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Unit 2 Review Short Answer 1. Find the value of x. Express your answer in simplest radical form. 30º x 3 6 2. The size of a TV screen is given by the length of its diagonal. The screen aspect ratio is the ratio of its width to its height. The screen aspect ratio of a standard TV screen is 4:3. What are the width and height of a 27" TV screen? y 24 60º x 5. Write the trigonometric ratio for cos X as a fraction and as a decimal rounded to the nearest hundredth. Y 15 9 height 27" X width 3. Find the missing side length. Tell if the side lengths form a Pythagorean triple. Explain. 25 Z 12 6. Use your calculator to find the trigonometric ratios sin 79 , cos 47 , and tan 77 . Round to the nearest hundredth. 7. Use your calculator to find the trigonometric ratios sin 49 , cos 50 , and tan 45 . Round to the nearest hundredth. 8. Find GH. Round to the nearest hundredth. 20 G 4. Find the values of x and y. Express your answers in simplest radical form. F 35° 18.4 in. H 9. Jessie is building a ramp for loading motorcycles onto a trailer. The trailer is 2.8 feet off of the ground. To avoid making it too difficult to push a motorcycle up the ramp, Jessie decides to make the angle between the ramp and the ground 15 . To the nearest hundredth of a foot, find the length of the ramp. 10. Find the sine and cosine of the acute angles in the right triangle. 1 1.3 cm 0.5 cm 2 3 1.2 cm 15. Use your calculator to find the angle measures to the nearest tenth of a degree. 16. Find to the nearest hundredth. B B 2 53 45 A C 4 17. Classify each angle in the diagram as an angle of elevation or an angle of depression. A 28 3 11. Find the sine and cosine of the acute angles in the right triangle. B 169 119 A 120 12. Write cos 16° in terms of the sine. 13. Write sin 74° in terms of the cosine. 14. Use the trigonometric ratio to determine which angle of the triangle is . 4 2 1 18. 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. 19. 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. 20. A pilot flying at an altitude of 1.8 km sights the runway directly in front of her. The angle of depression to the beginning of the runway is 31 . The angle of depression to the end of the runway is 23. What is the length of the runway? Round to the nearest tenth of a kilometer. Unit 2 Review Answer Section SHORT ANSWER 1. ANS: x= Pythagorean Theorem Substitute 3 for a, 6 for b, and x for c. Simplify. Find the positive square root. Simplify the radical. PTS: 1 DIF: 2 REF: 1af8a14a-4683-11df-9c7d-001185f0d2ea OBJ: 9-1.1 Using the Pythagorean Theorem STA: MCC9-12.G.SRT.8 LOC: MTH.C.10.05.10.05.01.001 | MTH.C.11.03.02.05.02.002 TOP: 9-1 The Pythagorean Theorem KEY: Pythagorean Theorem | side length DOK: DOK 1 2. ANS: width: 21.6 in., height: 16.2 in. Let 3x be the height in inches. Then 4x is the width of the TV screen. Pythagorean Theorem Substitute 4x for a, 3x for b, and 27 for c. Multiply and combine like terms. Divide both sides by 25. Find the positive square root. in. Width: Height: in. in. PTS: 1 DIF: 2 REF: 1afadc96-4683-11df-9c7d-001185f0d2ea OBJ: 9-1.2 Application NAT: NT.CCSS.MTH.10.9-12.G.SRT.8 STA: MCC9-12.G.SRT.8 LOC: MTH.C.10.05.10.05.01.001 | MTH.C.11.03.02.05.02.002 TOP: 9-1 The Pythagorean Theorem KEY: Pythagorean Theorem | side length DOK: DOK 1 3. ANS: The missing side length is 15. The side lengths form a Pythagorean triple because they are nonzero whole numbers that satisfy the equation . Pythagorean Theorem Substitute 20 for a and 25 for c. Multiply and subtract 400 from both sides. Find the positive square root. The side lengths are nonzero whole numbers that satisfy the equation triple. PTS: OBJ: LOC: TOP: DOK: 4. ANS: , so they form a Pythagorean 1 DIF: 1 REF: 1afd3ef2-4683-11df-9c7d-001185f0d2ea 9-1.3 Identifying Pythagorean Triples STA: MCC9-12.A.REI.4b MTH.C.11.03.02.05.02.002 | MTH.C.11.03.02.05.02.004 9-1 The Pythagorean Theorem KEY: Pythagorean Theorem | side length | Pythagorean triple DOK 2 , Hypotenuse Divide both sides by 2. PTS: OBJ: LOC: TOP: DOK: 5. ANS: 1 DIF: 2 REF: 1b046606-4683-11df-9c7d-001185f0d2ea 9-2.3 Finding Side Lengths in a 30-60-90 Triangle STA: MCC9-12.G.SRT.6 MTH.C.11.03.02.05.03.001 | MTH.C.11.03.02.05.03.002 9-2 Applying Special Right Triangles KEY: special right triangles | 30-60-90 DOK 2 cos X = cos X = The cosine of an is . PTS: 1 DIF: 1 REF: 1bc0c06a-4683-11df-9c7d-001185f0d2ea OBJ: 10-1.1 Finding Trigonometric Ratios NAT: NT.CCSS.MTH.10.9-12.G.SRT.6 STA: MCC9-12.G.SRT.6 LOC: MTH.C.14.02.01.002 | MTH.C.14.02.02.004 TOP: 10-1 Trigonometric Ratios KEY: trigonometric ratio | trigonometry | cosine DOK: DOK 2 6. ANS: sin 79 = 0.98, cos 47 = 0.68, tan 77 = 4.33 Make sure your calculator is in degree mode. sin 79 = 0.98, cos 47 = 0.68, tan 77 = 4.33 PTS: 1 DIF: 1 REF: 1bc349d6-4683-11df-9c7d-001185f0d2ea OBJ: 10-1.3 Calculating Trigonometric Ratios NAT: NT.CCSS.MTH.10.K-12.5.1 TOP: 10-1 Trigonometric Ratios KEY: trigonometric ratio | trigonometry | cosine | sine | tangent DOK: DOK 2 7. ANS: sin 49 = 0.75, cos 50 = 0.64, tan 45 = 1 Make sure your calculator is in degree mode. sin 49 = 0.75, cos 50 = 0.64, tan 45 = 1 PTS: 1 DIF: 1 REF: 1bc349d6-4683-11df-9c7d-001185f0d2ea OBJ: 10-1.3 Calculating Trigonometric Ratios NAT: NT.CCSS.MTH.10.K-12.5.1 TOP: 10-1 Trigonometric Ratios KEY: trigonometric ratio | trigonometry | cosine | sine | tangent DOK: DOK 2 8. ANS: GH = 22.46 in. GH is the length of the hypotenuse of the triangle. You are given FH, which is adjacent to . Since the adjacent side and hypotenuse are involved, use the cosine ratio. Write a trigonometric ratio. Substitute the given values. Multiply both sides by GH and divide by cos 35 . in Simplify the expression. PTS: 1 DIF: 2 REF: 1bc58522-4683-11df-9c7d-001185f0d2ea OBJ: 10-1.4 Using Trigonometric Ratios to Find Lengths STA: MCC9-12.G.SRT.8 LOC: MTH.C.14.02.03.001 TOP: 10-1 Trigonometric Ratios KEY: trigonometric ratio | trigonometry | side length DOK: DOK 2 9. ANS: 10.82 feet B 2.8 ft A C Write a trigonometric ratio. Substitute the given values. Multiply both sides by AB and divide by sin 15 . feet PTS: OBJ: STA: TOP: DOK: 10. ANS: 1 DIF: 2 REF: 1bc7e77e-4683-11df-9c7d-001185f0d2ea 10-1.5 Problem-Solving Application NAT: NT.CCSS.MTH.10.9-12.G.SRT.8 MCC9-12.G.SRT.8 LOC: MTH.C.14.02.03.001 10-1 Trigonometric Ratios KEY: trigonometric ratio | trigonometry | side length DOK 2 45 ; cos A = 53 28 sin B = ; cos B = 53 sin A = Simplify the expression. 28 53 45 53 PTS: OBJ: STA: KEY: 11. ANS: 1 DIF: 1 REF: 91632d45-6ab2-11e0-9c90-001185f0d2ea 10-1-Ext.1 Finding the Sine and Cosine of Acute Angles NAT: NT.CCSS.MTH.10.9-12.G.SRT.7 MCC9-12.G.SRT.7 TOP: 10-1-Ext Trigonometric Ratios and Complementary Angles right triangle trigonometry | sine | cosine | tangent DOK: DOK 2 119 ; cos A = 169 120 sin B = ; cos B = 169 sin A = 120 169 119 169 PTS: 1 DIF: 1 REF: 91632d45-6ab2-11e0-9c90-001185f0d2ea OBJ: 10-1-Ext.1 Finding the Sine and Cosine of Acute Angles NAT: NT.CCSS.MTH.10.9-12.G.SRT.7 STA: MCC9-12.G.SRT.7 TOP: 10-1-Ext Trigonometric Ratios and Complementary Angles KEY: right triangle trigonometry | sine | cosine | tangent DOK: DOK 2 12. ANS: sin 74° PTS: 1 DIF: 2 REF: 91635455-6ab2-11e0-9c90-001185f0d2ea OBJ: 10-1-Ext.2 Writing Sine in Cosine Terms and Cosine in Sine Terms NAT: NT.CCSS.MTH.10.9-12.G.SRT.7 STA: MCC9-12.G.SRT.7 TOP: 10-1-Ext Trigonometric Ratios and Complementary Angles KEY: right triangle trigonometry | cosine | sine DOK: DOK 1 13. ANS: cos 16° PTS: OBJ: NAT: TOP: KEY: 14. ANS: 2 1 DIF: 2 REF: 91635455-6ab2-11e0-9c90-001185f0d2ea 10-1-Ext.2 Writing Sine in Cosine Terms and Cosine in Sine Terms NT.CCSS.MTH.10.9-12.G.SRT.7 STA: MCC9-12.G.SRT.7 10-1-Ext Trigonometric Ratios and Complementary Angles right triangle trigonometry | cosine | sine DOK: DOK 1 Sine is the ratio of the opposite leg to the hypotenuse. 1.2 is the length of the leg opposite 1.3 is the length of the hypotenuse. 0.5 is the length of the leg adjacent 1.3 is the length of the hypotenuse. Since PTS: OBJ: LOC: KEY: 15. ANS: , 2 is . . A. 1 DIF: 2 REF: 1bc80e8e-4683-11df-9c7d-001185f0d2ea 10-2.1 Identifying Angles from Trigonometric Ratios STA: MCC9-12.G.SRT.8 MTH.C.14.02.03.002 | MTH.C.14.02.001 TOP: 10-2 Solving Right Triangles trigonometric ratio | trigonometry DOK: DOK 2 = 44.4°, = 72.5°, = 88.5° Change your calculator to degree mode. Use the inverse trigonometric functions on your calculator to find each angle measure. PTS: OBJ: NAT: LOC: TOP: DOK: 16. ANS: 1 DIF: 1 REF: 1bca49da-4683-11df-9c7d-001185f0d2ea 10-2.2 Calculating Angle Measures from Trigonometric Ratios NT.CCSS.MTH.10.9-12.F.TF.7 STA: MCC9-12.G.SRT.8 MTH.C.14.04.01.002 | MTH.C.14.04.02.002 | MTH.C.14.04.03.002 10-2 Solving Right Triangles KEY: trigonometric ratio | trigonometry | inverse trigonometric ratio DOK 1 = 0.45 By the Pythagorean Theorem, . PTS: 1 DIF: 2 REF: 1bccac36-4683-11df-9c7d-001185f0d2ea OBJ: 10-2.3 Solving Right Triangles STA: MCC9-12.G.SRT.8 LOC: MTH.C.14.02.02.002 TOP: 10-2 Solving Right Triangles KEY: trigonometric ratio | trigonometry | solve right triangles DOK: DOK 2 17. ANS: Angles of elevation: 1, 3 Angles of depression: 2, 4 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. PTS: OBJ: LOC: KEY: 18. ANS: 402 m 1 DIF: 1 REF: 1bd170ee-4683-11df-9c7d-001185f0d2ea 10-3.1 Classifying Angles of Elevation and Depression STA: MCC9-12.G.SRT.8 MTH.C.11.02.04.10.001 | MTH.C.11.02.04.10.002 TOP: 10-3 Angles of Elevation and Depression angle of elevation | angle of depression | trigonometry DOK: DOK 1 B 146.2 m 20º A x Use the side opposite the tangent ratio. and x, and the side adjacent to Multiply both sides by x and divide both sides by to write . Simplify. PTS: 1 DIF: 2 REF: 1bd197fe-4683-11df-9c7d-001185f0d2ea OBJ: 10-3.2 Finding Distance by Using Angle of Elevation NAT: NT.CCSS.MTH.10.9-12.G.SRT.8 STA: MCC9-12.G.SRT.8 LOC: MTH.C.14.02.03.001 TOP: 10-3 Angles of Elevation and Depression KEY: angle of elevation | angle of depression | trigonometry DOK: DOK 2 19. ANS: 1,120 ft R 15º 300 ft 15º S x By the Alternate Interior Angles Theorem, m . From the sketch, . So PTS: 1 DIF: 2 REF: 1bd3d34a-4683-11df-9c7d-001185f0d2ea OBJ: 10-3.3 Finding Distance by Using Angle of Depression NAT: NT.CCSS.MTH.10.9-12.G.SRT.8 STA: MCC9-12.G.SRT.8 LOC: MTH.C.14.02.03.001 TOP: 10-3 Angles of Elevation and Depression KEY: angle of elevation | angle of depression | trigonometry DOK: DOK 2 20. ANS: 1.2 km A 23° 1.8 km D 23° C B Step 1 Draw a sketch. Let B and C represent the beginning and end of the runway. Let CB be the length of the runway. Step 2 Find . By the Alternate Interior Angles Theorem, m In , So Step 3 Find . By the Alternate Interior Angles Theorem, m In , Step 4 Find . . So . So the runway is about 1.2 km long. PTS: 1 DIF: 2 REF: 1bd3fa5a-4683-11df-9c7d-001185f0d2ea . OBJ: STA: TOP: KEY: 10-3.4 Application NAT: NT.CCSS.MTH.10.9-12.G.SRT.8 MCC9-12.G.SRT.8 LOC: MTH.C.14.02.03.001 10-3 Angles of Elevation and Depression angle of elevation | angle of depression | trigonometry DOK: DOK 2