Reteaching Trigonometric Functions 7 - George … · · 2015-06-08You have learned about right...

Name Date Class

© Saxon. All rights reserved. 153 Saxon Geometry

You have learned about right triangles and been introduced to trigonometric functions. Now you will observe and compare some of the values of the trigonometric functions as the angle measures vary between 0° and 90°.

The sine and cosine ratios for a right triangle are as follows.

xC A

B

sin x � opposite

__________ hypotenuse

� BC ___ AB

cos x � adjacent

__________ hypotenuse

� AC ___ AB

Given a 30°�60°�90° triangle, find the sine and cosine for x if x � 30�. Then, find the sine and cosine if x � 60�.

For x � 30�: For x � 60�:

sin 30� � 1 __ 2 sin 60� � �

� 3 ___

2

cos 30� � ��

3 ___ 2

cos 60� � 1 __ 2

What do you notice about the sine and cosine of these two angles? They switch values.

Practice

1. Sketch a 45°�45°�90° triangle with legs that are 1 unit long.

Find sin 45° and cos 45°.

sin 45� � ��

2 ___ 2 ; cos 45� � �

� 2 ___

2

What do you notice about the sine and cosine of a 45° angle?

They have the same value.

Use your calculator for problems 2–4

2. Find the sine and cosine of 20°. sin 20° � 0.34; cos 20° � 0.94 3. Find the sine and cosine of 70°. sin 70° � 0.94; cos 70° � 0.34 4. What do you notice about the values of sine and cosine for 20° and 70°?

They switch values. 5. Both 20° and 70° angles and 30° and 60° angles are complementary. What

is the relationship between the sine and the cosine of complementary angles?

The sine of an angle equals the cosine of its complementary angle.

6. In �XYZ, �Z is a right angle, m�X � 30�, and YZ � 5. How are YZ and XZ related? How are YZ and XY related? Determine XZ and XY. Then, give exact values for sin 30° and cos 30°.

XZ � YZ ��

3 , XZ � 5 ��

3 ; XY � 2YZ, XY � 10; sin 30� � 1 __ 2 ; cos 30� � �

� 3 ___

2

INV

7Reteaching

Trigonometric Functions


Reteachingcontinued

The tangent ratio for a right triangle is as follows.

xC A

B

tan x � opposite

_______ adjacent

� BC ___ AC

The tangent of an angle is also the quotient of the sine and the cosine of the angle.

tan x � sin x _____ cos x

Find the tangent of 30° by dividing sin 30° by cos 30°.

sin 30� � 1 __ 2

and cos 30� � ��

3 ___ 2

tan 30� � 1 __ 2 ___

��

3 ___ 2 � � 1 __

2 � � 2 ___

��

3 � � 1 ___

��

3

Do not forget to rationalize the denominator.

� 1 ___ �

� 3 � � �

� 3 ___

��

3 � � �

� 3 ___

3

Therefore, tan 30� � ��

3 ___ 3 .

PracticeComplete the steps to find the tan 60° by dividing the sin 60° by the cos 60°.

7. sin 60� � ��

3 ___ 2 and cos 60� � 1 __

2

tan 60� � �

� 3 ___

2

______

1 __ 2 � � �

� 3 ___

2 � (2) � �

� 3

Use the information you learned from problems 2 and 3 for problems 8 and 9.

8. Find the tangent of 20° by dividing sin 20° by cos 20°. tan 20� � 0.36

9. Find the tangent of 70° by dividing sin 70° by cos 70°. tan 70� � 2.75

10. In �PQR, �R is a right angle, m�P � 45�, and PR � 4. Use the Converse of the Isosceles Triangle Theorem to determine how PR and RQ related. What is the length of RQ? Use the Pythagorean theorem to find PQ. Then, find and use sin 45° and cos 45° to find tan 45°.

PR and RQ are congruent, so RQ � 4; PQ � 4 ��

2 ;

sin 45� � cos 45� � ��

2 ___ 2 , so tan 45� � 1

INV

7

Name Date Class


You have worked with plotting points on a coordinate grid. Now you will translate figures on a coordinate grid both in one dimension and in two dimensions.

Translation in One Dimension

A translation, also known as a slide, is an isometry (a transformation that does not change the size or shape of the object being translated).

A translation in one dimension changes either the x-values or the y-values, but not both. A translation left or right on the coordinate grid will change only the x-values of the original figure. A translation up or down on the coordinate grid will change only the y-values of the original figure.

Find the coordinates of the vertices of the image of triangle ABC after a translation 4 units to the right. Show the preimage and image on the same coordinate grid.

Triangle ABC has coordinates A (2, 5), B (2, 2), and C (5, 2).Apply the rule (x, y) → (x � 4, y).

A (2, 5) → A�(2 � 4, 5) � A�(6, 5)

B (2, 2) → B�(2 � 4, 2) � B�(6, 2)

C (5, 2) → C�(5 � 4, 2) � C�(9, 2)

Triangle A�B�C�has coordinates A�(6, 5), B�(6, 2), and C �(9, 2).

PracticeComplete the steps to find the coordinates of the image of square JKLM after a translation of 3 units down. 1. Apply the rule (x, y) → ( x , y � 3 ). J (2, 6) → J�(2, 6 �3 ) � J�(2, 3 )

K (6, 6) → K�(6, 6 �3 ) � K�(6, 3 ) L (6, 2) → L�(6, 2 �3 ) � L�(6, �1 ) M (2, 2) → M�(2, 2 �3 ) � M�(2, �1 )Find the coordinates of each image after each indicated translation. 2. 5 units left 3. 6 units up

F�( �7 , 2 ) X�( 1 , 11 ) G�( �2 , 1 ) Y�( 4 , 12 ) H�( �3 , �3 ) Z �( �1 , 7 ) J�( �8 , �2 )

x

y

4

2

8

10

6

2 4 6 8 10O

BA

C

x

y

4

2

8

10

6

2 4 6 8 10

Change y-valueChange x-value

O

BA

C

4

2

8

10

6

2 4 6 8 10

x

y

OB

A

C B'

A'

C'

x

y

4

2

8

10

6

2 4 6 8 10O

M

J K

L

x

y4

2

-4 -2 2 4O

J

FG

H

-2

-4

x

y

4

2

8

6

2-2 4 6 8OZ

XY

-2

71ReteachingTranslations


Reteachingcontinued

Translation in Two Dimensions

A translation in two dimensions causes both the x-values and the y-values to change. For example, a translation of 3 units right and 2 units down can be written using the rule (x, y) → (x � 3, y � 2).

Find the coordinates of the vertices of the image of triangle ABC after a translation of 4 units right and 2 units up. Show both the preimage and the image on the same coordinate grid.

Triangle ABC has coordinates A (3, 7), B (3, 3), and C (5, 3).

Apply the rule (x, y) → (x � 4, y � 2).

A (3, 7) → A�(3 � 4, 7 � 2) � A�(7, 9)

B (3, 3) → B�(3 � 4, 3 � 2) � B�(7, 5)

C (5, 3) → C�(5 � 4, 2) � C�(9, 5)

Triangle A�B�C�has coordinates A�(7, 9), B�(7, 5), and C�(9, 5)

PracticeComplete the steps to find the coordinates of the image of rectangle WXYZ after a translation of 2 units left and 3 units up.

4. Apply the rule (x, y) → ( x � 2 , y � 3 ).

W (4, 6) → W�(4, �2 , 6 � �3 ) � W�(2, 9 )

X (6, 6) → X�(6, �2 , 6 � �3 ) � X�(4, 9 )

Y (6, 2) → Y�(6, �2, 2 � �3 ) � Y�(4, 5 )

Z (4, 2) → Z�(4, �2 , 2 � �3 ) � Z�(2, 5 )

Find the coordinates of each image after each indicated translation.

5. 3 unitsright, 6. 5 units left,

1 unit down 2 units up

J�( 1, 1 ) D�( �10 , �2 )

K�( 4 , 3 ) E�( �2 , �1 ) L�( 8, 0 ) F�( �5 , �2 ) M�( 4 , �4 )

x

y

4

2

8

10

6

2 4 6 8 10O

B

A

C

B'

A'

C'

x

y

-4

4

6

2

-4 -2 2 4 6O

M

J

K

L

-2

x

y

-4

-6

4

6

2

-4-6 -2 2 4 6O

FD

E-2

x

y

4

2

8

10

6

2 4 6 8 10O

Z

W X

Y

71

Name Date Class


You have worked with tangents and circles. Now you will solve problems involving common tangents.

Common Tangents: Internal

A common tangent is a tangent that is common to two circles. A common tangent is internally tangent if it intersects the segment that joins the centers of two circles. Two segments tangent from the same exterior point are congruent.

Line segments _

KL and _

PQ are common tangents to the circles. Find the lengths of

_ KL and

_ PQ .

Step 1: Solve for x.

KO � PO � 8

x � 3 � 8

x � 5

Step 2: Substitute x � 5 to find OQ or OL.

OQ � 2x � 2(5) � 10

OL � 3x � 5 � 3(5) � 5 � 10

Step 3: Find KL and PQ: KL � PQ � PO � OQ � 8 � 10 � 18.

PracticeComplete the steps to find each length.

1. 2.

AO � CO � 4 OM � OK � 4

x � 4 x � 4 OD � 4 � 5 � 9 JO � 2( 4 ) � 1 � 7 CD � CO � OD JK � JO � OK

CD � 4 � 9 � 13 JK � 7 � 4 � 11 AB � CD � 13 LM � JK � 11 Find each length.

3. 4.

PQ � RS � 8 AB � CD � 7

72ReteachingTangents and Circles, Part 2

b

a

X

Y

P

L

A

B

QK

O2xx + 3

3x - 58

J

Kx + 5

2x + 1 BC

AD

O

x

4

J

K

x + 3

2x - 1

B

L

A

M

Ox

4

J K

2x + 4BC

A D

O

x 6S

6 - x2x - 1

YQ

Ox

5

X

R

P


Reteachingcontinued

Common Tangents: ExternalA common tangent is externally tangent if the tangent does not intersect the segment joining the centers of two circles. Again, two segments tangent from the same exterior point are congruent.

Line segments _

AB and _

CD are common tangents to the circles. Find the lengths of

_ AB and

_ CD .

AB � CD Given

3x � 2 � x � 4 Substitute.

2x � 6 Solve.

x � 3

AB � 3(3) � 2 � 7 Substitute.

CD � 3 � 4 � 7 Substitute.

PracticeComplete the steps to find each length.

5. 6.

AB � CD AB � CD

3x � 2 � 5x � 6 4x � 2x � 12

8 � 2x 2x � 12

x � 4 x � 6 AB � 3(4 ) � 2 � 14 AB � 4(6 ) � 24 CD � 5(4) � 6 � 14 CD � 2(6) � 12 � 24

Find each length.

7. 8.

OB � 44 JL � 34

OD � 44 JN � 34

b

a X Y

BA

a

b

CD

OX Y

3x - 2

x + 4

B

A

CD

E X Y

3x + 2

5x - 6B

A C

D

E

M

N

2x + 124x

BA

CD

O

2x + 2

x

14

M NM

N

K

L

J

2x

x - 2 10

A

B

72

Name Date Class


You have worked with angles and trigonometry. Now you will use angles of elevation and depression to solve problems.

Angle of Elevation

An angle of elevation is the angle formed by a horizontal line and a line of sight to a point above. You can use trigonometry to solve problems involving angle of elevation.When the angle of elevation to the sun is 48°, a tree casts a shadow that is 10 meters long. What is the height of the tree? Round to the nearest tenth of a meter.

Sketch a diagram to represent the given information.The height of the tree is represented by x. The angle of elevation is 48°, and the length of the shadow on the ground is 10 meters.To find x, use the tangent function.

tan 48� � opp

____ abj

Tangent Function

tan 48� � x ___ 10

Substitute.

10 tan 48� � x Simplify.

11.1 � x Solve.

The tree is about 11.1 meters tall.

PracticeComplete the steps to solve the problem.

1. A building casts a shadow 45 feet long. The building is 120 feet tall. What is the angle of elevation from the point on the ground where theshadow of the building ends? Round to the nearest tenth of a degree.

tan x � opp

____ abj

tan x � 120 ____ 45

� 2.67

x � 69.4�

Solve the problem.

2. Andrew is 15 feet from a house his brother is painting. He measures the angle of elevation from the ground to the top of his brother’s head as 52°. How high up the house is the top of his brother’s head?Round to the nearest foot. 19 ft

3. Find the angle of elevation to the top of a tree for a person standing 28.6 meters from the tree, assuming the person’s eye is 1.5 meters above the ground and the tree is 19.1 meters tall. Round to the nearest tenth of a degree. 31.6°

73ReteachingApplying Trigonometry: Angles of Elevation and Depression

angle of elevation

line of sight

10 m48º

x

120.0 ft

45.0 ft

x


Reteachingcontinued

Angle of Depression

An angle of depression is the angle formed by a horizontal line and a line of sight to a point below. You can use trigonometry to solve problems involving angle of depression.

Patricia is watching a parade from a 20-foot-high balcony.The angle of depression to the parade is 47�. What is thedistance between Patricia and the parade?

Sketch a diagram to represent the given information.Let A represent Patricia and let B represent the parade. Let x represent the distance between Patricia and the parade.m�B � 47� by the Alternate Interior Angles Theorem.Write a sine ratio using �B

sin 47� � 20 ___ x Sine Function

x sin 47� � 20 Multiply both sides by x.

x � 20 _______ sin 47�

Divide both sides by sin 47�.

x � 27.3 Simplify.

The distance between Patricia and the parade is about 27 feet.

PracticeComplete the steps to solve the problem. 4. A person swimming in a pool sees a quarter on the bottom of the

pool at an angle of depression of 42°. She is 6 feet above the bottom of the pool. How far from the quarter is she? Round to the nearest foot.

sin x � opp

____ hyp

sin 42� � 6

__ x

x sin 42� � 6 x � 6 ______

sin 42�

x � 9 ft

Solve the problem.

5. Paul is looking out a window that is 18 feet off the ground, and sees a mailbox near the street. The angle of depression to the mailbox is28°. What is the horizontal distance from Paul to the mailbox? Round to the nearest foot. 34 ft

6. Elise is parasailing. She is attached to the boat by a rope that is 300 feet long. The angle of depression from Elise to the boat is 29°. How high up in the air is Elise? Round to the nearest foot. 145 ft

73

angle of depression

line of sight

20 ftx

A

B

6 ft6 ft x

42˚

Name Date Class


You have worked with translations. Now you will reflect figures across a line on a coordinate grid.

Reflecting Across an Axis

A reflection is an isometry (a transformation that doesnot change the size or shape of the object beingtranslated). It involves a translation across a line,where each point and its image are the samedistance from the line of reflection.

When an object is reflected across the x-axis, the reflection maps � x, y � → � x, �y � . When an object is reflected across they-axis, the reflection maps � x, y � → (�x, y).

Reflect �ABC across the x-axis. Find the coordinates of the vertices of the reflection image.

The image is reflected by using � x, y � → � x, �y � .

A � 1, 1 � → A� � 1, �1 �

B � 1, 4 � → B� � 1, �4 �

C � 4, 4 � → C� � 4, �4 �

PracticeComplete the steps to find the coordinates ofeach reflection.

1. Reflect �JKL across the y-axis.

J � �4, 4 � → J� � 4 , 4 �

K � �1, 4 � → K� � 1, 4 �

L � �4, 2 � → L� � 4, 2 �

Find the coordinates for each reflection.

2. Reflect WXYZ across the y-axis. 3. Reflect �MNO across the x-axis.

W� � 5, 3 � M� � �3, �2 �

X� � 3, 3 � N� � 2, �2 �

Y� � 3, �2 � O� � 2, 0 �

Z� � 5, �2 �

74

A

D

A'

D'

B

C

B'

C'

x

y

4

2

2-2 4 6 8O

B

A

C

B'

A'

C'

-2

-4

x

y

4

2

-2-4O

J

L

K

2 4

-2

x

y

4

2

-2-4-6O

W

Z Y

X

2 4

-2

x

y

4

2

-2-4-6O

M

O

N

2 4

-2

ReteachingReflections


Reteachingcontinued

Reflecting Across a Vertical or Horizontal Line

When reflecting an object across a vertical line x � a, the reflection maps � x, y � → � a � � x � a � , y � or � x, y � → � 2a � x, y � .When reflecting an object across a horizontal line y � b, the reflection maps � x, y � → � x, b � � y � b � � or � x , y � → � x, 2b � y � .

Reflect �ABC across the line x � 3. Find the coordinates of the vertices of the reflection image.

The image is reflected using � x, y � → � 3 � � x � 3 � , y � or � x, y � → � 6 � x, y � .A � 0, 4 � → A� � 6 � 0, 4 � � A� � 6 ,4 �

B � 2, 4 � → B� � 6 � 2, 4 � � B� � 4, 4 �

C � 2, 2 � → C� � 6 � 2, 2 � � C� � 4, 2 �

PracticeComplete the steps to find the coordinates of each reflection.

4. Reflect ABCD across the line y � 3.

A � �2, 2 � → A� � �2, 6 � 2 � � A� � �2, 4 �

B � 2, 2 � → B� � 2, 6 � 2 � � B� � 2, 4 �

C � 2, �2 � → C� � 2, 6 � � 2 � � �C� � 2, 8 �

D � �2, �2 � → D� � �2, 6 � � 2 � � � D� � �2, 8 �

Find the coordinates of each reflection.

5. Reflect �PQR across the line y � �2. 6. Reflect JKLM across the line x � �1.

P� � �7, �5 � J� � �3, 6 � Q� � �5, �8 � K� � �8, 6 � R� � �2, �5 � L� � �8, 2 �

M� � �3, 2 �

x

y

4

6

8

2

O

A B

CD

B' A'

D'C'

2 4 6 8 10

-2

x

y

4

6

2

O

A B

C

A'B'

C'

2 4 6

-2

x

y

4

2

-4O

A

D C

B

4

x

y

4

2

-6-8 -4 -2O

P R

Q

2x

y

4

6

8

2

-2 O

LM

KJ

2 4 6 8

-2

74

Name Date Class


You have worked with plotting figures on a coordinate plane. Now you will find the equation of a circle.

Equation of a Circle

The equation of a circle with center (h, k) and radius r is

(x � h) 2 � (y � k) 2 � r 2 .

Write the equation of �C with center C(2, –1) and radius 6.

(x � h) 2 � (y � k) 2 � r 2 Equation of a Circle

(x � 2) 2 � (y � (�1)) 2 � 6 2 Substitute.

(x � 2) 2 � (y � 1) 2 � 36 Simplify.

PracticeComplete the steps to write the equation of �C.

1. (x � h) 2 � (y � k) 2 � r 2

(x � �2) 2 � (y � �1)

2 � 3 2

(x � 2 ) 2 � (y � 1 ) 2 � 9

Write the equation of �C.

2. (x � 5) 2 � y 2 � 16 3. (x � 7) 2 � (y � 3) 2 � 9

75ReteachingWriting the equation of a circle

y4

2

O

r

2 4

(h, k)x

y

O

-4

-2

-2

x

y

6

-6 4OC

-4

x

y4

2

-4-6 OC

2

-2

-4

x

y4

2

OC

2 4 6 8 10

-2

-4

x

y

4

6

8

10

2

O

C

2 4 6 8 10


Reteachingcontinued

Equation of a Circle

You can also write the equation of a circle if you know the center and one point on the circle.

Write the equation of �C with center C(3, 7) that passes through (1, 7).

Step 1: Find the radius.

r � ��

( x 2 � x 1 ) 2 � ( y 2 � y 1 ) 2 Distance Formula

r � ��

(1 � 3) 2 � (7 � 7) 2 Substitute.

r � ��

4 � 2 Simplify.

Step 2: Use the equation of a circle.

(x � h) 2 � (y � k) 2 � r 2 Equation of a Circle

(x � 3) 2 � (y � 7) 2 � 2 2 Substitute.

(x � 3) 2 � (y � 7) 2 � 4 Simplify.

PracticeComplete the steps to find the equation of the circle.

4. �C with center C(4, 2) that passes 5. �B with center B(3, 1) that passes throughthrough (2, 2) (7, 1)

r � ��

( x 2 � x 1 ) 2 � ( y 2 � y 1 ) 2 r � ��

( x 2 � x 1 ) 2 � ( y 2 � y 1 ) 2

r � ��

(4 � 2) 2 � (2 � 2)

2 r � �

��

(7 � 3) 2 � (1 � 1)

2

r � ��

4 � 2 r � ��

16 � 4

(x � h) 2 � (y � k) 2 � r 2 (x � h) 2 � (y � k) 2 � r 2

(x � 4) 2 � (y � 2)

2 � 2

2 (x � 3)

2 � (y � 1)

2 � 4

2

(x � 4) 2 � (y � 2)

2 � 4 (x � 3)

2 � (y � 1)

2 � 16

Find the equation of the circle.

6. �T with center T(–1, 5) that passes 7. �C with center C(–4, –3) that passesthrough (4, 5) through (–4, 3)

(x � 1) 2 � (y � 5) 2 � 25 (x � 4) 2 � (y � 3) 2 � 36

75

x

y

4

6

8

2

O

C

2 4 6 8 10

10

Name Date Class


You have worked with reflections and rotations of geometric figures. Now you will work with symmetry.

Line Symmetry

A line of symmetry is a line that divides a plane figure into two congruent, reflected halves. Basically, you could put a mirror on the line of symmetry to see the entire object.

Line symmetry is a type of symmetry that describes a figure that can be reflected across a line so that the image coincides with the preimage. This is also called reflection symmetry.

one line of symmetry two lines of symmetry no line of symmetry

Write whether the figure below has line symmetry.If so, draw all lines of symmetry.

Step 1: Check for a line that divides the triangle into two congruent halves.

Step 2: If one exists, check for another line that divides the triangle into two congruent halves. Check for a third line.

Step 3: Draw the line, or lines, of symmetry.

PracticeWrite whether each figure has line symmetry. If so, draw all lines of symmetry.

1. 2. 3.

line symmetry no line of symmetry line symmetry

4. 5. 6.

line symmetry no line of symmetry no line of symmetry

76ReteachingSymmetry


Reteachingcontinued

Rotational Symmetry

A figure has rotational symmetry if it can be rotated about a point by an angle between 0° and 360° so that the image coincides with the original figure.

The smallest angle through which the figure is rotated to coincide with itself is called the angle of rotational symmetry.

The number of times that you can get an identical figure when repeating the degree of rotation is called the order of the rotational symmetry.

Below are some examples of figures with and without rotationally symmetry.

angle: 180° 120° no rotational

order: 2 3 symmetry

Write whether the figure below has rotational symmetry.If so, give the angle of rotational symmetry and the order of symmetry.

Step 1: Pick up your paper and rotate it clockwise 90°.

Step 2: Is the rotated figure congruent to the original? Does the figure have rotational symmetry? yes

Step 3: Write the angle and order of rotational symmetry for this figure. angle: 90°; order: 4

PracticeWrite whether each figure of symmetry has rotational symmetry. If so, give the angle of rotational symmetry and the order.

7. 8. 9.

10. 11. 12.

76

order: 2yes; angle: 180°; no rotational

symmetryyes; angle: 90°; order: 4

no rotational symmetry order: 2

yes; angle: 180°; order: 3yes; angle: 120°;

Name Date Class


You have found the lateral area, surface area, and volume of regular pyramids. Now you will find the lateral area, surface area and volume of cones.

Lateral and Surface Area of a ConeLateral Area: The lateral area of a cone with radius r and slant height l is LA � � rl.

Surface Area: The surface area of a cone with lateral area LA and base area B is SA � LA � B, where B � � r 2 .

Find the lateral and surface area of the cone. Round to the nearest tenth.

Step 1: Find LA and B using r � 3 and l � 5.

LA � � rl� � (3)(5)

� 47.1 mm 2

Step 2: Find SA using LA � 47.1 mm 2 and B � 28.3 mm 2 .

SA � LA � B � (47.1) � (28.3) � 75.4 mm 2

The surface area of the cone is approximately 75.4 mm2.

PracticeComplete the steps to find the lateral and surface area of the cone. Round to the nearest tenth.

1. LA � � rl � � (3)(8) � 75.4 in. 2

B � � r 2 � � (3) 2 � 28.3 in. 2

SA � LA � B

� (75.4) � (28.3)

� 103.7 i n. 2 Find the lateral and surface area of each cone. Round to the nearest tenth.

2. 15 cm6 cm 3.

5 ft4 ft

LA � 287.7 cm 2 LA � 62.8 ft 2

SA � 395.8 cm 2 SA � 113.1 ft 2

77ReteachingFinding Surface Areas and Volumes of Cones

5 mm

3 mm

l

r base

slant height

B � � r 2 � � (3) 2 � 28.3 mm 2

3 in.

8 in.


Reteachingcontinued

Volume of a ConeThe volume of a right cone with base area B and

height h is V � 1 __ 3 Bh, where B � � r 2 .

Find the volume of the cone. Round to the nearest tenth.

Step 1: Find the base area, B, using r � 10.

B � � r 2

� � (10) 2

� 314.2 km 2

Step 2: Find the volume using B � 314.2 km 2 and h � 24.

V � 1 __ 3 Bh

� 1 __ 3 (314.2)(24)

� 2513.6 km 3

The volume of the cone is approximately 2513.6 k m 3 .

PracticeComplete the steps to find the volume of the cone.Round to the nearest tenth.

4. B � � r 2 � � (4) 2 � 50.3 ft 2

V � 1 __ 3 Bh

� 1 __ 3 (50.3)(12)

� 201.2 ft 3

Find the volume of each cone. Round to the nearest tenth.

5. 3 cm

11 cm

6.

18 mm27 mm

V � 103.7 cm 3 V � 9161.1 mm 3

77

r

h

24 km

10 km

4 ft

12 ft

Name Date Class


Now you will rotate figures on a coordinate plane.

Rotating About the Origin

�ABC has vertices A(�6, 2), B(�3, 5), and C(�2, 2). Graph �ABC and its rotation 90° counterclockwise about the origin.

Step 1: Graph �ABC.

Step 2: Using the rule above, (x, y) becomes (�y, x).

A(�6, 2) → A�(�2, �6)

B(�3, 5) → B�(�5, �3)

C(�2, 2) → C�(�2, �2)

Step 3: Graph �A�B�C�.

Practice

1. �XY Z has vertices X(4, �4), Y(8, �4), and Z(8, �7).

Using the rule, (x, y ). becomes (�x, �y).

X(4, �4) → X �(�4, 4 )

Y(8, �4) → Y �( �8, 4)

Z(8, �7) → Z �( �8, 7 )

�X�Y�Z� has vertices X �(4, �4), Y �( 8 , �4), and Z �(�8, 7 ).

Rotate each figure by the specified angle counterclockwise about the origin. List the coordinates of the vertices of the rotated figure.

2. 270°

x

y

4

6

2

2 4 6 8-4-6-8 -2

-2

O

B

C

A'

B'

C'D'

A D

3. 90°

x

y4

2

2 4 6-4-6 -2

-4

-2

O O

M

N

O'

N'M'

A�(2, 6), B�(4, 4), C�(3, 3), D�(2, 4)

M�(�2, �2), N�(3, �2), O�(�5, 0)

78ReteachingRotations

x

L

y

4

6

2

2 4 6-4-6 -2

-4

-2

O

M

NO

L'

M'N'

O'

x

y

4

2

2 4 6 8-4-6-8 -2

-4

-6

-2

OA

B

C

B'

A'

C'

x

y

4

2

2 4-4 -2

-4

-2

OX

Y

Z

Z'

Y' X'

When a figure is rotated about the origin on a coordinate plane, some simple rules apply for certain rotations.

If a point (x, y ) is rotated 90°, then (x, y) becomes (�y, x).

If a point (x, y ) is rotated 180°, then (x, y) becomes (�x, �y).

If a point (x, y ) is rotated 270°, then (x, y) becomes (y, �x).

In this picture, quadrilateral LMNO has been rotated 270° about the origin to produce quadrilateral L�M�N�O�.


Reteachingcontinued

Rotating About a Chosen Point

Triangle DEF has vertices D(�4, 4), E(�3, 2), F(�2, 5). Graph the triangle and its rotation 180° counterclockwise about the point R(�4, 1).

Step 1: Graph �DEF.

Step 2: Draw a segment from the center of rotation, R(�4, 1), to point D.

Step 3: Use a protractor to measure a 180° angle counterclockwise from

_ RD .

Step 4: Draw _

RD� . Use a compass or a ruler to make sure D and D� are the same distance from R.

Step 5: Repeat these steps for points E and F.

PracticeComplete the steps to rotate ABCD 90° counterclockwise about R(0, �1).

4. Draw a segment from R(0, �1) to point A.

Measure a 90° angle from _

RA . Draw

_ RA� .

Make A� the same distance from R as A is from R. Repeat for points B, C, and D.

Rotate each figure by the specified angle counterclockwise about the specified point.

5. 45° about R(3, 0) 6. 180° about R(�3, 3)

78

x

y

4

6

8

2

2 4 6 8-4-6-8 -2

-4

O

C'

B'

A'A

R

B

C

45°

x

y

4

6

8

2

2 4 6 8-6-8 -2

-4

-6

-8

-2

OE'

D'F' R

D

R180°

F

E

x

y

4

6

8

2

4 6 8-4-6-8

-2

O

R

D C

A B

D'

A'

C'

B'

x

y

4

6

4

6

2

2 4 6-4-6 -2

-2

OD'

C'

A'

B'D C

A B

90° R

x

y

4

6

8

2

2 4 6 8-4-6-8 -2

-2

O

Y

RX Z

Y'

X'

Z'

When a figure is rotated about a chosen point on a coordinate plane, a protractor must be used to measure the angle of rotation. The rotated vertices will be the same distance from the center of rotation.

In this picture, �ABC was rotated 45° counterclockwise about the point R(0, �3) to produce �A�B�C�.

Name Date Class


You have worked with angles interior to circles. Now you will work with angles exterior to circles.

Intercepting by Two Secants or Two Tangents

An exterior angle of a circle has its vertex outside the circle with each ray intersecting the circle at least once. Below are two examples of exterior angles.

intercepting by two tangents intercepting by two secants

The measure of an exterior angle is equal to half the difference of the intercepted arcs.

m�1 � 1 __ 2 (mEHG � mEG) m�2 � 1 __

2 (mJN � mKM)

Find the value of x.

Step 1: Since mPVR � mPR � 360�, mPVR � 142 � 360�, and mPVR � 218�.

Step 2: Substitute mPVR � 218� and mPR � 142� into the formula and solve.

x � � 1 __ 2 (mPVR � mPR)

� 1 __ 2

(218� � 142�)

x � � 38�

PracticeFind the value of x.

1.

108°35°

A

B

DE

Cx°

2.

232°

G

x° H

J

3.

29°

75°

M x°

P

Q

L

N

35� � 1 __ 2

(108� � x); x � 38� x � 1 __ 2 (232� � 128�); x � 52� x � 1 __

2 (75� � 29�); x � 23�

4. What is the measure of the angle exterior to a circle made by two secants if the shorter arc is 52� and the longer arc is 135�? 41.5�

5. What is the measure of the angle exterior to a circle made by two tangents if there is an arc that is 267�? 87�

79ReteachingAngles Exterior to Circles

2

MN

L

K

J

1

E

H

F

G

142°

R

Q

V

P

x°


Reteachingcontinued

Intercepting by a Secant and a Tangent

It is also possible for an exterior angle to intersect by a secant and a tangent. The measure of an exterior angle is still equal to half the difference of the intercepted arcs.

m�1 � 1 __ 2 (mAD � mBD)

Find the value of x.

Step 1: Write the formula in terms of the given exterior angle.

x � � 1 __ 2 (mTU � mSU )

Step 2: Substitute mSU � 74� and mTU � 74� into the formula and solve.

x � � 1 __ 2 (mTU � mSU )

� 1 __ 2 (140� � 74�)

x � � 33�

PracticeFind the value of x.

6.

70° 204°

R

S

U

T

x°

7.

44°63°

A

x°

Y

X

F

8. 79°

173°

x°A

B

C

x � 1 __ 2 (204� � 70�); x � 67� 44� � 1 __

2 (x � 63�); x � 151� x � 1 __

2 (173� � 79�); x � 47�

9. What is the measure of the angle exterior to a circle made by one tangent and one secant if the shorter arc length between the tangent and secant is 12° and the longer arc length is 64°? 26�

10. If the measure of the angle exterior to a circle made by one tangent and one secant is 54° and the shorter arc length between the tangent and secant is 42°, what is the longer arc length? 150�

11. What is the measure of the angle exterior to a circle made by one tangent and one secant if the shorter arc length between the tangent and secant is 67° and the longer one is 172°? 52.5�

1CB

A

D

79

74°

140°

R

S

U

T

x°

Name Date Class


You have found the surface area and the volume of cylinders. Now you will find the surface area and the volume of spheres.

Surface Area of a Sphere

The surface area of a sphere with radius r is SA � 4� r 2 .

Find the surface area of the sphere below.

Round to the nearest tenth.

Step 1: Find r by dividing the diameter, d � 20 m, by two.

r � 20 ___ 2

� 10 m

Step 2: Substitute r � 10 m into the formula and solve.

SA � 4� r 2

� 4� (10) 2

� 1,256.6 m 2

PracticeFind the surface area of each sphere. Round each answer to the nearest tenth.

1. Find the surface area of sphere S.

The diameter of sphere S is 10 centimeters . This means r, the radius of the sphere, is 5 centimeters . r 2 � 25 SA � 4� r 2 � 314.2 cm 2

2.

8 ft

3.

3 m

4.

36 m

SA � 804.2 ft 2 SA � 131.1 m 2 SA � 4,071.5 m 2

80ReteachingFinding Surface Area and Volumes of Spheres

r

20 m

10 cm

S


Reteachingcontinued

Volume of a Sphere

The volume of a sphere with radius r is V � 4 __ 3 � (13) 3 .

Find the volume of the sphere below. Round to the nearest tenth.

Step 1: Find r by dividing the diameter, d � 26 ft, by two.

r � 26 ___ 2

� 13 ft

Step 2: Substitute r � 13 ft into the formula and solve.

V � 4 __ 3 � r 3

� 4 __ 3

� (13) 3

� 9,202.8 ft 3

PracticeFind the volume of each sphere. Round each answer to the nearest tenth.

5. The diameter of this sphere is 16 centimeters .

The radius, r, is 8 centimeters .

r 3 � 512

4 __ 3 � r 3 � 2144.7 cm 3

Answer each question. Round each answer to the nearest tenth.

6. What is the volume of the sphere to the right?

1436.8 i n 3

7. What is the volume of a sphere with diameter 42 centimeters?

38,792.4 c m 3

8. If the surface area of a sphere is 64� mm 2 , what is the volume of the sphere?

268.1 m m 3

80

r

26 ft

16 cm

7 in.

Reteaching Trigonometric Functions 7 - George … · · 2015-06-08You have learned about right...

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