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Areas for Any Triangle Trigonometric Ratios for Obtuse Angles

9.7: Objective. Areas for Any Triangle Trigonometric Ratios for Obtuse Angles Law of Sines and Law of Cosines for Any Triangles. h. Proof: Draw an altitude from B to side AC. Now, sin A = h/c or c sin A = h and from A Δ = ½(base)(height) A Δ = ½ bc sin A.

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Areas for Any Triangle Trigonometric Ratios for Obtuse Angles

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  1. 9.7: Objective Areas for Any Triangle Trigonometric Ratios for Obtuse Angles Law of Sines and Law of Cosines for Any Triangles

  2. h Proof: Draw an altitude from B to side AC. Now, sin A = h/c or c sin A = h and from AΔ = ½(base)(height) A Δ = ½ bc sin A Note: You need two sides and the included angle

  3. It can be shown with triangle compositions or with a unit circle that the following relationships are true for the trigonometric ratios for sine and cosine for obtuse angles between 90 and 180: The proof for this will be left to a later time… For now, we will use a calculator to find values for the trigonometric ratios of obtuse angles. HONORS HANDOUT ON UNIT CIRCLE

  4. 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

  5. Extra 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

  6. Solving triangles that are not right triangles Use 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)*. (SsA OK, sSA Ambiguous)

  7. Solving triangles that are not right triangles The Law of Sines cannot be used if you know two side lengths and the included angle (SAS) or if you know all three side lengths (SSS), Instead, you can apply the Law of Cosines.

  8. Example 3A: 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 3B: 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. Extra Example 3A 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°.

  11. Extra Example 3B 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.

  12. Example 4A: 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

  13. Example 4B: 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

  14. Example 4B 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.

  15. Extra Example 4A 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

  16. Extra Example 4B 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

  17. Extra Example 4B 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.

  18. 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.

  19. 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 5: Sailing Application

  20. Example 5 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

  21. Example 5 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.

  22. 31 m Example 6 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.

  23. Example 6 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

  24. Example 6 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.

  25. 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

  26. 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°

  27. 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°

  28. 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 Proof for Law of Sines: Use the altitude of a triangle to find a relationship between the triangle’s side lengths.

  29. Proof for The Law of Cosines First we'll divide the triangle into two right triangles by drawing a line that passes through B and is perpendicular to b. B Now we have a right triangle with c as the hypotenuse. And so according to the Pythagorean Theorem C A b

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