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Unit 8 – Right Triangle Trigonometry

Similar Right Triangles

LESSON/HOMEWORK

LESSON VIDEO

EDITABLE LESSON

EDITABLE KEY

The Trigonometric Ratios

Trigonometry and the Calculator

Solving for Missing Sides of a Right Triangle

Trigonometric Applications

More Trigonometry Applications

Unit Review

Unit #8 Review – Right Triangle Trigonometry

UNIT REVIEW

EDITABLE REVIEW

Unit 8 Assessment Form A

EDITABLE ASSESSMENT

Unit 8 Assessment Form B

Unit 8 Assessment Form C

Unit 8 Assessment Form D

Unit 8 Exit Tickets

Unit 8 Mid-Unit Quiz (Through Lesson #4) – Form A

Unit 8 Mid-Unit Quiz (Through Lesson #4) – Form B

Unit 8 Mid-Unit Quiz (Through Lesson #4) – Form C

Unit 8 Mid-Unit Quiz (Through Lesson #4) – Form D

U08.AO.01 – Terminology Warm-Up for the Trigonometric Ratios (Before Lesson 2)

EDITABLE RESOURCE

U08.AO.02 – Right Triangle Trigonometry Practice

U08.AO.03 – Multi-Step Right Triangle Trigonometry Practice

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Chapter 2: Trigonometric Ratios

Exercises: 2.2 Right Triangle Trigonometry

Exercises homework 2.2.

  • Use measurements to calculate the trigonometric ratios for acute angles #1-10, 57-60
  • Use trigonometric ratios to find unknown sides of right triangles #11-26
  • Solve problems using trigonometric ratios #27-34, 41-46
  • Use trig ratios to write equations relating the sides of a right triangle #35-40
  • Use relationships among the trigonometric ratios #47-56, 61-68

Suggested Homework Problems

Here are two right triangles with a [latex]65 °[/latex] angle.

  • Measure the sides [latex]AB[/latex] and [latex]BC[/latex] with a ruler. Use the lengths to estimate [latex]\sin 65°{.}[/latex]
  • Measure the sides [latex]AD[/latex] and [latex]DE[/latex] with a ruler. Use the lengths to estimate [latex]\sin 65°{.}[/latex]
  • Use your calculator to look up [latex]\sin 65°{.}[/latex] Compare your answers. How close were your estimates?

Use the figure in Problem 1 to calculate two estimates each for the cosine and tangent of [latex]65 °{.}[/latex] Compare your estimates to your calculator’s values for [latex]\cos 65°[/latex] and [latex]\tan 65°{.}[/latex]

Here are two right triangles with a [latex]40 °[/latex] angle.

  • Measure the sides [latex]AB[/latex] and [latex]AC[/latex] with a ruler. Use the lengths to estimate [latex]\cos 40°{.}[/latex]
  • Measure the sides [latex]AD[/latex] and [latex]AE[/latex] with a ruler. Use the lengths to estimate [latex]\cos 40°{.}[/latex]
  • Use your calculator to look up [latex]\cos 40°{.}[/latex] Compare your answers. How close were your estimates?

Use the figure in Problem 2 to calculate two estimates each for the cosine and tangent of [latex]40 °{.}[/latex] Compare your estimates to your calculator’s values for [latex]\sin 40°[/latex] and [latex]\tan 40°{.}[/latex]

Exercise Group

For the right triangles in Problems 5–10,

  • Find the length of the unknown side.
  • Find the sine, cosine, and tangent of [latex]\theta \text{.}[/latex] Round your answers to four decimal places.

For Problems 11–16,

  • Sketch and label the sides of a right triangle with angle [latex]\theta\text{.}[/latex]
  • Sketch and label another right triangle with angle [latex]\theta[/latex] and longer sides.

[latex]\cos \theta = \dfrac{3}{5}[/latex]

[latex]\tan \theta = \dfrac{7}{2}[/latex]

[latex]\tan \theta = \dfrac{11}{4}[/latex]

[latex]\sin \theta = \dfrac{4}{9}[/latex]

[latex]\sin \theta = \dfrac{1}{9}[/latex]

[latex]\cos \theta = \dfrac{7}{8}[/latex]

For Problems 17–22, use one of the three trigonometric ratios to find the unknown side of the triangle. Round your answer to hundredths.

For Problems 23–26, sketch and label a right triangle with the given properties.

One angle is [latex]40°{,}[/latex] the side opposite that angle is 8 inches

One angle is [latex]65°{,}[/latex] the side adjacent to that angle is 30 yards

One angle is [latex]28°{,}[/latex] the hypotenuse is 56 feet

One leg is [latex]15[/latex] meters, the hypotenuse is [latex]18[/latex] meters

For Problems 27–34,

  • Sketch a right triangle that illustrates the situation. Label your sketch with the given information.
  • Choose the appropriate trig ratio and write an equation, then solve the problem.

To measure the height of cloud cover, airport controllers fix a searchlight to shine a vertical beam on the clouds. The searchlight is [latex]120[/latex] yards from the office. A technician in the office measures the angle of elevation to the light on the cloud cover at [latex]54.8°{.}[/latex] What is the height of the cloud cover?

To measure the distance across a canyon, Evel first sights an interesting rock directly opposite on the other side. He then walks [latex]200[/latex] yards down the rim of the canyon and sights the rock again, this time at an angle of [latex]18.5°[/latex] from the canyon rim. What is the width of the canyon?

A salvage ship is searching for the wreck of a pirate vessel on the ocean floor. Using sonar, they locate the wreck at an angle of depression of [latex]36.2°{.}[/latex] The depth of the ocean at their location is [latex]260[/latex] feet. How far should they move so that they are directly above the wrecked vessel?

Ramps for wheelchairs should be no steeper than an angle of [latex]6°{.}[/latex] How much horizontal distance should be allowed for a ramp that rises [latex]5[/latex] feet in height?

The radio signal from a weather balloon indicates that it is [latex]1500[/latex] meters from a meteorologist on the ground. The angle of elevation to the balloon is [latex]48°{.}[/latex] What is the balloon’s altitude?

According to Chinese legend, around 200 BC, the general Han Xin used a kite to determine the distance from his location to an enemy palace. He then dug a secret tunnel which emerged inside the palace. When the kite was directly above the palace, its angle of elevation was [latex]27°[/latex] and the string to the kite was [latex]1850[/latex] feet long. How far did Han Xin’s troops have to dig?

A cable car on a ski lift traverses a horizontal distance of [latex]1800[/latex] meters at an angle of [latex]38°{.}[/latex] How long is the cable?

Zelda is building the loft on her summer cottage. At its central point, the height of the loft is [latex]8[/latex] feet, and the pitch of the roof should be [latex]24°{.}[/latex] How long should the rafters be?

For Problems 35–40, use a trig ratio to write an equation for [latex]x[/latex] in terms of [latex]\theta{.}[/latex]

For Problems 41–44, find the altitude of the triangle. Round your answer to two decimal places.

For Problems 45 and 46, find the length of the chord [latex]AB{.}[/latex] Round your answer to two decimal places.

For Problems 47–50, fill in the table.

[latex]~~~~[/latex] sin cos tan
[latex]\theta[/latex] [latex]~~~~[/latex] [latex]~~~~[/latex] [latex]~~~~[/latex]
[latex]\phi[/latex] [latex]~~~~[/latex] [latex]~~~~[/latex] [latex]~~~~[/latex]
  • In each of the figures for Problems 47-50, what is the relationship between the angles [latex]\theta[/latex] and [latex]\phi{?}[/latex]
  • Study the tables for Problems 47-50. What do you notice about the values of sine and cosine for the angles [latex]\theta[/latex] and [latex]\phi{?}[/latex] Explain why this is true.

There is a relationship between the tangent, the sine, and the cosine of any angle. Study the tables for Problems 47-50 to discover this relationship. Write your answer as an equation.

  • Use the figure to explain what happens to [latex]\tan \theta[/latex] as [latex]\theta[/latex] increases, and why.
  • Use the figure to explain what happens to [latex]\cos \theta[/latex] as [latex]\theta[/latex] increases, and why.
[latex]\theta[/latex] [latex]~~0 °[/latex] [latex]~10 °[/latex] [latex]~20 °[/latex] [latex]~30 °[/latex] [latex]~40 °[/latex] [latex]~50 °[/latex] [latex]~60 °[/latex] [latex]~70 °[/latex] [latex]~80 °[/latex]
[latex]\tan \theta[/latex] [latex]~~~[/latex] [latex]~~~[/latex] [latex]~~~[/latex] [latex]~~~[/latex] [latex]~~~[/latex] [latex]~~~[/latex] [latex]~~~[/latex] [latex]~~~[/latex] [latex]~~~[/latex]
[latex]\theta[/latex] [latex]~81 °[/latex] [latex]~82 °[/latex] [latex]~83 °[/latex] [latex]~84 °[/latex] [latex]~85 °[/latex] [latex]~86 °[/latex] [latex]~87 °[/latex] [latex]~88 °[/latex] [latex]~89 °[/latex]
[latex]\tan \theta[/latex] [latex]~~~[/latex] [latex]~~~[/latex] [latex]~~~[/latex] [latex]~~~[/latex] [latex]~~~[/latex] [latex]~~~[/latex] [latex]~~~[/latex] [latex]~~~[/latex] [latex]~~~[/latex]
  • What happens to [latex]\tan \theta[/latex] as [latex]\theta[/latex] increases?
  • What value does your calculator give for [latex]\tan 90°{?}[/latex] Why?

Explain why it makes sense that [latex]\sin 0° = 0[/latex] and [latex]\sin 90° = 1{.}[/latex] Use a figure to illustrate your explanation.

Explain why it makes sense that [latex]\cos 0° = 1[/latex] and [latex]\cos 90° = 0{.}[/latex] Use a figure to illustrate your explanation

For Problems 57–60, explain why the trigonometric ratio is not correct.

[latex]\sin \theta = \dfrac{5}{9}[/latex]

[latex]\tan \theta = \dfrac{4}{7}[/latex]

[latex]\cos \theta = \dfrac{21}{20}[/latex]

[latex]\sin \theta = \dfrac{8}{10}[/latex]

For Problems 61–64, sketch and label a right triangle, then fill in the blank.

If [latex]\sin \theta = 0.2358\text{,}[/latex]  then [latex]\cos (90^{o} - \theta)=\underline{\qquad} \text{,}[/latex]

  • If [latex]\cos \alpha = \dfrac{3}{11} \text{,}[/latex] then [latex]\underline{\qquad} (90° - \alpha) = \dfrac{3}{11}\text{.}[/latex]
  • If [latex]\sin 42° = n\text{,}[/latex] then [latex]\cos \underline{\qquad} = n\text{.}[/latex]
  • If [latex]\cos 13° = z\text{,}[/latex] then [latex]\sin \underline{\qquad} = z\text{.}[/latex]
  • If [latex]\cos \beta = \dfrac{2}{\sqrt{7}}{,}[/latex] then [latex]\sin (90° - \beta) =\underline{\qquad} {.}[/latex]
  • If [latex]\sin \phi = 0.693{,}[/latex] then [latex](90° - \phi) = 0.693{.}[/latex]
  • If [latex]\cos 87° = p{,}[/latex] then [latex]\sin \underline{\qquad} = p{.}[/latex]
  • If [latex]\sin 59° =w{,}[/latex] then [latex]\cos \underline{\qquad} = w{.}[/latex]
  • If [latex]\sin \phi = \dfrac{5}{13}[/latex] and [latex]\cos \phi = \dfrac{12}{13}{,}[/latex] then [latex]\tan \phi =\underline{\qquad} {.}[/latex]
  • If [latex]\cos \beta = \dfrac{1}{\sqrt{10}}{,}[/latex] and [latex]\sin \beta = \dfrac{3}{\sqrt{10}}{,}[/latex] then [latex]\tan \beta = \underline{\qquad}{.}[/latex]
  • If [latex]\tan B = \dfrac{2}{\sqrt{5}}[/latex] and [latex]\cos B = \dfrac{\sqrt{5}}{3}{,}[/latex] then [latex]\sin B =\underline{\qquad}{.}[/latex]
  • If [latex]\sin W = \sqrt{\dfrac{3}{7}}[/latex] and [latex]\tan W = \dfrac{\sqrt{3}}{2}{,}[/latex] then [latex]\cos W =\underline{\qquad}{.}[/latex]
  • If [latex]\cos \theta = \dfrac{2}{\sqrt{10}}[/latex] and [latex]\sin \theta = \sqrt{\dfrac{3}{5}}{,}[/latex] then [latex]\tan \theta = \underline{\qquad}{.}[/latex]
  • If [latex]\sin \alpha = \dfrac{\sqrt{2}}{4}{,}[/latex] and [latex]\cos \alpha =\dfrac{\sqrt{14}}{4}{,}[/latex] then [latex]\tan \alpha = \underline{\qquad}{.}[/latex]
  • If [latex]\tan A = \dfrac{\sqrt{7}}{3}[/latex] and [latex]\cos A = \dfrac{3}{4}{,}[/latex] then [latex]\sin A =\underline{\qquad}{.}[/latex]
  • If [latex]\sin V = \sqrt{\dfrac{10}{5}}[/latex] and [latex]\tan V = \dfrac{2}{5}{,}[/latex] then [latex]\cos V =\underline{\qquad}{.}[/latex]

Explain why the cosine of a [latex]73°[/latex] angle is always the same, no matter what size triangle the angle is in. Illustrate your explanation with a sketch.

[latex]\theta[/latex] [latex]~~0 °[/latex] [latex]~15 °[/latex] [latex]~30 °[/latex] [latex]~45 °[/latex] [latex]~60 °[/latex] [latex]~75 °[/latex] [latex]~90 °[/latex]
[latex]\cos \theta[/latex] [latex]~~~[/latex] [latex]~~~[/latex] [latex]~~~[/latex] [latex]~~~[/latex] [latex]~~~[/latex] [latex]~~~[/latex] [latex]~~~[/latex]
  • If you plotted the points in your table, would they lie on a straight line? Why or why not?
  • What is the slope of the line through the origin and point [latex]P{?}[/latex]
  • What is the tangent of the angle [latex]\theta{?}[/latex]
  • On the same grid, sketch an angle whose tangent is [latex]\dfrac{8}{5}.[/latex]
  • Use your calculator to complete the table. Round your answers to hundredths.
[latex]\theta[/latex] [latex]~14 °[/latex] [latex]~22 °[/latex] [latex]~35 °[/latex] [latex]~42 °[/latex] [latex]~58 °[/latex] [latex]~78 °[/latex]
[latex]\tan \theta[/latex] [latex]~~~[/latex] [latex]~~~[/latex] [latex]~~~[/latex] [latex]~~~[/latex] [latex]~~~[/latex] [latex]~~~[/latex]
  • Use the values of tan [latex]\theta[/latex] to sketch all the angles listed in the table. Locate the vertex of each angle at the origin and the initial side along the positive [latex]x[/latex]-axis.

Trigonometry Copyright © 2024 by Bimal Kunwor; Donna Densmore; Jared Eusea; and Yi Zhen. All Rights Reserved.

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Right Triangle Trigonometry Calculator

Table of contents

The right triangle trigonometry calculator can help you with problems where angles and triangles meet: keep reading to find out:

  • The basics of trigonometry;
  • How to calculate a right triangle with trigonometry;
  • A worked example of how to use trigonometry to calculate a right triangle with steps;

And much more!

Basics of trigonometry

Trigonometry is a branch of mathematics that relates angles to the length of specific segments . We identify multiple trigonometric functions: sine, cosine, and tangent, for example. They all take an angle as their argument, returning the measure of a length associated with the angle itself. Using a trigonometric circle , we can identify some of the trigonometric functions and their relationship with angles.

Trigonometric circle

As you can see from the picture, sine and cosine equal the projection of the radius on the axis, while the tangent lies outside the circle. If you look closely, you can identify a right triangle using the elements we introduced above: let's discover the relationship between trigonometric functions and this shape.

Right triangles trigonometry calculations

Consider an acute angle in the trigonometric circle above: notice how you can build a right triangle where:

  • The radius is the hypotenuse; and
  • The sine and cosine are the catheti of the triangle.

α \alpha α is one of the acute angles, while the right angle lies at the intersection of the catheti (sine and cosine)

Let this sink in for a moment: the length of the cathetus opposite from the angle α \alpha α is its sine , sin ⁡ ( α ) \sin(\alpha) sin ( α ) ! You just found an easy and quick way to calculate the angles and sides of a right triangle using trigonometry.

The complete relationships between angles and sides of a right triangle need to contain a scaling factor, usually the radius (the hypotenuse). Identify the opposite and adjacent . We can then write:

By switching the roles of the legs, you can find the values of the trigonometric functions for the other angle.

Taking the inverse of the trigonometric functions , you can find the values of the acute angles in any right triangle.

Using the three equations above and a combination of sides, angles, or other quantities, you can solve any right triangle . The cases we implemented in our calculator are:

  • Solving the triangle knowing two sides ;
  • Solving the triangle knowing one angle and one side ; and
  • Solving the triangle knowing the area and one side .

Example of right triangle trigonometry calculations with steps

Take a right triangle with hypotenuse c = 5 c = 5 c = 5 and an angle α = 38 ° \alpha=38\degree α = 38° . Surprisingly enough, this is enough data to fully solve the right triangle! Follow these steps:

  • Calculate the third angle: β = 90 ° − α \beta = 90\degree - \alpha β = 90° − α .
  • sin ⁡ ( α ) = 0.61567 \sin(\alpha) = 0.61567 sin ( α ) = 0.61567 .
  • o p p o s i t e = sin ⁡ ( α ) ⋅ h y p o t e n u s e = 0.61567 ⋅ 5 = 3.078 \mathrm{opposite} = \sin(\alpha)\cdot\mathrm{hypotenuse} = 0.61567 \cdot 5 = 3.078 opposite = sin ( α ) ⋅ hypotenuse = 0.61567 ⋅ 5 = 3.078 .
  • a d j a c e n t = 0.788 ⋅ 5 = 3.94 \mathrm{adjacent} = 0.788\cdot 5 = 3.94 adjacent = 0.788 ⋅ 5 = 3.94 .

More trigonometry and right triangles calculators (and not only)

If you liked our right triangle trigonometry calculator, why not try our other related tools? Here they are:

  • The trigonometry calculator ;
  • The cosine triangle calculator ;
  • The sine triangle calculator ;
  • The trig triangle calculator ;
  • The trig calculator ;
  • The sine cosine tangent calculator ;
  • The tangent ratio calculator ; and
  • The tangent angle calculator .

How do I apply trigonometry to a right triangle?

To apply trigonometry to a right triangle, remember that sine and cosine correspond to the legs of a right triangle . To solve a right triangle using trigonometry:

  • sin(α) = opposite/hypotenuse ; and
  • cos(α) = adjacent/hypotenuse .
  • By taking the inverse trigonometric functions , we can find the value of the angle α .
  • You can repeat the procedure for the other angle.

What is the hypotenuse of a triangle with α = 30° and opposite leg a = 3?

The length of the hypotenuse is 6 . To find this result:

  • Calculate the sine of α : sin(α) = sin(30°) = 1/2 .
  • Apply the following formula: sin(α) = opposite/hypotenuse hypotenuse = opposite/sin(α) = 3 · 2 = 6 .

Can I apply right-triangle trigonometric rules in a non-right triangle?

Not directly: to apply the relationships between trigonometric functions and sides of a triangle, divide the shape alongside one of the heights lying inside it. This way, you can split the triangle into two right triangles and, with the right combination of data, solve it!

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Side length a

Side length b

Side length c

Trigonometry in a right triangle with angles and sides marked.

© Omni Calculator

7.2 Right Triangle Trigonometry

Learning objectives.

In this section you will:

  • Use right triangles to evaluate trigonometric functions.
  • Find function values for 30° ( π 6 ) , 45° ( π 4 ) , 30° ( π 6 ) , 45° ( π 4 ) , and 60° ( π 3 ) . 60° ( π 3 ) .
  • Use equal cofunctions of complementary angles.
  • Use the definitions of trigonometric functions of any angle.
  • Use right-triangle trigonometry to solve applied problems.

Mt. Everest, which straddles the border between China and Nepal, is the tallest mountain in the world. Measuring its height is no easy task and, in fact, the actual measurement has been a source of controversy for hundreds of years. The measurement process involves the use of triangles and a branch of mathematics known as trigonometry. In this section, we will define a new group of functions known as trigonometric functions, and find out how they can be used to measure heights, such as those of the tallest mountains.

Using Right Triangles to Evaluate Trigonometric Functions

Figure 1 shows a right triangle with a vertical side of length y y and a horizontal side has length x . x . Notice that the triangle is inscribed in a circle of radius 1. Such a circle, with a center at the origin and a radius of 1, is known as a unit circle .

We can define the trigonometric functions in terms an angle t and the lengths of the sides of the triangle. The adjacent side is the side closest to the angle, x . (Adjacent means “next to.”) The opposite side is the side across from the angle, y . The hypotenuse is the side of the triangle opposite the right angle, 1. These sides are labeled in Figure 2 .

Given a right triangle with an acute angle of t , t , the first three trigonometric functions are listed.

A common mnemonic for remembering these relationships is SohCahToa, formed from the first letters of “ underline S end underline ine is underline o end underline pposite over underline h end underline ypotenuse, underline C end underline osine is underline a end underline djacent over underline h end underline ypotenuse, underline T end underline angent is underline o end underline pposite over underline a end underline djacent.”

For the triangle shown in Figure 1 , we have the following.

Given the side lengths of a right triangle and one of the acute angles, find the sine, cosine, and tangent of that angle.

  • Find the sine as the ratio of the opposite side to the hypotenuse.
  • Find the cosine as the ratio of the adjacent side to the hypotenuse.
  • Find the tangent as the ratio of the opposite side to the adjacent side.

Evaluating a Trigonometric Function of a Right Triangle

Given the triangle shown in Figure 3 , find the value of cos α . cos α .

The side adjacent to the angle is 15, and the hypotenuse of the triangle is 17.

Given the triangle shown in Figure 4 , find the value of sin t . sin t .

Reciprocal Functions

In addition to sine, cosine, and tangent, there are three more functions. These too are defined in terms of the sides of the triangle.

Take another look at these definitions. These functions are the reciprocals of the first three functions.

When working with right triangles, keep in mind that the same rules apply regardless of the orientation of the triangle. In fact, we can evaluate the six trigonometric functions of either of the two acute angles in the triangle in Figure 5 . The side opposite one acute angle is the side adjacent to the other acute angle, and vice versa.

Many problems ask for all six trigonometric functions for a given angle in a triangle. A possible strategy to use is to find the sine, cosine, and tangent of the angles first. Then, find the other trigonometric functions easily using the reciprocals.

Given the side lengths of a right triangle, evaluate the six trigonometric functions of one of the acute angles.

  • If needed, draw the right triangle and label the angle provided.
  • Identify the angle, the adjacent side, the side opposite the angle, and the hypotenuse of the right triangle.
  • sine as the ratio of the opposite side to the hypotenuse
  • cosine as the ratio of the adjacent side to the hypotenuse
  • tangent as the ratio of the opposite side to the adjacent side
  • secant as the ratio of the hypotenuse to the adjacent side
  • cosecant as the ratio of the hypotenuse to the opposite side
  • cotangent as the ratio of the adjacent side to the opposite side

Evaluating Trigonometric Functions of Angles Not in Standard Position

Using the triangle shown in Figure 6 , evaluate sin α , cos α , tan α , sec α , csc α , and cot α . sin α , cos α , tan α , sec α , csc α , and cot α .

Another approach would have been to find sine, cosine, and tangent first. Then find their reciprocals to determine the other functions.

Using the triangle shown in Figure 7 ,evaluate sin t , cos t , tan t , sec t , csc t , and cot t . sin t , cos t , tan t , sec t , csc t , and cot t .

Finding Trigonometric Functions of Special Angles Using Side Lengths

It is helpful to evaluate the trigonometric functions as they relate to the special angles—multiples of 30° , 60° , 30° , 60° , and 45° . 45° . Remember, however, that when dealing with right triangles, we are limited to angles between 0°  and 90° . 0°  and 90° .

Suppose we have a 30° , 60° , 90° 30° , 60° , 90° triangle, which can also be described as a π 6 , π 3 , π 2 π 6 , π 3 , π 2 triangle. The sides have lengths in the relation s , 3 s , 2 s . s , 3 s , 2 s . The sides of a 45° , 45° , 90° 45° , 45° , 90° triangle, which can also be described as a π 4 , π 4 , π 2 π 4 , π 4 , π 2 triangle, have lengths in the relation s , s , 2 s . s , s , 2 s . These relations are shown in Figure 8 .

We can then use the ratios of the side lengths to evaluate trigonometric functions of special angles.

Given trigonometric functions of a special angle, evaluate using side lengths.

  • Use the side lengths shown in Figure 8 for the special angle you wish to evaluate.
  • Use the ratio of side lengths appropriate to the function you wish to evaluate.

Evaluating Trigonometric Functions of Special Angles Using Side Lengths

Find the exact value of the trigonometric functions of π 3 , π 3 , using side lengths.

Find the exact value of the trigonometric functions of π 4 , π 4 , using side lengths.

Using Equal Cofunction of Complements

If we look more closely at the relationship between the sine and cosine of the special angles, we notice a pattern. In a right triangle with angles of π 6 π 6 and π 3 , π 3 , we see that the sine of π 3 , π 3 , namely 3 2 , 3 2 , is also the cosine of π 6 , π 6 , while the sine of π 6 , π 6 , namely 1 2 , 1 2 , is also the cosine of π 3 . π 3 .

See Figure 9 .

This result should not be surprising because, as we see from Figure 9 , the side opposite the angle of π 3 π 3 is also the side adjacent to π 6 , π 6 , so sin ( π 3 ) sin ( π 3 ) and cos ( π 6 ) cos ( π 6 ) are exactly the same ratio of the same two sides, 3 s 3 s and 2 s . 2 s . Similarly, cos ( π 3 ) cos ( π 3 ) and sin ( π 6 ) sin ( π 6 ) are also the same ratio using the same two sides, s s and 2 s . 2 s .

The interrelationship between the sines and cosines of π 6 π 6 and π 3 π 3 also holds for the two acute angles in any right triangle, since in every case, the ratio of the same two sides would constitute the sine of one angle and the cosine of the other. Since the three angles of a triangle add to π , π , and the right angle is π 2 , π 2 , the remaining two angles must also add up to π 2 . π 2 . That means that a right triangle can be formed with any two angles that add to π 2 π 2 —in other words, any two complementary angles. So we may state a cofunction identity : If any two angles are complementary, the sine of one is the cosine of the other, and vice versa. This identity is illustrated in Figure 10 .

Using this identity, we can state without calculating, for instance, that the sine of π 12 π 12 equals the cosine of 5 π 12 , 5 π 12 , and that the sine of 5 π 12 5 π 12 equals the cosine of π 12 . π 12 . We can also state that if, for a given angle t , cos t = 5 13 , t , cos t = 5 13 , then sin ( π 2 − t ) = 5 13 sin ( π 2 − t ) = 5 13 as well.

Cofunction Identities

The cofunction identities in radians are listed in Table 1 .

Given the sine and cosine of an angle, find the sine or cosine of its complement.

  • To find the sine of the complementary angle, find the cosine of the original angle.
  • To find the cosine of the complementary angle, find the sine of the original angle.

Using Cofunction Identities

If sin t = 5 12 , sin t = 5 12 , find cos ( π 2 − t ) . cos ( π 2 − t ) .

According to the cofunction identities for sine and cosine, we have the following.

If csc ( π 6 ) = 2 , csc ( π 6 ) = 2 , find sec ( π 3 ) . sec ( π 3 ) .

Using Trigonometric Functions

In previous examples, we evaluated the sine and cosine in triangles where we knew all three sides. But the real power of right-triangle trigonometry emerges when we look at triangles in which we know an angle but do not know all the sides.

Given a right triangle, the length of one side, and the measure of one acute angle, find the remaining sides.

  • For each side, select the trigonometric function that has the unknown side as either the numerator or the denominator. The known side will in turn be the denominator or the numerator.
  • Write an equation setting the function value of the known angle equal to the ratio of the corresponding sides.
  • Using the value of the trigonometric function and the known side length, solve for the missing side length.

Finding Missing Side Lengths Using Trigonometric Ratios

Find the unknown sides of the triangle in Figure 11 .

We know the angle and the opposite side, so we can use the tangent to find the adjacent side.

We rearrange to solve for a . a .

We can use the sine to find the hypotenuse.

Again, we rearrange to solve for c . c .

A right triangle has one angle of π 3 π 3 and a hypotenuse of 20. Find the unknown sides and angle of the triangle.

Using Right Triangle Trigonometry to Solve Applied Problems

Right-triangle trigonometry has many practical applications. For example, the ability to compute the lengths of sides of a triangle makes it possible to find the height of a tall object without climbing to the top or having to extend a tape measure along its height. We do so by measuring a distance from the base of the object to a point on the ground some distance away, where we can look up to the top of the tall object at an angle. The angle of elevation of an object above an observer relative to the observer is the angle between the horizontal and the line from the object to the observer's eye. The right triangle this position creates has sides that represent the unknown height, the measured distance from the base, and the angled line of sight from the ground to the top of the object. Knowing the measured distance to the base of the object and the angle of the line of sight, we can use trigonometric functions to calculate the unknown height.

Similarly, we can form a triangle from the top of a tall object by looking downward. The angle of depression of an object below an observer relative to the observer is the angle between the horizontal and the line from the object to the observer's eye. See Figure 12 .

Given a tall object, measure its height indirectly.

  • Make a sketch of the problem situation to keep track of known and unknown information.
  • Lay out a measured distance from the base of the object to a point where the top of the object is clearly visible.
  • At the other end of the measured distance, look up to the top of the object. Measure the angle the line of sight makes with the horizontal.
  • Write an equation relating the unknown height, the measured distance, and the tangent of the angle of the line of sight.
  • Solve the equation for the unknown height.

Measuring a Distance Indirectly

To find the height of a tree, a person walks to a point 30 feet from the base of the tree. She measures an angle of 57° 57° between a line of sight to the top of the tree and the ground, as shown in Figure 13 . Find the height of the tree.

We know that the angle of elevation is 57° 57° and the adjacent side is 30 ft long. The opposite side is the unknown height.

The trigonometric function relating the side opposite to an angle and the side adjacent to the angle is the tangent. So we will state our information in terms of the tangent of 57° , 57° , letting h h be the unknown height.

The tree is approximately 46 feet tall.

How long a ladder is needed to reach a windowsill 50 feet above the ground if the ladder rests against the building making an angle of 5 π 12 5 π 12 with the ground? Round to the nearest foot.

Access these online resources for additional instruction and practice with right triangle trigonometry.

  • Finding Trig Functions on Calculator
  • Finding Trig Functions Using a Right Triangle
  • Relate Trig Functions to Sides of a Right Triangle
  • Determine Six Trig Functions from a Triangle
  • Determine Length of Right Triangle Side

7.2 Section Exercises

For the given right triangle, label the adjacent side, opposite side, and hypotenuse for the indicated angle.

When a right triangle with a hypotenuse of 1 is placed in a circle of radius 1, which sides of the triangle correspond to the x - and y -coordinates?

The tangent of an angle compares which sides of the right triangle?

What is the relationship between the two acute angles in a right triangle?

Explain the cofunction identity.

For the following exercises, use cofunctions of complementary angles.

cos ( 34° ) = sin ( ___° ) cos ( 34° ) = sin ( ___° )

cos ( π 3 ) = sin ( ___ ) cos ( π 3 ) = sin ( ___ )

csc ( 21° ) = sec ( ___° ) csc ( 21° ) = sec ( ___° )

tan ( π 4 ) = cot ( ___ ) tan ( π 4 ) = cot ( ___ )

For the following exercises, find the lengths of the missing sides if side a a is opposite angle A , A , side b b is opposite angle B , B , and side c c is the hypotenuse.

cos B = 4 5 , a = 10 cos B = 4 5 , a = 10

sin B = 1 2 , a = 20 sin B = 1 2 , a = 20

tan A = 5 12 , b = 6 tan A = 5 12 , b = 6

tan A = 100 , b = 100 tan A = 100 , b = 100

sin B = 1 3 , a = 2 sin B = 1 3 , a = 2

a = 5 , ∡ A = 60° a = 5 , ∡ A = 60°

c = 12 , ∡ A = 45° c = 12 , ∡ A = 45°

For the following exercises, use Figure 14 to evaluate each trigonometric function of angle A . A .

sin A sin A

cos A cos A

tan A tan A

csc A csc A

sec A sec A

cot A cot A

For the following exercises, use Figure 15 to evaluate each trigonometric function of angle A . A .

For the following exercises, solve for the unknown sides of the given triangle.

For the following exercises, use a calculator to find the length of each side to four decimal places.

b = 15 , ∡ B = 15° b = 15 , ∡ B = 15°

c = 200 , ∡ B = 5° c = 200 , ∡ B = 5°

c = 50 , ∡ B = 21° c = 50 , ∡ B = 21°

a = 30 , ∡ A = 27° a = 30 , ∡ A = 27°

b = 3.5 , ∡ A = 78° b = 3.5 , ∡ A = 78°

Find x . x .

A radio tower is located 400 feet from a building. From a window in the building, a person determines that the angle of elevation to the top of the tower is 36° , 36° , and that the angle of depression to the bottom of the tower is 23° . 23° . How tall is the tower?

A radio tower is located 325 feet from a building. From a window in the building, a person determines that the angle of elevation to the top of the tower is 43° , 43° , and that the angle of depression to the bottom of the tower is 31° . 31° . How tall is the tower?

A 200-foot tall monument is located in the distance. From a window in a building, a person determines that the angle of elevation to the top of the monument is 15° , 15° , and that the angle of depression to the bottom of the monument is 2° . 2° . How far is the person from the monument?

A 400-foot tall monument is located in the distance. From a window in a building, a person determines that the angle of elevation to the top of the monument is 18° , 18° , and that the angle of depression to the bottom of the monument is 3° . 3° . How far is the person from the monument?

There is an antenna on the top of a building. From a location 300 feet from the base of the building, the angle of elevation to the top of the building is measured to be 40° . 40° . From the same location, the angle of elevation to the top of the antenna is measured to be 43° . 43° . Find the height of the antenna.

There is lightning rod on the top of a building. From a location 500 feet from the base of the building, the angle of elevation to the top of the building is measured to be 36° . 36° . From the same location, the angle of elevation to the top of the lightning rod is measured to be 38° . 38° . Find the height of the lightning rod.

Real-World Applications

A 33-ft ladder leans against a building so that the angle between the ground and the ladder is 80° . 80° . How high does the ladder reach up the side of the building?

A 23-ft ladder leans against a building so that the angle between the ground and the ladder is 80° . 80° . How high does the ladder reach up the side of the building?

The angle of elevation to the top of a building in New York is found to be 9 degrees from the ground at a distance of 1 mile from the base of the building. Using this information, find the height of the building.

The angle of elevation to the top of a building in Seattle is found to be 2 degrees from the ground at a distance of 2 miles from the base of the building. Using this information, find the height of the building.

Assuming that a 370-foot tall giant redwood grows vertically, if I walk a certain distance from the tree and measure the angle of elevation to the top of the tree to be 60° , 60° , how far from the base of the tree am I?

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Access for free at https://openstax.org/books/algebra-and-trigonometry/pages/1-introduction-to-prerequisites
  • Authors: Jay Abramson
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Unit 7: Tight Triangles & Trigonometry Homework 3: Similar Right Triangles & Geometric Mean by

Unit 7: Tight Triangles & Trigonometry Homework 3: Similar Right Triangles & Geometric Mean By

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Related Questions

Can someone explain to me what I did wrong for this question, I have a test tomorrow

Please open the image attached to find the work and solution for the 1st problem.

You are told that you will have to wait for 5 hours in a line with a group of other people. Determine if: You know the number of minutes you have to wait. You know how many people have to wait. For each statement, if you answer yes draw an input-output diagram and write a statement that describes the way one quantity depends on another. If you answer no give an example of 2 outputs that are possible for the same input.

You know the number of minutes you have to wait is equal to 300 minutes.

Measurement is the method of comparing the properties of a quantity or object using a standard quantity.

Measurement is essential to determine the quantity of any object

Given that you are told that you will have to wait for 5 hours in a line with a group of other people.

So time you have to wait in the line = 5 hours

But you need the time in minutes.

1 hour = 60 minutes

5 hours = 60 × 5 minutes

             = 300 minutes

Hence you will have to wait for 300 minutes in the line.

Learn more about Measurements here :

https://brainly.com/question/29824090

Becky went out to eat with another friend. They ordered total of $97.89 for lunch. The sales tax of the city is 8.65% and they plan to leave 15% of tips. How much will each person pay if they evenly divide the bill? 15. Sale tax Round to the nearest cent= 16. Tips Round to the nearest cent= 17. Total = 18. Amount each person pays Round to the nearest cent =

17. $121.04

Step-by-step explanation:

15. Sale tax Round to the nearest cent=

8.65% of $97.89 = 0.0865 × $97.89 = $8.47

16. Tips Round to the nearest cent=

15% of $97.89 = 0.15 × $97.89 = $14.68

17. Total =

$97.89 + $8.47 + $14.68 = $121.04

18. Amount each person pays Round to the nearest cent =

$121.04/2 = $60.52

Solve this quadratic equation x^2+5x+3=0

The solutions of the quadratic equation will be -0.697 and -4.303.

Let the equation be ax² + bx + c = 0. Then the roots of the equation will be given as,

[tex]\rm x = \dfrac{-b \pm \sqrt{b^2 - 4 a c }}{2a}[/tex]

The quadratic equation is given below.

x² + 5x + 3 = 0

The zeroes of the quadratic equation are given by the formula method . Then we have

x = [- 5 ± √(5² - 4 × 1 × 3)] / (2 × 1)

x = (- 5 ± √13) / 2

x = (- 5 ± 3.6055) / 2

x = (- 5 + 3.6055) / 2, (- 5 - 3.6055) / 2

x = -0.697, -4.303

More about the roots of the equation link is given below.

https://brainly.com/question/12029673

What is an equation of the line that passes through the point (-8,0)(−8,0) and is parallel to the line x+2y=14x+2y=14?

percent of decrease from 30 to 22

From 30 to 2 is a decrease of 8

8 is what percent of 30 ?      8/30 x 100% = 26.7 %

brady has purchased a home for $299,000. he made a 20% down payment and financed the remaining amount. the intangible tax is 0.2%. which of the following is the total amount of the intangible tax? (2 points) $478.40 $598.00 $47,840.00 $59,800.00

The intangible tax consists of $47,840.00 in total.

First, calculate the amount of the down payment : 20% of $299,000 is $59,800.

Then, calculate the remaining amount that was financed : $299,000 - $59,800 = $239,200.

Next, calculate the intangible tax rate: 0.2% of $239,200 is 0.002.

Multiply the intangible tax rate by the remaining amount that was financed: 0.002 x $239,200 = $478.40.

Finally, multiply the intangible tax rate by the down payment amount: 0.2% of $59,800 is 0.002.

Multiply the intangible tax rate by the down payment amount: 0.002 x $59,800 = $47,840.

Therefore, the total amount of the intangible tax is $47,840.00.

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A hotel buyer is ordering new towels for all of the rooms in the hotel. Hand towels cost $2 a piece and bath towels cost $5. The total expenditure must be under $4,900. Select the inequality in standard form that describes this situation x = the number of hand towels ordered y = the number of bath towels ordered

x=100,   y=100

4,900/7= 700

500+200=700

Tonya has 39 packages. Each package weighs 58 pounds. Choose the best estimate of the total weight of the packages.

To pass algebra II requires an average of at least 70 on four tests. A student has scores of 80, 62, and 73. What possible scores on the fourth test would guarantee this student a passing score in the class?

Answer: Let x be the student's score on the fourth test. The average of the four scores must be at least 70, so we can write the equation:

(80 + 62 + 73 + x) / 4 >= 70

Expanding and simplifying the left-hand side:

(215 + x) / 4 >= 70

Multiplying both sides by 4 to isolate x:

215 + x >= 280

Subtracting 215 from both sides:

So, the student must score at least 65 on the fourth test in order to guarantee a passing average of 70. Any score higher than 65 will also guarantee a passing average.

Suppose H (+) = (2x + 4)° Find two functions / and g such that (/ 9g) (x) = H (x). Neither function can be the identity function. (There may be more than one correct answer.) f(x)= g(x) =

Suppose H (x)=(2x+4)⁶ . Find two functions ƒ and g such that (f°g)(x) = H (x). One of the possible solution is

f(x) = (2x + 4)^3

g(x ) = x^2

(f°g)(x) = f(g(x)) = f(x^2) = (2x^2 + 4)^3

One possible solution is to set f(x) = (2x + 4)^3 and g(x) = x^2.

Then, (f°g)(x) = f(g(x)) = f(x^2) = (2x^2 + 4)^3

And, ( 2x^2 + 4)^3 = (2(x^2) + 4)^3 = (2x^2 + 4)^3 = H(x).

Note that this is just one of the possible solutions. There could be other functions ƒ and g that would satisfy the equation (f°g)(x) = H (x).

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You might have noticed that the numbers, or frequencies, on an old radio dial are not evenly spaced. There is, however, some pattern to the placement of these frequencies. You will analyze this data and answer the following questions. frequency reading on the radio dial (kilohertz) 53 60 70 80 100 120 140 170 distance from the left end of the radio dial (centimeters) 1.54 2.17 2.95 3.62 4.75 5.67 6.45 7.43 1. a. What is the independent variable and what are its units? b. What is the dependent variable and what are its units? 2. Make a scatter plot of the data. Describe the data set based on the scatter plot. b. What do you expect to happen to the distance from the left end of the radio as the frequency gets higher? c. What do you expect to happen to the distance from the left end of the radio as the frequency gets smaller?

1a. The independent variable is the frequency reading on the radio dial and its unit is kilohertz (SI Unit is Hertz or hz)

b. The dependent variable is the distance from the left end of radio dial and its unit is centimeter  (SI Unit is metre)

A variable that is independent is precisely what it sounds like. It is a stand-alone variable t hat is unaffected by the other variables you are attempting to assess. Age, for instance, could be an independent variable.

2. The required scatter plot of frequency reading on the radio dial  (khz) vs distance from the left end of the radio dial (cm) was obtained.

a. The data set indicates that the present variable is increasing with an increase in value with concave upward . The increase is steeper for lower frequencies, while for higher frequencies, it is moderate .

b. As the frequency gets higher, the distance from the left end of the radio dial will increase .

c. As the frequency gets smaller , the distance from the left end of the radio dial will decrease .

d. The values for a is 4.98802 and b is -18.2939

The best curve fit for data is y= 4.98802 ln (x) - 18.2939

4. a. To find a station at frequency 93.3 kHz the distance from the left end of the dial is 4.2319 cm or by rounding off it is 4.23 cm

b. When the radio dial is tuned to be 4 cm from the left end of the radio it will pick up a radio station at a frequency of 89.062 kHz or by rounding off 89 kHz

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i'm stuck can someone help

gotta find the area?

2ft × 3ft = 6ft

and also 12ft divided by 2 is 6

So there for the missing ft is 6ft.

if a company charges x dollars per item, it finds that it can sell 1400-4x of them. each item costs $7 to produce

Rearranging and combining terms, we get: -4x^2 + 34x - 9800 < 0

To determine if the company is making a profit, we need to compare the revenue from selling 1400-4x items at x dollars per item to the cost of producing the items.

The revenue from selling the items is given by the expression x * (1400-4x).

The cost of producing the items is given by the expression 7 * (1400-4x).

We can set up an equation to find out if the company is making a profit:

x * (1400-4x) - 7 * (1400-4x) > 0.

Expanding the equation, we get:

x * 1400 - 4x^2 - 7 * 1400 + 28x > 0

Rearranging and combining terms, we get:

-4x^2 + 34x - 9800 < 0

This is a quadratic inequality , and we can use the quadratic formula to find the values of x that satisfy the inequality. We can then determine if there is a range of values of x such that the company is making a profit.

However, to make this solution simpler, we can observe that if x is very large, the company will make a loss because the cost of producing the items will exceed the revenue from selling them. On the other hand, if x is very small, the company will also make a loss because the revenue from selling the items will not cover the cost of producing them. Hence, there must be a specific value of x such that the company is making exactly $0 profit. To find this value of x, we can set the expression x * (1400-4x) - 7 * (1400-4x) equal to 0, and then solve for x.

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The lateral height of a cone is 4 inches and the area of the base of the cone is 49π in². It requires 2.5 minutes to paint the cone. The area of the base is doubled. How long will it take to paint this cone if it can be painted at the same rate? Use π≈3.14. Enter your answer, rounded to the nearest tenth, in the box.

It will take approximately 3.54 minutes to paint the cone if it can be painted at the same rate . Rounded to the nearest tenth , this is 3.5 minutes .

What is the lateral area?

The lateral surface of an object is all of the sides of the object, excluding its base and top (when they exist). The lateral surface area is the area of the lateral surface . This is to be distinguished from the total surface area , which is the lateral surface area together with the areas of the base and top .

Let's call the radius of the cone "r". The area of the base of the cone is 49π in², so:

r² * π = 49π

The lateral surface area of the cone is given by:

A = π * r * l = π * 7 * 4 = 28π

Using π ≈ 3.14, the lateral surface area is approximately:

A ≈ 3.14 * 7 * 4 = 88.56 in²

It takes 2.5 minutes to paint the cone, so the rate at which the cone can be painted is given by:

R = A / t = 88.56 / 2.5 = 35.424 in²/min

When the area of the base is doubled, the new radius is:

r' = √(2 * r²) = √(2 * 7²) = √(2 * 49) = √(98) = 7 * √(2)

The new lateral surface area is:

A' = π * r' * l = π * (7 * √(2)) * 4 = 28 * √(2)π

Using π ≈ 3.14, the new lateral surface area is approximately:

A' ≈ 3.14 * (7 * √(2)) * 4 = 126.12 in²

The time it takes to paint the new cone is given by:

t' = A' / R = 126.12 / 35.424 = 3.54 min

Hence, it will take approximately 3.54 minutes to paint the cone if it can be painted at the same rate . Rounded to the nearest tenth , this is 3.5 minutes .

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Let p be the proposition “All politicians are rich.”, q be “Mike is a politician.” , and r be “Mike is rich.” Express r ∨(~ q → r)

The expression r ∨ (~q → r) can be expressed in English as "Mike is rich or if Mike is not a politician, then Mike is rich."

The first part, "Mike is rich," represents the truth value of the proposition r. The second part, "q → r," is a conditional statement that can be read as "if not q (Mike is not a politician), then r (Mike is rich)." The negation symbol () in this conditional statement represents "not."

Note: The underline should be crossing out what it underlines

Therefore, the entire expression can be read as "Mike is rich or if Mike is not a politician, then Mike is rich."

Justin prepared 987 ads for mailing. To prepare each ad, it took him about 7 seconds (sec) to put each ad into an envelope and 8 seconds to seal, label, and stamp each envelope. Which is closest to the total amount of time it took Justin to prepare the ads?

The total amount of time it took Justin to prepare the ads is closest to option D) 15,000 sec.

Arithmetic operations are the basic part of mathematics which combines the basic operations like addition, subtraction, multiplication and division.

Given that,

Total number of ads prepared for mailing = 987

To prepare each ad,

Time taken to put an ad into an envelope = 7 seconds

Tome taken to seal , label , and stamp each envelope = 8 seconds

Total time taken to prepare an ad = 7 + 8 = 15 seconds

Total time taken for 987 ads = 987 × 15

                                                = 14,805, closest to 15,000 seconds

Hence the time taken to prepare 987 ads is closest to 15,000 seconds.

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Suppose P is drawn from the Uniform[0, 1] distribution, and then conditional on P, another random variable X is drawn from a Bernoulli(P) distribution. (a) Use the tower law to compute P(P ≤ t, X = 1) where t ∈ [0, 1] is some constant value. (b) Compute the CDF of the conditional distribution of P given X = 1. (c) Now suppose that, after P is drawn, we instead draw twice from the Bernoulli(P) distribution to produce random variables X and Y , rather than drawing only once to produce X as before. (We assume that, conditional on the value of P, the random variables X and Y are drawn independently.) Calculate P(X ≠Y ).

Using Bernoulli distribution P(X ≠Y | P ≤ 1/2) = 1 and P(X ≠Y | P > 1/2) = 1/2, the probability of random variable X is P(X ≠Y ) = 1/2.

(a)Using the pinnacle regulation, P(P ≤ t, X = 1) can be determined as P(P ≤ t|X = 1)P(X = 1) = P(P ≤ t)P(X = 1).

(b) The CDF of the contingent appropriation of P given X = 1 is F(t|X = 1) = P(P ≤ t|X = 1) = P(P ≤ t)P(X = 1).

(c) To work out P(X ≠Y ), we can involve the standard of absolute likelihood as P(X ≠Y ) = P(X ≠Y | P ≤ 1/2)P(P ≤ 1/2) + P(X ≠Y | P > 1/2)*P(P > 1/2).

The Bernoulli distribution is a probability distribution that is used to describe a random variable having two outcomes (success or failure).

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here’s The rest of The answer choices for question 12 b and d

Question Marc is building a gazebo and wants to brace each corner by placing a 8-inch wooden bracket diagonally as shown. How far below the corner should he fasten the bracket if he wants the distances from the corner to each end of the bracket to be equal? (Round to the nearest tenth of an inch.)

The required distance from the corner to each end of the bracket is 5.7 inches.

A right triangle is defined as a triangle in which one angle is a right angle or two sides are perpendicular .

Since the distance from the corner to each end is equal.

So ABC make an isosceles right triangle, where

AB = x inches

AC = x inches

BC = 8 inches

According to the Pythagoras theorem , we have

AB² + AC² = BC²

x² + x² = 8²

Round to the nearest tenth , and we get

Thus, the distance from the corner to each end is 5.7 inches.

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Please Please help me

The number line that represents the expression :

-5 + (-2.5)

Is the one in option B.

Here we have some number lines , and we want to see which one represents the expression below:

First, we start by moving 5 units down, with that we can discard option C.

Then we add -2.5, that is just equivalent to subtracting 2.5, so here we need to move another 2.5 units down.

Then the correct option is B, where you can see that both arrows go downwards.

The first one for 5 units and the second for 2.5 units.

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What is the area of these 2d shapes

The areas of the 2D shapes are:

To find the area of each 2D shape , decompose it and find the area of each of the shapes it is composed of.

1. Area of triangle = 1/2(base * height) = 1/2(8 * 2)

2. Area of the 2D shape = area of triangle + area of rectangle = 1/2(7 * 4) + 7 * 3

3. Area of the 2D shape = area of triangle + area of rectangle = 1/2(16 * 5) + 16 * 6

4. Area of the 2D shape = area of rectangle 1 + area of rectangle 2 = 8 * 7 + 12 * 8

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3 1 point The linear function, C, is represented by C(t) = -3.5t + 27.5. Which equation best represents the inverse of function C? t= t= C(t)+27.5 -3.5 Previous C(t)-27.5 -3.5 O C(t)-3.5 t= 27.5 O None of the equations are the inverse of function C.

The inverse function of C(t) is C⁻¹ (t) = (27.5 - t) / 3.5, then no available option is the inverse function of C (E).

From the case, we know that function C is:

C(t) = -3.5t + 27.5

To find the inverse function of C(t), we need to subtitute C(t) with y and try to find the function of t:

y = -3.5t + 27.5

3.5t = 27.5 - y

t = (27.5 - y)/3.5

We subtitute t with C⁻¹ (t) and y with t:

C⁻¹ (t) = (27.5 - t) / 3.5

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Bishop Corporation reports taxable income of $700,000 on its tax return. Given the following information from the corporation's records, determine Bishop's net income per its financial accounting records. Deduction for federal income taxes per books $240,000 Depreciation claimed on the tax return 135.000 Depreciation reported on the financial accounting books 75,000 Life insurance proceeds on death of a corporate officer 100,000 A $660.000 B. $560,000 C. $520,000 D. $620,000

Bishop Corporation reports taxable income of $700,000 on its tax return. Given the following information from the corporation's records, Bishop's net income per its financial accounting records $620,000.

The federal income tax provision is not deductible for determining taxable income, but it is deductible while determining book income . The proceeds from the life insurance are not taxable but are counted as income in the books.

The calculation of net income per book is as follows:

Taxable income + Life insurance proceeds - Provision for federal income + Depreciation on tax return - Depreciation per books

⇒ 700,000 + 100,000 - 240,000 + 135,000 - 75000

Therefore , net income per book is $620,000.

Net income can either be added to retained earnings by the company or given as a dividend to ordinary stockholders . Net earnings and net profit are frequently used as synonyms for net income because profit and earnings are used interchangeably for income (depending on usage in the UK and the US as well).

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If angle A=−260∘, what is the radian measure of A? Enter an exact fraction that contains the π symbol.

To convert degrees to radians, multiply by

, since a full circle is

Cancel the common factor of

Tap for more steps...

Move the negative in front of the fraction.

Mai says f(x) is always greater than g(x) for the same value of x. Is this true? Explain how you know.

if f(x) > g(x)

it depends on the equation. this can be true

You are in charge of production of a very in-demand part. Your clients have an expectation that the probability of a defective part is under 10%. The table below summarizes part production at five of your facilities. Find the following probabilities. Factory A Factory B Factory C Factory D Factory E Total Rejected Parts 120 125 210 712 120 H115 - Principles of Mathematics Assignment 6.3 - Journal 6 1287 Good Parts 3380 2575 4790 3988 3480 18213 Total 3500 2700 5000 4700 3600 19500 Write your answer as a percent rounded to one decimal place. 1. What is the probability that a randomly selected part is defective?

The probability that a randomly selected part is defective is given as follows:

A probability is calculated as the division of the desired number of outcomes by the total number of outcomes.

For this problem, the outcomes are given as follows:

Hence the probability that a randomly selected part is defective is given as follows:

p = 1287/19500 = 0.066 = 6.6%.

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The probability that someone will win a certain game is p=0.41 . Let X be the random variable that represents the number of wins in 868 attempts at this game. Assume that the outcomes of all games are independent. What is the mean number of wins when someone plays the game 868 times? (Round your answer to 2 places after the decimal point, if necessary.) = What is the standard deviation for the number of wins when someone plays the game 868 times? (Round your answer to 2 places after the decimal point, if necessary.) = Use the range rule of thumb (the " " rule) to find the usual minimum and maximum values for x . That is, find the usual minimum and maximum number of wins when this game is played 868 times. (Round your answers to 2 places after the decimal point, if necessary.) usual minimum value = usual maximum value =

The probability that someone will win a certain game is p=0.41

usual minimum value = 102.9

usual maximum value = 139.1

Binomial is the name given to the algebraic expression with just two terms. It has two terms and is a polynomial. It is commonly refereed as the total and difference among two or even more monomials. It is a polynomial's most basic form. The binomial distribution is a discrete probability distribution with just two outcomes in an experiment: success or failure. The distribution of binary data from a finite sample is described by the binomial distribution. As a result, it provides the likelihood that r occurrences will occur in n trials.

This will be a binomial distribution with parameters:

n = 378, p = 0.32

μ = np = 378 × (0.32) = 121.0

σ = √npq = √{378 × 0.32 × 0.68} = 9.07

Usual minimum value = 121 - 2(9.07) = 102.9

Usual maximum value = 121 + 2(9.07) = 139.1

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he data set shows the number of tickets sold each day for 10 days. 159, 162, 199, 200, 204, 208, 215, 235, 261, 294 which statement is true? responses only 294 is an outlier. only 294 is an outlier. only 159 is an outlier. only 159 is an outlier. both 159 and 162 are outliers. both 159 and 162 are outliers. both 159 and 294 are outliers.

I think that answer is only 294 is an outlier.

outlier is always maximum or minimum which has a big diffrence between other numbers. in this situation it is 294

Please answer the question below

If 2 large solar panels = 12 small solar panels

then ? = 6 small solar panels

therefore 6/12 × 2 large solar panels

= 1 large solar panels.

if 2 large solar panels = 18 minutes

than 5 large solar panels = ?

therefore = 5/2 × 180 minutes

=450 minutes

if 2 large solar panels = 180 minutes

than 1 large solar panels = ?

therefore = 1/2 × 180 minutes

= 90 minutes.

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    s it possible for the interior angles of a triangle to be in the ration of 1:2:6, but is it possible for the exterior angles of a triangle to be in Categorize these resources as renewable or nonrenewable: Renewable Resources: - Lumber - Solar - Wind Nonrenewable Resources: - Mineral - Gasoline -

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    To solve a right triangle using trigonometry: Identify an acute angle in the triangle α. For this angle: sin(α) = opposite/hypotenuse; and. cos(α) = adjacent/hypotenuse. By taking the inverse trigonometric functions, we can find the value of the angle α. You can repeat the procedure for the other angle.

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    Solution. The triangle with the given information is illustrated on the right. The third side, which in this case is the "adjacent" side, can be found by using the Theorem of Pythagoras a2 + b2 = c2. Always remember that in the formula, c is the length of the hypotenuse. From x2 + 52 = 92 we obtain x2 = 81 − 25 = 56.

  17. PDF RIGHT TRIANGLE TRIGONOMETRY

    Right Triangle Trigonometry Special Right Triangles Examples Find x and y by using the theorem above. Write answers in simplest radical form. 1. Solution: The length of the shorter leg is 6. Since the length of the hypotenuse is twice the length of the shorter leg, x =2 6 12.⋅= The length of the longer leg is 3 times

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    Use right-triangle trigonometry to solve applied problems. Mt. Everest, which straddles the border between China and Nepal, is the tallest mountain in the world. Measuring its height is no easy task and, in fact, the actual measurement has been a source of controversy for hundreds of years. The measurement process involves the use of triangles ...

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    In Mathematics, especially in the context of a geometry or trigonometry course at the high school level, special right triangles refer to the 45-45-90 and 30-60-90 triangles, which have set ratios for their sides. When forming a right triangle with one leg representing the horizontal distance and one leg the vertical distance, and finding the ...

  22. Unit 7: Tight Triangles &amp; Trigonometry Homework 3: Similar Right

    What is the right triangle? A right triangle is defined as a triangle in which one angle is a right angle or two sides are perpendicular. Since the distance from the corner to each end is equal. So ABC make an isosceles right triangle, where. AB = x inches. AC = x inches. BC = 8 inches. According to the Pythagoras theorem, we have. AB² + AC² ...

  23. 2.2: Solving Right Triangles.

    Practice each skill in the Homework Problems listed. 1 Solve a right triangle #1-16, 63-74. 2 Use inverse trig ratio notation #17-34. 3 Use trig ratios to find an angle #17-22, 35-38. 4 Solve problems involving right triangles #35-48. 5 Know the trig ratios for the special angles #49-62, 75-78.

  24. Unit 8: Right Triangles &amp; Trigonometry Homework 2: Special Right

    There are two "special triangles" in geometry and trigonometry. They are the 30°-60°-90° right triangle that is half of an equilateral triangle, and the 45°-45°-90° isosceles right triangle that is half a square (cut by the diagonal). The side ratios of these special triangles are relatively easy to remember. It is useful to memorize them. __