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CH. 6 – ADDITIONAL TOPICS IN TRIGONOMETRY 6.1 – Law of Sines In this chapter, ΔABC has vertices A, B, and C, and the sides opposite those vertices are a, b, and c, respectively. a b c Law of Sines: 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 Recall: Angles in a Δ sum to 180°! a b A c B C a b c sin A sin B sin 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 B c 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! 27.4 c sin 28.7 sin102.3 27.4sin102.3 c sin 28.7 c 55.75 ft To find a, use Law of Sines! 27.4 a sin 28.7 sin 49 27.4sin 49 a sin 28.7 a 43.06 ft C Ex: Solve the triangle. Can’t find anything but A right now! To find A, use Law of Sines! 22 12 sin 42 sin A Cross-multiply to get… A 12sin 42 A sin 1 22 To find C, subtract angles from 180°! 22sin A 12sin 42 12 in 22 in 42° B c A 21.41 180 – 42 – 21.41 = 116.59°! To find c, use Law of Sines! 22 c 22sin116.59 c sin 42 sin116.59 sin 42 c 29.40 in To get the best answer, try not to round until the end! C SPECIAL CASE #1 Ex: Solve the triangle. Can’t find anything but A right now! To find A, use Law of Sines! 11 31 sin 75 sin A A 75° c Cross-multiply to get… 11sin A 31sin 75 31 in 11 in 31sin 75 A sin 11 1 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. Final answer: NO SOLUTION! B C SPECIAL CASE #2 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 46sin 31 A sin 1 A 54.78 29 BUT WAIT! You just remembered that there is a 2nd 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! 29sin A 46sin31 B C SPECIAL CASE #2 (CONT’D) Triangle #1: A = 54.78° 29 in To find C, subtract the angles from 180! 46 in 180 – 31 – 54.78 = 94.21°! To find c, use Law of Sines! 29 c sin 31 sin 94.21 A 29sin 94.21 c sin 31 31° c c 56.15 in B C SPECIAL CASE #2 (CONT’D) Triangle #2: A = 125.22° 29 in To find C, subtract the angles from 180! 46 in 180 – 31 – 125.22 = 23.78°! To find c, use Law of Sines! 29 c sin 31 sin 23.78 29sin 23.78 c sin 31 c c 22.71 in When do we know if there will be 2 triangles formed? 1. 2. A 31° 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! B 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