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8-5

8-5. Law of Sines and Law of Cosines. Warm Up. Lesson Presentation. Lesson Quiz. Holt McDougal Geometry. Holt Geometry. Warm Up 1. What is the third angle measure in a triangle with angles measuring 65° and 43°? Find each value. Round trigonometric

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8-5

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  1. 8-5 Law of Sines and Law of Cosines Warm Up Lesson Presentation Lesson Quiz Holt McDougal Geometry Holt Geometry

  2. Warm Up 1. What is the third angle measure in a triangle with angles measuring 65° and 43°? Find each value. Round trigonometric ratios to the nearest hundredth and angle measures to the nearest degree. 2. sin 73° 3. cos 18° 4. tan 82° 5. sin-1 (0.34) 6. cos-1 (0.63) 7. tan-1 (2.75) 72° 0.96 0.95 7.12 20° 51° 70°

  3. Objective Use the Law of Sines and the Law of Cosines to solve triangles.

  4. In this lesson, you will learn to solve any triangle. To do so, you will need to calculate trigonometric ratios for angle measures up to 180°. You can use a calculator to find these values.

  5. Example 1: Finding Trigonometric Ratios for Obtuse Angles Use your calculator to find each trigonometric ratio. Round to the nearest hundredth. A. tan 103° B. cos 165° C. sin 93° tan 103°  –4.33 cos 165°  –0.97 sin 93°  1.00

  6. In ∆ABC, let h represent the length of the altitude from C to From the diagram, , and By solving for h, you find that h = b sin A and h = a sin B. So b sin A = a sin B, and . You can use another altitude to show that these ratios equal You can use the altitude of a triangle to find a relationship between the triangle’s side lengths.

  7. You can use the Law of Sines to solve a triangle if you are given • two angle measures and any side length (ASA or AAS) or • two side lengths and a non-included angle measure (SSA).

  8. Example 2A: Using the Law of Sines Find the measure. Round lengths to the nearest tenth and angle measures to the nearest degree. FG Law of Sines Substitute the given values. FG sin 39° = 40 sin 32° Cross Products Property Divide both sides by sin 39.

  9. Example 2B: Using the Law of Sines Find the measure. Round lengths to the nearest tenth and angle measures to the nearest degree. mQ Law of Sines Substitute the given values. Multiply both sides by 6. Use the inverse sine function to find mQ.

  10. The Law of Sines cannot be used to solve every triangle. If you know two side lengths and the included angle measure or if you know all three side lengths, you cannot use the Law of Sines. Instead, you can apply the Law of Cosines.

  11. You can use the Law of Cosines to solve a triangle if you are given • two side lengths and the included angle measure (SAS) or • three side lengths (SSS).

  12. Helpful Hint The angle referenced in the Law of Cosines is across the equal sign from its corresponding side.

  13. Example 3A: Using the Law of Cosines Find the measure. Round lengths to the nearest tenth and angle measures to the nearest degree. XZ XZ2 = XY2 + YZ2 – 2(XY)(YZ)cos Y Law of Cosines Substitute the given values. = 352 + 302 – 2(35)(30)cos 110° XZ2 2843.2423 Simplify. Find the square root of both sides. XZ 53.3

  14. Example 3B: Using the Law of Cosines Find the measure. Round lengths to the nearest tenth and angle measures to the nearest degree. mT RS2 = RT2 + ST2 – 2(RT)(ST)cos T Law of Cosines Substitute the given values. 72 = 132 + 112 – 2(13)(11)cos T 49 = 290 – 286 cosT Simplify. Subtract 290 both sides. –241 = –286 cosT

  15. Example 3B Continued Find the measure. Round lengths to the nearest tenth and angle measures to the nearest degree. mT –241 = –286 cosT Solve for cosT. Use the inverse cosine function to find mT.

  16. Helpful Hint Do not round your answer until the final step of the computation. If a problem has multiple steps, store the calculated answers to each part in your calculator.

  17. A sailing club has planned a triangular racecourse, as shown in the diagram. How long is the leg of the race along BC? How many degrees must competitors turn at point C? Round the length to the nearest tenth and the angle measure to the nearest degree. Example 4: Sailing Application

  18. Example 4 Continued Step 1 Find BC. BC2 = AB2 + AC2 – 2(AB)(AC)cos A Law of Cosines Substitute the given values. = 3.92 + 3.12 – 2(3.9)(3.1)cos 45° Simplify. BC2 7.7222 Find the square root of both sides. BC 2.8 mi

  19. Example 4 Continued Step 2 Find the measure of the angle through which competitors must turn. This is mC. Law of Sines Substitute the given values. Multiply both sides by 3.9. Use the inverse sine function to find mC.

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