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1/2/2017 In this chapter, ΔABC has vertices A, B, and C, and the sides opposite those vertices are a, b, and c, respectively. Law of Sines: a b c sin A sin B sin C This law can be used to find all angles and sides of an oblique (not right) triangle You must know 3 angle or side measures to use this law, and 2 of the knowns must be an opposite side/angle pair C CH. 6 – ADDITIONAL TOPICS IN TRIGONOMETRY Recall: Angles in a Δ sum to 180°! A C 102.3° Ex: Solve the triangle. 27.4 ft To find A, subtract angles from 180°! a 180 – 102.3 – 28.7 = 49°! 28.7° To find c, use Law of Sines! A 180 – 42 – 21.41 = 116.59°! A 75° c Cross-multiply to get… 31sin 75 A sin 1 11 ERR: DOMAIN! Your calculator is telling you that there is no angle measure for A that will create a triangle. Why? Because in reality, 11 inches is not long enough to reach down to the 3rd side. c 29.40 in To get the best answer, try not to round until the end! C SPECIAL CASE #2 31 in 11 in a 43.06 ft Can’t find anything but A right now! 11sin A 31sin 75 A 21.41 To find c, use Law of Sines! 22 c 22sin116.59 c sin 42 sin116.59 sin 42 C To find A, use Law of Sines! 11 31 sin 75 sin A B c 27.4sin 49 a sin 28.7 Ex: Solve the triangle. c 55.75 ft SPECIAL CASE #1 42° A 12sin 42 A sin 1 22 To find C, subtract angles from 180°! To find a, use Law of Sines! 27.4 a sin 28.7 sin 49 Cross-multiply to get… 12 in 22 in 22sin A 12sin 42 Hint: Use the given info instead of determined info when possible in case you made a mistake! Hint: Don’t evaluate or round until you have the answer! Hint: We’re in degrees! 27.4sin102.3 c sin 28.7 Can’t find anything but A right now! To find A, use Law of Sines! 22 12 sin 42 sin A B c 27.4 c sin 28.7 sin102.3 Ex: Solve the triangle. B c C a b c sin A sin B sin C a b 6.1 – Law of Sines B Ex: Solve the triangle. To find A, use Law of Sines! 29 46 sin 31 sin A Cross-multiply to get… 46 in 29 in A 31° c B 46sin 31 A sin 1 29sin A 46sin 31 A 54.78 29 BUT WAIT! You just remembered that there is a 2 nd quadrant angle with the same sine, mainly 180-A = 125.22°! Therefore, there are 2 separate triangles that exist with the given measurements! You must solve both triangles! Final answer: NO SOLUTION! 1 1/2/2017 C SPECIAL CASE #2 (CONT’D) Triangle #1: A = 54.78° 29 in To find C, subtract the angles from 180! 46 in 29 c sin 31 sin 94.21 c 29sin 94.21 c sin 31 c 56.15 in Triangle #2: A = 125.22° 29 in To find C, subtract the angles from 180! 31° A To find c, use Law of Sines! 180 – 31 – 54.78 = 94.21°! C SPECIAL CASE #2 (CONT’D) B 180 – 31 – 125.22 = 23.78°! To find c, use Law of Sines! 29 c sin 31 sin 23.78 A 29sin 23.78 c sin 31 31° c B c 22.71 in When do we know if there will be 2 triangles formed? 1. 2. 46 in Only in the ASS condition Only when the first angle found is greater than the given angle Basically, consider a 2nd angle whenever you do sin-1! AREA OF A TRIANGLE 1 1 1 Area ab sin C ac sin B bc sin A 2 2 2 C a b A c B 2