University:
Missouri State University
Course:
MTH 103 | Intermediate Algebra
Academic year:

2020
Views:

248

Pages:

2
Author:

Eden Robbins

Pre-Calculus/Trig 3 Law of Sines Worksheet Solve for the missing sides and angles in each triangle. Name Key ***For SSA cases, determine if a)no triangle can be formed, b) only one triangle can be formed, or c)two triangles can be formed. If one or two triangles can be formed, solve for all angle and sides in each triangle. 1. a = 13.7 A = 25.433 deg B = 78 deg bin 26 * 43 deg = 13 * 75i * n deg * 15 deg 63121 iz = (13.7sin 75 deg)/(5in * 25 * 43) (5.725 * 0.43 deg)/13.7 = underline 3.075^ b B = 78 deg a = 13.7 7657 b C • 3 25 43 113.750 C 13.7 59 7657^ 4 3201 35 * 43 ^ 3 c + 31.03 (5in)/13 * (225.43 deg)/13.7 = (5in * 76.57)/c 2. A = 38 deg B = 63 deg c = 15 B = 63 deg c = 15 a 79° A = 38 ^ sigma b C 15-338 a = (15 * 5in * 35 deg)/(5mk * 79 deg) underline sin 19^ 2 15 * sin 35 deg * 6m a = 9.41 b-sin = 15.5763 b = 15.602 k= 15 5 19 ^ a 510^ a (5in * 79 deg)/15 = (sin 43 deg)/(4pi) 3. b = 795.1 c = 775.6 B = 51.85 deg B 7756 775. a 18.05 50 DIL a = 989.3 5in * 51.87 ^ 2 795 1 = 5in C 115 x0 5in * 51.55' * (705 - 1) = (5in * 99 - 0')/C_{n} c =129.9^ prime prime c = 50.1 deg A 7951 775.1 A 4. a = 125 A = 25 deg b = 150 B 30 A C 25- 125 b = 30.5 deg 6149.5 aligned s ln 25^ kappa in f\\ 12s = 15omega 149.5125 C 124 5.5 C 150 ISO 2428 2.n25 125 = = - alpha + 29.1 deg c A A = 25.43 deg (5in * 25 deg)/42.5 = (5in * 5 * 5 deg)/6 c = 28.3 25 ^ 6 Law of Sines. Solve for the missing sides and angles in each triangle. ***For SSA cases, determine if a)no triangle can be formed, b) only one triangle can be formed, or c)two triangles can be formed. If one or two triangles can be formed, solve for all angle and sides in each triangle. 1. b = 41 A = 33 deg B = 29 deg 118 C 41 A 2. a = 8.2 B = 41 deg C = 90 deg alpha = (41 plus/minus m * 23)/(5m2a) 0.5m * 29 ^ 2 = 41 * 5m / (35 ^ 2) alpha = 4lo * 0Le (sin 29 ^ 2)/(4i) = (5m * 33 ^ 2)/(c*omega) B 8.2 C tan 41": 82 b = 7.13 C ט SCATA 3. a = 12 b = 15 A = 55 deg C 12 Sin 49 C c = 10.87 c = 5.2 5m overline 149 z A 15 C ambigious Case 4. b = 15 c = 13 c = 50 deg A 13 679 deg 6 15 50 C amagnons cree 5. n * 50 deg 13 / 5.n 167.9^ Cm 15 sui55: 125 m issins sin 12 no trangle can be formed is * sin 5 * x deg - 15sin s (5105v)/13 = (sin 13)/15 B = 62I ^ 2 155.m50^ 13 = plus/minus m * 5 13 15 1175 12. t ^ 1 50 A 15 B = 180' - 42.2 * 0.1 ^ 3 = 117.9 ^ 31 alpha = 3 * 5t * 6 (Sm * 55 ^ 2)/12 = sin theta i5 55 deg 33 deg 2n ^ 4 (5.0124 ^ 2)/0 5.052/13 41 deg