Chapter 8 practice

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Name: ________________________ Class: ___________________ Date: __________ ID: A

Chapter 8 Practice problems

Multiple Choice

Identify the choice that best completes the statement or answers the question.

____ 1. Write a similarity statement comparing the three triangles in the diagram.

a. GFJ GHF JHF c. GFJ FHG FJH

b. GFJ GFHJFH d. GFJ GHF FHJ

____ 2. Find the geometric mean of the pair of numbers 2 and 8.

a. 8 c. 5

b. 16 d. 4

____ 3. Find x, y, and z.

a. x = 5, y = 5 2, z = 5 7 c. x = 3, y = 4, z = 153

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Name: ________________________ ID: A

____ 4. To estimate the height of a radio tower, Jody steps 25 feet away from the center of the tower’s base, until his line of sight to the top of the tower and his line of sight to the center of its base form a right angle. His eyes are 6 feet above the ground. How tall is the radio tower to the nearest foot?

a. 98 feet c. 110 feet

b. 104 feet d. 31 feet

____ 5. Write the trigonometric ratio for cos X as a fraction and as a decimal rounded to the nearest hundredth.

a. cos X = 12

9 1.33 c. cos X =

12 15 0.80 b. cos X = 9

15 0.60 d. cos X =

9

12 0.75

____ 6. Use a special right triangle to write tan 60 as a fraction.

a. 3

1 c.

2 1

b. 1

3 d.

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Name: ________________________ ID: A

____ 8. Find GH. Round to the nearest hundredth.

a. GH = 32.08 in. c. GH = 22.46 in. b. GH = 15.07 in. d. GH = 26.28 in.

____ 9. Find the sine and cosine of the acute angles in the right triangle.

a. sin A = 45

53; cos A = 28 53 sin B = 28₅₃; cos B = 45₅₃

c. sin A = 28

53; cos A = 45 53 sin B = 45₅₃; cos B = 28₅₃

b. sin A = 45

28 ; cos A = 45 53 sin B = 45

53 ; cos B = 45 28

d. sin A = 28

53; cos A = 45 28 sin B = 45

28; cos B = 28 53

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Name: ________________________ ID: A

____ 11. The coordinates of the vertices of RPQ are R(2,1),P(2,2), andQ(2,1). Find mP.

a. mP = 53 c. mP = 93

b. mP = 37 d. mP = 42

____ 12. Classify each angle in the diagram as an angle of elevation or an angle of depression.

a. Angles of elevation: 1, 3 Angles of depression: 2, 4

c. Angles of elevation: 1, 4 Angles of depression: 2, 3 b. Angles of elevation: 2, 4

Angles of depression: 1, 3

d. Angles of elevation: 2, 3 Angles of depression: 1, 4

____ 13. The largest Egyptian pyramid is 146.5 m high. When Rowena stands far away from the pyramid, her line of sight to the top of the pyramid forms an angle of elevation of 20 with the ground. What is the horizontal distance between the center of the pyramid and Rowena? Round to the nearest meter.

a. 402 m c. 156 m

b. 427 m d. 65 m

____ 14. An eagle 300 feet in the air spots its prey on the ground. The angle of depression to its prey is 15. What is the horizontal distance between the eagle and its prey? Round to the nearest foot.

a. 1,120 ft c. 310 ft

b. 1,159 ft d. 723 ft

____ 15. Find AB. Round to the nearest tenth.

a. AB = 13.8 c. AB = 33.8

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Name: ________________________ ID: A

____ 16. Find AC. Round to the nearest tenth.

a. AC = 17.5 c. AC = 16.6

b. AC = 306.1 d. AC = 10.3

____ 17. A dam needs a supporting beam. The dam leans at an 80 angle and is 200 ft tall. If the base of the supporting beam is placed 75 feet from the base of the dam and the beam extends to the top of the dam, how long must the beam be?

a. 201.0 ft c. 170.8 ft

b. 40,415.6 ft d. 221.2 ft

____ 18. Write the vector AB in component form.

a. 5, 3 c. 5, 3

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Name: ________________________ ID: A

____ 19. Draw the vector 6, 3 on the coordinate plane. Find its magnitude to the nearest tenth. a.

6.7

c.

45.0 b.

5.2

d.

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Name: ________________________ ID: A

____ 20. A current’s velocity is given by the vector 3, 1 . Draw the vector on the coordinate plane. Find the direction of the vector to the nearest tenth of a degree.

a.

18.4 c. 30.0

b.

71.6 d. 19.5

Numeric Response

1. A hiking trail has a slope of ₃₂7 . What is the measure of the angle that the trail makes with a horizontal line? Round to the nearest degree.

2. TM has an initial point of (5, 5) and a terminal point of (7, 3). Find the magnitude of TM to the nearest tenth.

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ID: A

Chapter 8 Practice problems

Answer Section

MULTIPLE CHOICE 1. ANS: D

Sketch the three right triangles with the angles of the triangles in corresponding positions.

The altitude to the hypotenuse of a right triangle forms two triangles that are similar to each other and to the original triangle, so GFJ GHF FHJ.

Feedback

A All corresponding long legs must match up. B All right angles must align.

C Sketch the triangles in similar orientations to help you align corresponding parts. D Correct!

PTS: 1 DIF: Average REF: 1bb99956-4683-11df-9c7d-001185f0d2ea OBJ: 8-1.1 Identifying Similar Right Triangles LOC: MTH.C.11.08.03.05.002 TOP: 8-1 Similarity in Right Triangles KEY: similarity | right triangles | altitude

MSC: DOK 2 2. ANS: D

Let x be the geometric mean.

x2 _₍₂₎₍₈₎_₁₆ _{Definition of geometric mean}

x4 Find the positive square root.

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ID: A

3. ANS: B

52 _₁₀_x _{5 is the geometric mean of}_x_{and 10.}

x2.5 Simplify.

y2 __(12.5)(2.5) y is the geometric mean of the hypotenuse and the segment of the

hypotenuse adjacent to y.

y  125 4 

5 5

2 Simplify. Find the positive square root.

z2 __(12.5)(10) z is the geometric mean of the hypotenuse and the segment of the

hypotenuse adjacent to z.

z 125 5 5 Simplify. Find the positive square root.

Feedback

A The length of the altitude to the hypotenuse of a right triangle is the geometric mean of the two segments of the hypotenuse.

B Correct!

C The leg of a right triangle is the geometric mean of the hypotenuse and the segment of the hypotenuse adjacent to that leg.

D Create an proportion like this to solve for geometric mean x with regards to parts a and

b: x/a = b/x.

PTS: 1 DIF: Average REF: 1bbbfbb2-4683-11df-9c7d-001185f0d2ea OBJ: 8-1.3 Finding Side Lengths in Right Triangles LOC: MTH.C.11.08.03.05.006 TOP: 8-1 Similarity in Right Triangles KEY: similarity | right triangles | altitude to the hypotenuse MSC: DOK 2

4. ANS: C

Let x be the height of the radio tower above eye level.

252 _₆_x _{25 is the geometric mean of}_x_{and 6.}

6256x Square 25.

x104.17 Divide both sides by 6.

104.176110.17 Add 6 to find the height of the tower.

The radio tower is about 110 feet tall.

Feedback

A The altitude to the hypotenuse of a right triangle is the geometric mean of the two segments of the hypotenuse.

B Solve for the total height of the tower, not the height above Jody. C Correct!

D Use a proportion with geometric mean to solve.

PTS: 1 DIF: Average REF: 1bbe5e0e-4683-11df-9c7d-001185f0d2ea OBJ: 8-1.4 Application LOC: MTH.C.11.08.03.05.006

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ID: A

5. ANS: C

cos X = 12

15 0.80 The cosine of an  is

adjacent leg hypotenuse.

Feedback

A Cosine is the ratio of the adjacent side to the hypotenuse. B Sine is the ratio of the opposite side to the hypotenuse. C Correct!

D Tangent is the ratio of the opposite to the adjacent side.

PTS: 1 DIF: Basic REF: 1bc0c06a-4683-11df-9c7d-001185f0d2ea

OBJ: 8-2.1 Finding Trigonometric Ratios NAT: NT.CCSS.MTH.10.9-12.G.SRT.6 STA: MI.MIGLC.MTH.06.9-12.G1.3.1 | MI.MIGLC.MTH.06.9-12.G1.3.3

LOC: MTH.C.14.02.01.002 | MTH.C.14.02.02.004 TOP: 8-2 Trigonometric Ratios KEY: trigonometric ratio | trigonometry | cosine MSC: DOK 2

6. ANS: A

Draw and label a 30–60–90 triangle.

The tangent of an angle is opposite leg_{adjacent leg}.

tan60  s _s3  ₁3 s 3

Feedback A Correct!

B The tangent of an angle is the ratio opposite leg/ adjacent leg.

C Use the ratios of the sides of a 30°–60°–90° triangle to create the fraction. D The tangent of an angle is the ratio opposite leg/ adjacent leg.

PTS: 1 DIF: Average REF: 1bc322c6-4683-11df-9c7d-001185f0d2ea OBJ: 8-2.2 Finding Trigonometric Ratios in Special Right Triangles

NAT: NT.CCSS.MTH.10.9-12.F.TF.3 | NT.CCSS.MTH.10.9-12.G.SRT.6 STA: MI.MIGLC.MTH.06.9-12.G1.3.1 | MI.MIGLC.MTH.06.9-12.G1.3.3

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ID: A

7. ANS: D

Make sure your calculator is in degree mode. sin 79 = 0.98, cos 47 = 0.68, tan 77 = 4.33

Feedback

A Change your calculator to degree mode. B Change your calculator to degree mode. C Switch the first and second answers. D Correct!

PTS: 1 DIF: Basic REF: 1bc349d6-4683-11df-9c7d-001185f0d2ea

OBJ: 8-2.3 Calculating Trigonometric Ratios NAT: NT.CCSS.MTH.10.K-12.5.1 TOP: 8-2 Trigonometric Ratios KEY: trigonometric ratio | trigonometry | cosine | sine | tangent MSC: DOK 2

8. ANS: C

GH is the length of the hypotenuse of the triangle. You are given FH, which is adjacent to H. Since the adjacent side and hypotenuse are involved, use the cosine ratio.

cosH adj. leg_hyp.  FH_GH Write a trigonometric ratio.

cos 35=18.4_GH Substitute the given values.

GH _{cos 35}18.4 Multiply both sides by GH and divide by cos 35.

GH22.46 in Simplify the expression.

Feedback

A The hypotenuse and adjacent side are involved, so a cosine ratio is appropriate. B The cosine of an angle is adjacent leg / hypotenuse.

C Correct!

D The hypotenuse is involved, so a tangent ratio is not appropriate.

PTS: 1 DIF: Average REF: 1bc58522-4683-11df-9c7d-001185f0d2ea OBJ: 8-2.4 Using Trigonometric Ratios to Find Lengths

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ID: A

9. ANS: A

Feedback A Correct!

B Check the definitions of sine and cosine. C Check the definitions of sine and cosine. D Check the definitions of sine and cosine.

PTS: 1 DIF: Basic REF: 91632d45-6ab2-11e0-9c90-001185f0d2ea

OBJ: 8-2-Ext.1 Finding the Sine and Cosine of Acute Angles NAT: NT.CCSS.MTH.10.9-12.G.SRT.7 TOP: 8-2-Ext. Trigonometric Ratios and Complementary Angles

KEY: right triangle trigonometry | sine | cosine | tangent MSC: DOK 2 10. ANS: A

By the Pythagorean Theorem, AB2 __AC2__BC2 _₄2_₂2 __20.

SoAB 20 4.47.

sinA 2

4.47 0.45

Feedback A Correct!

B Use the Pythagorean Theorem to find AB. Sin A = BC/AB. C Sin A = BC/AB.

D Sin A = BC/AB.

PTS: 1 DIF: Average REF: 1bccac36-4683-11df-9c7d-001185f0d2ea OBJ: 8-3.3 Solving Right Triangles

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ID: A

11. ANS: A

From the figure, QR = 4 and PR = 3. tanP QR_PR 4

3, so tan

1 4 3

 

 _ 53.

Feedback A Correct!

B Find the complement of this angle.

C Angle A is the arctangent or inverse tangent of (4/3). D Angle A is the arctangent or inverse tangent of (4/3).

PTS: 1 DIF: Average REF: 1bccd346-4683-11df-9c7d-001185f0d2ea OBJ: 8-3.4 Solving a Right Triangle in the Coordinate Plane

LOC: MTH.C.14.04.03.001 | MTH.C.14.04.03.002 TOP: 8-3 Solving Right Triangles KEY: trigonometric ratio | trigonometry | coordinate geometry | inverse trigonometric ratios

MSC: DOK 2 12. ANS: A

1 and 3 are formed by a horizontal line and a line of sight to a point above the line. They are angles of elevation.

2 and 4 are formed by a horizontal line and a line of sight to a point below the line. They are angles of depression.

Feedback A Correct!

B An angle of elevation is formed from a horizontal line and a point above the line. C An angle of elevation is formed from a horizontal line and a point above the line. D An angle of elevation is formed from a horizontal line and a point above the line.

PTS: 1 DIF: Basic REF: 1bd170ee-4683-11df-9c7d-001185f0d2ea OBJ: 8-4.1 Classifying Angles of Elevation and Depression

LOC: MTH.C.11.02.04.10.001 | MTH.C.11.02.04.10.002 TOP: 8-4 Angles of Elevation and Depression

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ID: A

13. ANS: A

tan20  146.2_x Use the side opposite _{tangent ratio.} A and x, and the side adjacent to A to write the

x _tan20146.2 Multiply both sides by x and divide both sides by tan20.

x402 m Simplify.

Feedback A Correct!

B Divide 146.2 by tan 20. C Divide 146.2 by tan 20.

D Change your calculator to degree mode.

PTS: 1 DIF: Average REF: 1bd197fe-4683-11df-9c7d-001185f0d2ea

OBJ: 8-4.2 Finding Distance by Using Angle of Elevation NAT: NT.CCSS.MTH.10.9-12.G.SRT.8 STA: MI.MIGLC.MTH.06.9-12.G1.3.1 | MI.MIGLC.MTH.06.9-12.G1.3.3

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ID: A

14. ANS: A

By the Alternate Interior Angles Theorem, mS15. From the sketch, tan15  300_x . So

x _{tan15 }300 1,120 ft.

Feedback A Correct!

B Divide 300 by tan 15. C Divide 300 by tan 15. D Divide 300 by tan 15.

PTS: 1 DIF: Average REF: 1bd3d34a-4683-11df-9c7d-001185f0d2ea

OBJ: 8-4.3 Finding Distance by Using Angle of Depression NAT: NT.CCSS.MTH.10.9-12.G.SRT.8 STA: MI.MIGLC.MTH.06.9-12.G1.3.1 | MI.MIGLC.MTH.06.9-12.G1.3.3

LOC: MTH.C.14.02.03.001 TOP: 8-4 Angles of Elevation and Depression KEY: angle of elevation | angle of depression | trigonometry MSC: DOK 2

15. ANS: A sinB

AC 

sinC

AB Law of Sines

sin50° 12 

sin62°

AB Substitute the given values. ABsin50 12sin62 Cross Products Property

AB 12sin62_{sin50 }13.8 Divide both sides by sin50.

Feedback A Correct!

B Multiply 12 by sin 62, then divide by sin 50. C Change your calculator to degree mode. D Multiply 12 by sin 62, then divide by sin 50.

PTS: 1 DIF: Average REF: 1bd89802-4683-11df-9c7d-001185f0d2ea OBJ: 8-5.2 Using the Law of Sines

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ID: A

16. ANS: A

AC2 __AB2__BC2_₂ _AB

 BCcosB Law of Cosines

AC2 _₁₀2_₁₂2__{2(10)(12) cos(105)} _{Substitute the given values.}

AC2 __306.1166 _Simplify.

AC17.5 Take the square root of both sides.

Feedback A Correct!

B Take the square root of this answer. C Change your calculator to degree mode. D Use the Law of Cosines.

PTS: 1 DIF: Average REF: 1bd8bf12-4683-11df-9c7d-001185f0d2ea OBJ: 8-5.3 Using the Law of Cosines

NAT: NT.CCSS.MTH.10.9-12.G.SRT.10 | NT.CCSS.MTH.10.9-12.G.SRT.11 STA: MI.MIGLC.MTH.06.9-12.G1.3.2 LOC: MTH.C.14.06.01.002

TOP: 8-5 Law of Sines and Law of Cosines KEY: law of cosines | trigonometry MSC: DOK 2

17. ANS: A

AC2 __AB2__BC2_₂ _AB

 BCcosB Law of Cosines

AC2 _₇₅2_₂₀₀2 __{2(75)(200) cos(80)} _{Substitute the given values.}

AC2 __40,415.5547 _Simplify.

AC201.0 ft Take the square root of both sides.

Feedback A Correct!

B Take the square root of this answer. C Use the Law of Cosines.

D Change your calculator to degree mode.

PTS: 1 DIF: Average REF: 1bdafa5e-4683-11df-9c7d-001185f0d2ea OBJ: 8-5.4 Application

NAT: NT.CCSS.MTH.10.9-12.G.SRT.10 | NT.CCSS.MTH.10.9-12.G.SRT.11 STA: MI.MIGLC.MTH.06.9-12.G1.3.2 LOC: MTH.C.14.06.01.002

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ID: A

18. ANS: A

The horizontal change from A to B is 5 units. The vertical change from A to B is –3 units. So the component of AB is 5, 3 .

Feedback A Correct!

B The first number is the horizontal component. C The horizontal component is 5.

D The first number is the horizontal component.

PTS: 1 DIF: Average REF: 1bdd83ca-4683-11df-9c7d-001185f0d2ea

OBJ: 8-6.1 Writing Vectors in Component Form NAT: NT.CCSS.MTH.10.9-12.N.VM.2 STA: MI.MIGLC.MTH.06.9-12.L1.2.3 LOC: MTH.C.10.09.05.03.001

TOP: 8-6 Vectors KEY: vector | component form MSC: DOK 2 19. ANS: A

Draw the vector with base at (0, 0) and tip at (6, 3). To find the magnitude, use the Distance Formula.

6, 3 

   (60)2 (30)2  45 6.7

Feedback A Correct!

B When using the Distance Formula, add the squares of 6 and 3. The square of 3 is 9. C Take the square root of this answer.

D Use the Distance Formula with points (0,0) and (6, 3).

PTS: 1 DIF: Average REF: 1bdfbf16-4683-11df-9c7d-001185f0d2ea OBJ: 8-6.2 Finding the Magnitude of a Vector

STA: MI.MIGLC.MTH.06.9-12.L1.2.3 | MI.MIGLC.MTH.06.9-12.G1.1.5 | MI.MIGLC.MTH.06.9-12.G1.4.2 | MI.MIGLC.MTH.06.9-12.G2.3.4

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ID: A

20. ANS: A

Draw the vector and its components, creating ABC.

A is the angle formed by the vector and the x-axis, and tanA 1

3. Therefore mAtan

1 1 3

 

 _ 18.4.

Feedback A Correct!

B Find the inverse tangent of 1 3.

C Change to degree mode on your calculator.

D Find the inverse tangent of 1 3.

PTS: 1 DIF: Average REF: 1be22172-4683-11df-9c7d-001185f0d2ea OBJ: 8-6.3 Finding the Direction of a Vector LOC: MTH.C.14.10.001 TOP: 8-6 Vectors KEY: vector | direction MSC: DOK 2

NUMERIC RESPONSE 1. ANS: 12

PTS: 1 DIF: Basic REF: 1be94886-4683-11df-9c7d-001185f0d2ea NAT: NT.CCSS.MTH.10.9-12.G.SRT.8 | NT.CCSS.MTH.10.9-12.F.TF.7