony-Smith, and Andrea Honeycutt Mathis, which is under a CC BY 4.0 Licence . Adapted by Izabela Mazur. See the Copyright page for more information.

Introductory Algebra Copyright © 2021 by Izabela Mazur is licensed under a Creative Commons Attribution 4.0 International License , except where otherwise noted.

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9.5 Solving Trigonometric Equations

Learning objectives.

• 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.

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 2 π . 2 π . In other words, every 2 π 2 π units, the y- values repeat. If we need to find all possible solutions, then we must add 2 π k , 2 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 :

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.

Solving a Linear Trigonometric Equation Involving the Cosine Function

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

From the unit circle , we know that

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

where k k is an integer.

Solving a Linear Equation Involving the Sine Function

Find all possible exact solutions for the equation sin t = 1 2 . sin t = 1 2 .

Solving for all possible values of t means that solutions include angles beyond the period of 2 π . 2 π . From Figure 2 , we can see that the solutions are t = π 6 t = π 6 and t = 5 π 6 . t = 5 π 6 . But the problem is asking for all possible values that solve the equation. Therefore, the answer is

Given a trigonometric equation, solve using algebra .

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

Solve the Linear Trigonometric Equation

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

Use algebraic techniques to solve the equation.

Solve exactly the following linear equation on the interval [ 0 , 2 π ) : 2 sin x + 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 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 2 π . 2 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.

Solving a Problem Involving a Single Trigonometric Function

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

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

Solving a Trigonometric Equation Involving Cosecant

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

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

As sin θ = − 1 2 , sin θ = − 1 2 , notice that all four solutions are in the third and fourth quadrants.

Solving an Equation Involving Tangent

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

Recall that the tangent function has a period of π . π . On the interval [ 0 , π ) , [ 0 , π ) , and at the angle of π 4 , π 4 , the tangent has a value of 1. However, the angle we want is ( θ − π 2 ) . ( θ − π 2 ) . Thus, if tan ( π 4 ) = 1 , tan ( π 4 ) = 1 , then

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

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

Identify all Solutions to the Equation Involving Tangent

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

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

There are two angles on the unit circle that have a tangent value of −1 : θ = 3 π 4 −1 : θ = 3 π 4 and θ = 7 π 4 . θ = 7 π 4 .

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.

Using a Calculator to Solve a Trigonometric Equation Involving Sine

Use a calculator to solve the equation sin θ = 0.8 , sin θ = 0.8 , where θ θ is in radians.

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 − 1 sin − 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,

The solution is

The angle measurement in degrees is

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 ± 2 π k ≈ 2.2143 ± 2 π k

Using a Calculator to Solve a Trigonometric Equation Involving Secant

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

We can begin with some algebra.

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

Since π 2 ≈ 1.57 π 2 ≈ 1.57 and π ≈ 3.14 , π ≈ 3.14 , 1.8235 is between these two numbers, thus θ ≈ 1 .8235 θ ≈ 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 .

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 ± 2 π k θ ≈ 1.8235 ± 2 π k and θ ≈ 4.4597 ± 2 π k . θ ≈ 4.4597 ± 2 π k .

Solve cos θ = − 0.2. cos θ = − 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 x x 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.

Solving a Trigonometric Equation in Quadratic Form

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

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

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

Replace x x with cos θ , cos θ , and solve.

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 − 13 2 ) on a calculator, since the domain of the inverse cosine function is [ − 1 , 1 ] . [ − 1 , 1 ] . However, there is a second solution:

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

Solving a Trigonometric Equation in Quadratic Form by 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 + 3 = 0 , 0 ≤ θ ≤ 2 π .

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

Now set each factor equal to zero.

Next solve for θ : sin θ ≠ 3 2 , θ : sin θ ≠ 3 2 , as the range of the sine function is [ −1 , 1 ] . [ −1 , 1 ] . However, sin θ = 1 , sin θ = 1 , giving the solution θ = π 2 . θ = π 2 .

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

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

Solving a Trigonometric Equation Using Algebra

Solve exactly:

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

Set each factor equal to zero.

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

The solutions within the domain 0 ≤ θ < 2 π 0 ≤ θ < 2 π are θ = 0 , π , 7 π 6 , 11 π 6 . θ = 0 , π , 7 π 6 , 11 π 6 .

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.

We can see the solutions on the graph in Figure 3 . On the interval 0 ≤ θ < 2 π , 0 ≤ θ < 2 π , 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.

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

Solving a Trigonometric Equation Quadratic in Form

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

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

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

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 .

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

From the unit circle in Figure 2 , we see that cos x = 3 2 cos x = 3 2 when x = π 6 , 11 π 6 . x = π 6 , 11 π 6 .

Solving the Equation Using a Double-Angle Formula

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

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:

So, if cos θ = − 1 2 , cos θ = − 1 2 , then θ = 2 π 3 ± 2 π k θ = 2 π 3 ± 2 π k and θ = 4 π 3 ± 2 π k ; θ = 4 π 3 ± 2 π k ; if cos θ = 1 , cos θ = 1 , then θ = 0 ± 2 π k . θ = 0 ± 2 π k .

Solving an Equation Using an Identity

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

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

Our solutions are θ = 2 π 3 , 4 π 3 , π . θ = 2 π 3 , 4 π 3 , 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 ( 2 x ) sin ( 2 x ) or cos ( 3 x ) . 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 x . 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 compression of the graph leads us to believe there may be twice as many x -intercepts or solutions to sin ( 2 x ) = 0 sin ( 2 x ) = 0 compared to sin x = 0. sin x = 0. This information will help us solve the equation.

Solving a Multiple Angle Trigonometric Equation

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

We can see that this equation is the standard equation with a multiple of an angle. If cos ( α ) = 1 2 , cos ( α ) = 1 2 , we know α α is in quadrants I and IV. While θ = cos − 1 1 2 θ = cos − 1 1 2 will only yield solutions in quadrants I and II, we recognize that the solutions to the equation cos θ = 1 2 cos θ = 1 2 will be in quadrants I and IV.

Therefore, the possible angles are θ = π 3 θ = π 3 and θ = 5 π 3 . θ = 5 π 3 . So, 2 x = π 3 2 x = π 3 or 2 x = 5 π 3 , 2 x = 5 π 3 , which means that x = π 6 x = π 6 or x = 5 π 6 . x = 5 π 6 . Does this make sense? Yes, because cos ( 2 ( π 6 ) ) = cos ( π 3 ) = 1 2 . cos ( 2 ( π 6 ) ) = cos ( π 3 ) = 1 2 .

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

In quadrant I, 2 x = π 3 , 2 x = π 3 , so x = π 6 x = π 6 as noted. Let us revolve around the circle again:

so x = 7 π 6 . x = 7 π 6 .

One more rotation yields

x = 13 π 6 > 2 π , x = 13 π 6 > 2 π , so this value for x x is larger than 2 π , 2 π , so it is not a solution on [ 0 , 2 π ) . [ 0 , 2 π ) .

In quadrant IV, 2 x = 5 π 3 , 2 x = 5 π 3 , so x = 5 π 6 x = 5 π 6 as noted. Let us revolve around the circle again:

so x = 11 π 6 . x = 11 π 6 .

x = 17 π 6 > 2 π , x = 17 π 6 > 2 π , so this value for x x is larger than 2 π , 2 π , so it is not a solution on [ 0 , 2 π ) . [ 0 , 2 π ) .

Our solutions are x = π 6 , 5 π 6 , 7 π 6 , and  11 π 6 x = π 6 , 5 π 6 , 7 π 6 , and  11 π 6 . Note that whenever we solve a problem in the form of sin ( n x ) = c , sin ( n x ) = c , we must go around the unit circle n n 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, a 2 + b 2 = c 2 , a 2 + b 2 = c 2 , and model an equation to fit a situation.

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 .

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

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.

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

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.

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 4 a feet. See Figure 5 .

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

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:

Thus, the ladder touches the wall at a 15 a 15 feet from the ground.

Access these online resources for additional instruction and practice with solving trigonometric equations.

• Solving Trigonometric Equations I
• Solving Trigonometric Equations II
• Solving Trigonometric Equations III
• Solving Trigonometric Equations IV
• Solving Trigonometric Equations V
• Solving Trigonometric Equations VI

9.5 Section Exercises

Will there always be solutions to trigonometric function equations? If not, describe an equation that would not have a solution. Explain why or why not.

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

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

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

2 sin θ = − 2 2 sin θ = − 2

2 sin θ = 3 2 sin θ = 3

2 cos θ = 1 2 cos θ = 1

2 cos θ = − 2 2 cos θ = − 2

tan θ = −1 tan θ = −1

tan x = 1 tan x = 1

cot x + 1 = 0 cot x + 1 = 0

4 sin 2 x − 2 = 0 4 sin 2 x − 2 = 0

csc 2 x − 4 = 0 csc 2 x − 4 = 0

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

2 cos θ = 2 2 cos θ = 2

2 cos θ = −1 2 cos θ = −1

2 sin θ = −1 2 sin θ = −1

2 sin θ = − 3 2 sin θ = − 3

2 sin ( 3 θ ) = 1 2 sin ( 3 θ ) = 1

2 sin ( 2 θ ) = 3 2 sin ( 2 θ ) = 3

2 cos ( 3 θ ) = − 2 2 cos ( 3 θ ) = − 2

cos ( 2 θ ) = − 3 2 cos ( 2 θ ) = − 3 2

2 sin ( π θ ) = 1 2 sin ( π θ ) = 1

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

For the following exercises, find all exact solutions on [ 0 , 2 π ) . [ 0 , 2 π ) .

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

tan ( x ) − 2 sin ( x ) tan ( x ) = 0 tan ( x ) − 2 sin ( x ) tan ( x ) = 0

2 cos 2 t + cos ( t ) = 1 2 cos 2 t + cos ( t ) = 1

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

2 sin ( x ) cos ( x ) − sin ( x ) + 2 cos ( x ) − 1 = 0 2 sin ( x ) cos ( x ) − sin ( x ) + 2 cos ( x ) − 1 = 0

cos 2 θ = 1 2 cos 2 θ = 1 2

sec 2 x = 1 sec 2 x = 1

tan 2 ( x ) = −1 + 2 tan ( − x ) tan 2 ( x ) = −1 + 2 tan ( − x )

8 sin 2 ( x ) + 6 sin ( x ) + 1 = 0 8 sin 2 ( x ) + 6 sin ( x ) + 1 = 0

tan 5 ( x ) = tan ( x ) tan 5 ( x ) = tan ( x )

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

sin ( 3 x ) cos ( 6 x ) − cos ( 3 x ) sin ( 6 x ) = −0.9 sin ( 3 x ) cos ( 6 x ) − cos ( 3 x ) sin ( 6 x ) = −0.9

sin ( 6 x ) cos ( 11 x ) − cos ( 6 x ) sin ( 11 x ) = −0.1 sin ( 6 x ) cos ( 11 x ) − cos ( 6 x ) sin ( 11 x ) = −0.1

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

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

9 cos ( 2 θ ) = 9 cos 2 θ − 4 9 cos ( 2 θ ) = 9 cos 2 θ − 4

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

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

cos ( 6 x ) − cos ( 3 x ) = 0 cos ( 6 x ) − cos ( 3 x ) = 0

For the following exercises, solve exactly on the interval [ 0 , 2 π ) . [ 0 , 2 π ) . Use the quadratic formula if the equations do not factor.

tan 2 x − 3 tan x = 0 tan 2 x − 3 tan x = 0

sin 2 x + sin x − 2 = 0 sin 2 x + sin x − 2 = 0

sin 2 x − 2 sin x − 4 = 0 sin 2 x − 2 sin x − 4 = 0

5 cos 2 x + 3 cos x − 1 = 0 5 cos 2 x + 3 cos x − 1 = 0

3 cos 2 x − 2 cos x − 2 = 0 3 cos 2 x − 2 cos x − 2 = 0

5 sin 2 x + 2 sin x − 1 = 0 5 sin 2 x + 2 sin x − 1 = 0

tan 2 x + 5 tan x − 1 = 0 tan 2 x + 5 tan x − 1 = 0

cot 2 x = − cot x cot 2 x = − cot x

− tan 2 x − tan x − 2 = 0 − tan 2 x − tan x − 2 = 0

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

sin 2 x − cos 2 x − sin x = 0 sin 2 x − cos 2 x − sin x = 0

sin 2 x + cos 2 x = 0 sin 2 x + cos 2 x = 0

sin ( 2 x ) − sin x = 0 sin ( 2 x ) − sin x = 0

cos ( 2 x ) − cos x = 0 cos ( 2 x ) − cos x = 0

2 tan x 2 − sec 2 x − sin 2 x = cos 2 x 2 tan x 2 − sec 2 x − sin 2 x = cos 2 x

1 − cos ( 2 x ) = 1 + cos ( 2 x ) 1 − cos ( 2 x ) = 1 + cos ( 2 x )

sec 2 x = 7 sec 2 x = 7

10 sin x cos x = 6 cos x 10 sin x cos x = 6 cos x

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

4 cos 2 x − 4 = 15 cos x 4 cos 2 x − 4 = 15 cos x

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

8 cos 2 θ = 3 − 2 cos θ 8 cos 2 θ = 3 − 2 cos θ

6 cos 2 x + 7 sin x − 8 = 0 6 cos 2 x + 7 sin x − 8 = 0

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

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

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

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.

6 sin 2 x − 5 sin x + 1 = 0 6 sin 2 x − 5 sin x + 1 = 0

8 cos 2 x − 2 cos x − 1 = 0 8 cos 2 x − 2 cos x − 1 = 0

100 tan 2 x + 20 tan x − 3 = 0 100 tan 2 x + 20 tan x − 3 = 0

2 cos 2 x − cos x + 15 = 0 2 cos 2 x − cos x + 15 = 0

20 sin 2 x − 27 sin x + 7 = 0 20 sin 2 x − 27 sin x + 7 = 0

2 tan 2 x + 7 tan x + 6 = 0 2 tan 2 x + 7 tan x + 6 = 0

130 tan 2 x + 69 tan x − 130 = 0 130 tan 2 x + 69 tan x − 130 = 0

For the following exercises, use a calculator to find all solutions to four decimal places.

sin x = 0.27 sin x = 0.27

sin x = −0.55 sin x = −0.55

tan x = −0.34 tan x = −0.34

cos x = 0.71 cos x = 0.71

For the following exercises, solve the equations algebraically, and then use a calculator to find the values on the interval [ 0 , 2 π ) . [ 0 , 2 π ) . Round to four decimal places.

tan 2 x + 3 tan x − 3 = 0 tan 2 x + 3 tan x − 3 = 0

6 tan 2 x + 13 tan x = −6 6 tan 2 x + 13 tan x = −6

tan 2 x − sec x = 1 tan 2 x − sec x = 1

sin 2 x − 2 cos 2 x = 0 sin 2 x − 2 cos 2 x = 0

2 tan 2 x + 9 tan x − 6 = 0 2 tan 2 x + 9 tan x − 6 = 0

4 sin 2 x + sin ( 2 x ) sec x − 3 = 0 4 sin 2 x + sin ( 2 x ) sec x − 3 = 0

For the following exercises, find all solutions exactly to the equations on the interval [ 0 , 2 π ) . [ 0 , 2 π ) .

csc 2 x − 3 csc x − 4 = 0 csc 2 x − 3 csc x − 4 = 0

sin 2 x − cos 2 x − 1 = 0 sin 2 x − cos 2 x − 1 = 0

sin 2 x ( 1 − sin 2 x ) + cos 2 x ( 1 − sin 2 x ) = 0 sin 2 x ( 1 − sin 2 x ) + cos 2 x ( 1 − sin 2 x ) = 0

3 sec 2 x + 2 + sin 2 x − tan 2 x + cos 2 x = 0 3 sec 2 x + 2 + sin 2 x − tan 2 x + cos 2 x = 0

sin 2 x − 1 + 2 cos ( 2 x ) − cos 2 x = 1 sin 2 x − 1 + 2 cos ( 2 x ) − cos 2 x = 1

tan 2 x − 1 − sec 3 x cos x = 0 tan 2 x − 1 − sec 3 x cos x = 0

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

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

2 cos 2 x − sin 2 x − cos x − 5 = 0 2 cos 2 x − sin 2 x − cos x − 5 = 0

1 sec 2 x + 2 + sin 2 x + 4 cos 2 x = 4 1 sec 2 x + 2 + sin 2 x + 4 cos 2 x = 4

Real-World Applications

An airplane has only enough gas to fly to a city 200 miles northeast of its current location. If the pilot knows that the city is 25 miles north, how many degrees north of east should the airplane fly?

If a loading ramp is placed next to a truck, at a height of 4 feet, and the ramp is 15 feet long, what angle does the ramp make with the ground?

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

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

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

A woman is standing 8 meters away from a 10-meter tall building. At what angle is she looking to the top of the building?

Issa is standing 10 meters away from a 6-meter tall building. Travis is at the top of the building looking down at Issa. At what angle is Travis looking at Issa?

A 20-foot tall building has a shadow that is 55 feet long. What is the angle of elevation of the sun?

A 90-foot tall building has a shadow that is 2 feet long. What is the angle of elevation of the sun?

A spotlight on the ground 3 meters from a 2-meter tall man casts a 6 meter shadow on a wall 6 meters from the man. At what angle is the light?

A spotlight on the ground 3 feet from a 5-foot tall woman casts a 15-foot tall shadow on a wall 6 feet from the woman. At what angle is the light?

For the following exercises, find a solution to the following word problem algebraically. Then use a calculator to verify the result. Round the answer to the nearest tenth of a degree.

A person does a handstand with their feet touching a wall and their hands 1.5 feet away from the wall. If the person is 6 feet tall, what angle do their feet make with the wall?

A person does a handstand with her feet touching a wall and her hands 3 feet away from the wall. If the person is 5 feet tall, what angle do her feet make with the wall?

A 23-foot ladder is positioned next to a house. If the ladder slips at 7 feet from the house when there is not enough traction, what angle should the ladder make with the ground to avoid slipping?

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Angle Problems and Solving Equations

Related Topics: Lesson Plans and Worksheets for Grade 7 Lesson Plans and Worksheets for all Grades More Lessons for Grade 7 Common Core For Grade 7

Examples, videos, and solutions to help Grade 7 students learn how to use facts about supplementary, complementary, vertical, and adjacent angles in a multi-step problem write and solve simple equations for an unknown angle in a figure.

New York State Common Core Math Grade 7, Module 3, Lesson 11

Worksheets for Grade 7

Lesson 11 Student Outcomes

• Students use facts about supplementary, complementary, vertical, and adjacent angles in a multi-step problem to write and solve simple equations for an unknown angle in a figure.

Lesson 11 Classwork Opening Exercise a. In a complete sentence, describe the angle relationship in the diagram. Write an equation for the angle relationship shown in the figure and solve for x. Confirm your answers by measuring the angle with a protractor. b. CD and EF are intersecting lines. In a complete sentence, describe the angle relationship in the diagram. Write an equation for the angle relationship shown in the figure and solve for . Confirm your answers by measuring the angle with a protractor. c. In a complete sentence, describe the angle relationship in the diagram. Write an equation for the angle relationship shown in the figure and solve for y. Confirm your answers by measuring the angle with a protractor. d. The following figure shows three lines intersecting at a point. In a complete sentence, describe the angle relationship in the diagram. Write an equation for the angle relationship shown in the figure and solve for z. Confirm your answers by measuring the angle with a protractor. e. Write an equation for the angle relationship shown in the figure and solve for x. In a complete sentence, describe the angle relationship in the diagram. Find the measurements of EPB and CPA. Confirm your answers by measuring the angle with a protractor.

Example 1 The following figure shows three lines intersecting at a point. In a complete sentence, describe the angle relationship in the diagram. Write an equation for the angle relationship shown in the figure and solve for x. Confirm your answers by measuring the angle with a protractor.

Exercise 1 The following figure shows four lines intersecting at a point. In a complete sentence, describe the angle relationships in the diagram. Write an equation for the angle relationship shown in the figure and solve for x and y. Confirm your answers by measuring the angle with a protractor.

Example 2 In a complete sentence, describe the angle relationships in the diagram. You may label the diagram to help describe the angle relationships. Write an equation for the angle relationship shown in the figure and solve for x. Confirm your answers by measuring the angle with a protractor.

Exercise 2 In a complete sentence, describe the angle relationships in the diagram. Write an equation for the angle relationship shown in the figure and solve for x and y. Confirm your answers by measuring the angle with a protractor.

Example 3 In a complete sentence, describe the angle relationships in the diagram. Write an equation for the angle relationship shown in the figure and solve for x. Find the measures of JAH and GAF. Confirm your answers by measuring the angle with a protractor.

Exercise 3 In a complete sentence, describe the angle relationships in the diagram. Write an equation for the angle relationship shown in the figure and solve for x. Find the measure of JKG. Confirm your answers by measuring the angle with a protractor.

Example 4 In the accompanying diagram, DBE is four times the measure of FBG. a. Label DBE as y° and FBG as x°. Write an equation that describes the relationship between DBE and FBG b. Find the value of x. c. Find the measures of FBG, CBD, ABF, GBE, DBE. d. What is the measure of ABG? Identify the angle relationship used to get your answer.

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MODULE 6: GEOMETRY

Using properties of angles to solve problems, learning outcomes.

• Identify types of angles
• Find the supplement of an angle
• Find the complement of an angle
• Use properties of angles to find the measures of various angles in a diagram

Are you familiar with the phrase ‘do a [latex]180[/latex]?’ It means to make a full turn so that you face the opposite direction. It comes from the fact that the measure of an angle that makes a straight line is [latex]180[/latex] degrees. See the image below.

[latex]\angle A[/latex] is the angle with vertex at [latex]\text{point }A[/latex].

We measure angles in degrees, and use the symbol [latex]^ \circ[/latex] to represent degrees. We use the abbreviation [latex]m[/latex] to for the measure of an angle. So if [latex]\angle A[/latex] is [latex]\text{27}^ \circ [/latex], we would write [latex]m\angle A=27^ \circ[/latex].

We classify angles based on their measures.  The four types of angles are acute, right, obtuse, and straight.  An acute angle is an angle that measures between [latex]\text{0}^ \circ [/latex] and [latex]\text{90}^ \circ [/latex].  A right angle measures exactly [latex]\text{90}^ \circ [/latex]. An obtuse angle measures between [latex]\text{90}^ \circ [/latex] and [latex]\text{180}^ \circ [/latex].  A straight angle measures exactly [latex]\text{180}^ \circ [/latex].

Supplementary and Complementary Angles

If the sum of the measures of two angles is [latex]\text{180}^ \circ[/latex], then they are called supplementary angles. In the images below, each pair of angles is supplementary because their measures add to [latex]\text{180}^ \circ [/latex]. Each angle is the supplement of the other.

The sum of the measures of supplementary angles is [latex]\text{180}^ \circ [/latex].

The sum of the measures of complementary angles is [latex]\text{90}^ \circ[/latex].

If the sum of the measures of two angles is [latex]\text{180}^\circ [/latex], then the angles are supplementary .

If angle [latex]A[/latex] and angle [latex]B[/latex] are supplementary, then [latex]m\angle{A}+m\angle{B}=180^\circ[/latex].

If the sum of the measures of two angles is [latex]\text{90}^\circ[/latex], then the angles are complementary .

If angle [latex]A[/latex] and angle [latex]B[/latex] are complementary, then [latex]m\angle{A}+m\angle{B}=90^\circ[/latex].

In this section and the next, you will be introduced to some common geometry formulas. We will adapt our Problem Solving Strategy for Geometry Applications. The geometry formula will name the variables and give us the equation to solve.

In addition, since these applications will all involve geometric shapes, it will be helpful to draw a figure and then label it with the information from the problem. We will include this step in the Problem Solving Strategy for Geometry Applications.

Use a Problem Solving Strategy for Geometry Applications.

• Read the problem and make sure you understand all the words and ideas. Draw a figure and label it with the given information.
• Identify what you are looking for.
• Name what you are looking for and choose a variable to represent it.
• Translate into an equation by writing the appropriate formula or model for the situation. Substitute in the given information.
• Solve the equation using good algebra techniques.
• Check the answer in the problem and make sure it makes sense.
• Answer the question with a complete sentence.

The next example will show how you can use the Problem Solving Strategy for Geometry Applications to answer questions about supplementary and complementary angles.

An angle measures [latex]\text{40}^ \circ[/latex].

1. Find its supplement

2. Find its complement

2.
Step 1. the problem. Draw the figure and label it with the given information.
Step 2. what you are looking for.	The complement of a [latex]40°[/latex]angle. Write the appropriate formula for the situation and substitute in the given information. [latex]m\angle A+m\angle B=90°[/latex] Step 5. Solve the equation. [latex]c+40°=90°[/latex] [latex]c=50°[/latex] Step 6. Check: [latex]50°+40°\stackrel{?}{=}90°[/latex] In the following video we show more examples of how to find the supplement and complement of an angle. Did you notice that the words complementary and supplementary are in alphabetical order just like [latex]90[/latex] and [latex]180[/latex] are in numerical order? Two angles are supplementary. The larger angle is [latex]\text{30}^ \circ[/latex] more than the smaller angle. Find the measure of both angles. Step 1. the problem. Draw the figure and label it with the given information. Step 2. what you are looking for. The measures of both angles. Step 3. Choose a variable to represent it. The larger angle is 30° more than the smaller angle. Let [latex]a=[/latex] measure of smaller angle [latex]a+30=[/latex] measure of larger angle Step 4. Write the appropriate formula and substitute. [latex]m\angle A+m\angle B=180[/latex] Step 5. the equation. [latex](a+30)+a=180[/latex] [latex]2a+30=180[/latex] [latex]2a=150[/latex] [latex]a=75=[/latex] measure of smaller angle. [latex]a+30=[/latex] measure of larger angle. [latex]75+30[/latex] [latex]105[/latex] Step 6. [latex]m\angle A+m\angle B=180°[/latex] [latex]75°+105°\stackrel{?}{=}180°[/latex] [latex]180°=180°\quad\checkmark[/latex] Step 7. the question. The measures of the angle are [latex]75°[/latex]and [latex]105°[/latex]. Question ID 146497, 146496, 146495. Authored by : Lumen Learning. License : CC BY: Attribution Determine the Complement and Supplement of a Given Angle. Authored by : James Sousa (mathispower4u.com). Located at : https://youtu.be/ZQ_L3yJOfqM . License : CC BY: Attribution Prealgebra. Provided by : OpenStax. License : CC BY: Attribution . License Terms : Download for free at http://cnx.org/contents/[email protected] 201 IMAGES Math Example--Solving Equations--Solving Equations Using Angle Example 3 Using Angle Properties to Solve Equations How to Solve Equations Involving Angles and a Pair of Parallel Lines How to Solve Equations Involving Angles and 2 Pairs of Parallel Lines Using Angle Properties to Solve Equations VIDEO 006A Find the angle in the figure Algebra Tutorial Angles in circles word problem An amazingly powerful tool: solving a Calculus problem using linear algebra Using half angle formulas to solve trigonometric equations COMMENTS Using Properties of Angles to Solve Problems If the sum of the measures of two angles is 90∘ 90 ∘, then the angles are complementary. If angle A A and angle B B are complementary, then m∠A+m∠B =90∘ m ∠ A + m ∠ B = 90 ∘. In this section and the next, you will be introduced to some common geometry formulas. We will adapt our Problem Solving Strategy for Geometry Applications. PDF Equations with Angles Notes and Practice Key Information: sum is 90° Equation: 2x + 71 + 9= 90. Solution: x = 5. 6. (-2x - 6)°. 40°. Type of Angles: vertical Key Information: angles are congruent Equation: -2x - 6 = 40. Solution: x = -23. Using Angles to Solve Equations - Exit Slip Name: KEY Write an equation and solve for the missing value. Show your work. Study Guide Translate into an equation by writing the appropriate formula or model for the situation. Substitute in the given information. Solve the equation using good algebra techniques. Check the answer in the problem and make sure it makes sense. Answer the question with a complete sentence. 7.1.5: Using Equations to Solve for Unknown Angles To find an unknown angle measure, sometimes it is helpful to write and solve an equation that represents the situation. For example, suppose we want to know the value of x in this diagram. Figure 7.1.5.12. Using what we know about vertical angles, we can write the equation 3x + 90 = 144 to represent this situation. Then we can solve the equation. 9.4: Use Properties of Angles, Triangles, and the Pythagorean Theorem The geometry formula will name the variables and give us the equation to solve. In addition, since these applications will all involve geometric shapes, it will be helpful to draw a figure and then label it with the information from the problem. We will include this step in the Problem Solving Strategy for Geometry Applications. 9.3 Use Properties of Angles, Triangles, and the Pythagorean ... In the next few sections, we will apply our problem-solving strategies to some common geometry problems. Use the Properties of Angles. ... Solve the equation using good algebra techniques. Step 6. Check the answer in the problem and make sure it makes sense. Step 7. 9.4: Use Properties of Angles, Triangles, and the Pythagorean Theorem Use the Pythagorean Theorem. The Pythagorean Theorem is a special property of right triangles that has been used since ancient times. It is named after the Greek philosopher and mathematician Pythagoras who lived around 500 500 BCE.. Remember that a right triangle has a 90° Figure 9.12.. Figure 9.12 In a right triangle, the side opposite the 90° 90° angle is called the hypotenuse and each ... PDF Lesson 10: Angle Problems and Solving Equations Lesson 10: Angle Problems and Solving Equations. Students use vertical and adjacent angles and angles on a line and angles at a point in a multi-step problem to write and solve simple equations for an unknown angle in a figure. In Lessons 10 and 11, students apply their understanding of equations to unknown angle problems. 9.1 Use Properties of Angles, Triangles, and the Pythagorean Theorem See. ° angle is called the hypotenuse and each of the other sides is called a leg. The Pythagorean Theorem tells how the lengths of the three sides of a right triangle relate to each other. It states that in any right triangle, the sum of the squares of the two legs equals the square of the hypotenuse. Using Properties of Angles to Solve Problems Use a Problem Solving Strategy for Geometry Applications. Read the problem and make sure you understand all the words and ideas. Draw a figure and label it with the given information. Identify what you are looking for.; Name what you are looking for and choose a variable to represent it.; Translate into an equation by writing the appropriate formula or model for the situation. 9.5 Solving Trigonometric Equations 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. PDF Lesson 10: Angle Problems and Solving Equations Students use vertical angles, adjacent angles, angles on a line, and angles at a point in a multistep problem to write and solve simple equations for an unknown angle in a figure. Lesson Notes In Lessons 10 and 11, students apply their understanding of equations to unknown angle problems. The geometry topic Angle Problems and Solving Equations Exercise 1. The following figure shows four lines intersecting at a point. In a complete sentence, describe the angle relationships in the diagram. Write an equation for the angle relationship shown in the figure and solve for x and y. Confirm your answers by measuring the angle with a protractor. Example 2. Geometry Angle Relationships Step 1: Identify the angle relationship. The two angles are complementary angles because they make a right angle of 90 degrees. Step 2: Solve for the missing angle. The given angle is 15 degrees ... Equations: Perimeter/Angles Textbook Exercise The Corbettmaths Textbook Exercise on Equations involving Perimeter/Angles. Welcome; Videos and Worksheets; Primary; 5-a-day. 5-a-day GCSE 9-1; 5-a-day Primary ... Click here for Questions . forming, creating, perimeter, angles, solving. Textbook Exercise. Previous: Equations: Letters on Both Sides Textbook Exercise. Next: Forming Equations ... PDF Lesson 4: Solving for Unknown Angles Using Equations Lesson 4: Solving for Unknown Angles Using EquationsLe. Unknown Angles Using Equations Student Outcomes Students solve for unknown angles in word proble. ork Opening Exercise (5 minutes)Opening ExerciseThe complement of. n angle is four times the measurement of the angle. its complement.+ =Complementary . Lesson HOW TO solve problems on the angles of triangles x + 2x + 3x = 180. Simplify this equation step by step and solve it: 6x = 180, x = 30. So, the angle LA of the triangle is 30 . Then the angle LB is 2*30 = 60 , and the angle LC is 3*30 = 90 in accordance with the problem condition. You can check that the sum of the angles is 180 : 30 + 60 + 90 = 180 . Using Equations To Solve Problems This tutorial shows how to set up an equation to solve a problem involving supplementary angles. Concepts that are reviewed during this tutorial include:Com... Mathway They do not store directly personal information, but are based on uniquely identifying your browser and internet device. If you do not allow these cookies, you will experience less targeted advertising. Free math problem solver answers your algebra homework questions with step-by-step explanations. Using Properties of Angles to Solve Problems We use the abbreviation m m to for the measure of an angle. So if ∠A ∠ A is 27∘ 27 ∘, we would write m∠A= 27∘ m ∠ A = 27 ∘. We classify angles based on their measures. The four types of angles are acute, right, obtuse, and straight. An acute angle is an angle that measures between 0∘ 0 ∘ and 90∘ 90 ∘ . A right angle ... Khan Academy Khan Academy. If you're seeing this message, it means we're having trouble loading external resources on our website. If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked. Courses. Search. Donate Log in Sign up. Khan Academy If this problem persists, tell us. Our mission is to provide a free, world-class education to anyone, anywhere. Khan Academy is a 501(c)(3) nonprofit organization. Donate or volunteer today! Site Navigation. About. News; Impact; Our team; Our interns; Our content specialists; Our leadership; Our supporters; Our contributors; Our finances; assignment essay homework paper phd resume speech thesis