1 / 37

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

johnomartin
Download Presentation

8-5

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  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. Check It Out! Example 1 Use a calculator to find each trigonometric ratio. Round to the nearest hundredth. a. tan 175° b. cos 92° c. sin 160° tan 175°  –0.09 cos 92°  –0.03 sin 160°  0.34

  7. 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.

  8. 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).

  9. 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.

  10. 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.

  11. Check It Out! Example 2a Find the measure. Round lengths to the nearest tenth and angle measures to the nearest degree. NP Law of Sines Substitute the given values. NP sin 39° = 22 sin 88° Cross Products Property Divide both sides by sin 39°.

  12. Check It Out! Example 2b Find the measure. Round lengths to the nearest tenth and angle measures to the nearest degree. mL Law of Sines Substitute the given values. Cross Products Property 10 sin L = 6 sin 125° Use the inverse sine function to find mL.

  13. Check It Out! Example 2c Find the measure. Round lengths to the nearest tenth and angle measures to the nearest degree. mX Law of Sines Substitute the given values. Cross Products Property 7.6 sin X = 4.3 sin 50° Use the inverse sine function to find mX.

  14. Check It Out! Example 2d Find the measure. Round lengths to the nearest tenth and angle measures to the nearest degree. AC mA + mB + mC = 180° Prop of ∆. Substitute the given values. mA + 67° + 44° = 180° mA = 69° Simplify.

  15. Check It Out! Example 2D Continued Find the measure. Round lengths to the nearest tenth and angle measures to the nearest degree. Law of Sines Substitute the given values. AC sin 69° = 18 sin 67° Cross Products Property Divide both sides by sin 69°.

  16. 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.

  17. 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).

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

  19. 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

  20. 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

  21. 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.

  22. Check It Out! Example 3a Find the measure. Round lengths to the nearest tenth and angle measures to the nearest degree. DE DE2 = EF2 + DF2 – 2(EF)(DF)cos F Law of Cosines Substitute the given values. = 182 + 162 – 2(18)(16)cos 21° DE2 42.2577 Simplify. Find the square root of both sides. DE 6.5

  23. Check It Out! Example 3b Find the measure. Round lengths to the nearest tenth and angle measures to the nearest degree. mK JL2 = LK2 + KJ2 – 2(LK)(KJ)cos K Law of Cosines Substitute the given values. 82 = 152 + 102 – 2(15)(10)cos K 64 = 325 – 300 cosK Simplify. Subtract 325 both sides. –261 = –300 cosK

  24. Check It Out! Example 3b Continued Find the measure. Round lengths to the nearest tenth and angle measures to the nearest degree. mK –261 = –300 cosK Solve for cosK. Use the inverse cosine function to find mK.

  25. Check It Out! Example 3c Find the measure. Round lengths to the nearest tenth and angle measures to the nearest degree. YZ YZ2 = XY2 + XZ2 – 2(XY)(XZ)cos X Law of Cosines Substitute the given values. = 102 + 42 – 2(10)(4)cos 34° YZ2 49.6770 Simplify. Find the square root of both sides. YZ 7.0

  26. Check It Out! Example 3d Find the measure. Round lengths to the nearest tenth and angle measures to the nearest degree. mR PQ2 = PR2 + RQ2 – 2(PR)(RQ)cos R Law of Cosines Substitute the given values. 9.62 = 5.92 + 10.52 – 2(5.9)(10.5)cos R 92.16 = 145.06 – 123.9cosR Simplify. Subtract 145.06 both sides. –52.9 = –123.9 cosR

  27. Check It Out! Example 3d Continued Find the measure. Round lengths to the nearest tenth and angle measures to the nearest degree. mR –52.9 = –123.9 cosR Solve for cosR. Use the inverse cosine function to find mR.

  28. 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.

  29. 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

  30. 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

  31. 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.

  32. 31 m Check It Out! Example 4 What if…? Another engineer suggested using a cable attached from the top of the tower to a point 31 m from the base. How long would this cable be, and what angle would it make with the ground? Round the length to the nearest tenth and the angle measure to the nearest degree.

  33. Check It Out! Example 4 Continued Step 1 Find the length of the cable. AC2 = AB2 + BC2 – 2(AB)(BC)cos B Law of Cosines Substitute the given values. = 312 + 562 – 2(31)(56)cos 100° Simplify. AC2 4699.9065 Find the square root of both sides. AC68.6 m

  34. Check It Out! Example 4 Continued Step 2 Find the measure of the angle the cable would make with the ground. Law of Sines Substitute the given values. Multiply both sides by 56. Use the inverse sine function to find mA.

  35. Lesson Quiz: Part I Use a calculator to find each trigonometric ratio. Round to the nearest hundredth. 1. tan 154° 2. cos 124° 3. sin 162° –0.49 –0.56 0.31

  36. Lesson Quiz: Part II Use ΔABC for Items 4–6. Round lengths to the nearest tenth and angle measures to the nearest degree. 4. mB = 20°, mC = 31° and b = 210. Find a. 5. a = 16, b = 10, and mC = 110°. Find c. 6.a = 20, b = 15, and c = 8.3. Find mA. 477.2 21.6 115°

  37. Lesson Quiz: Part III 7. An observer in tower A sees a fire 1554 ft away at an angle of depression of 28°. To the nearest foot, how far is the fire from an observer in tower B? To the nearest degree, what is the angle of depression to the fire from tower B? 1212 ft; 37°

More Related