Question

Solve the triangle with the given parts. 25) 27 51° 260 A) y = 103°, a = 21.5, b = 12.1 C) y = 97º, a = 11.9, b = 21.1 B) Y = 103°, a = 60, b = 33.9 D) y = 103°, a = 12.1, b = 21.5 Solve the triangle with the given information. 26) B = 63.5° a = 12.20 C=7.80 A) b = 12.2, a = 75.1º, y = 41.4° C) b = 11.2, a = 77.6°, y = 38.9° B) b = 13.2, a=73.1°, Y= 37.49 D) No solution Find the area of the triangle. If necessary, round the answer to two decimal places. 27) 8 6 B a 5 A) 14.98 B) 4.86 C) 44.74 D) 195.03

Answer 1

28 TA a=3 ✓ 6-? C А B Let ABC be the tringle whose vertices are A, B and and corresponding angles les are <d, kB and <r. let, AB= 6=27, Be=a=? AG=6=? Given that, <d: 26', B=51° We know the sum of three angle tringle le is. 180° 180°-( 517269 103° any tringle the Sine cau b C SinA o SinB Sinc . le of any. vo For :: 9 - 103 gives Sin A Sinc * a= 27x Sin 260 Nowy sin 103 Again = 12:147 = 12.1 Cone doceimal b . places SinB sine = b = 27 Sin 51 = 21.5348 En 103 Honce, V-1039 a=[12.1 = 21.5 Cone decimal b=215 places) Ans: . D. V=130; a = 12:1, 6221159 А 6:7.8 b=? AB=6350 3 a 212-2 C Bea a = 12.2 ret Let Abe te the tringle whose vertices are wespectively A, B,C and corresponding angles are <a, TB and AB ses 7.8, aw AC=6=2 also, <BE <= 83.50 For any tring te the cosine law gires after 6² qac 62= ata?dacasa (2.29 767.89 = 2X12:2x7.8.636 3:55 124.7596 6 = [124.7596 = 11:1695 = 1102 cone decimay places) 1.6 -111.22 as B =>

2 di da Again as a = bite? ar 2 be (11.2) 7 (7.8)?~(12.232 -0-21 4285 2x11.2.7.8 as $'(0.21 0-214285) = 77.6263 Again, las y arebne2 2a6 *635) = (12-2) 7(11-2)=(7.872 +38-99 2x 12:2x112 ::y 38.99 - 77.6° 7 [v=180° (77.67 Ans: @ C. b-1102, 0= 71.6, V= 38.99 .C 27) (6²6 ma B c=5 Perimeter of the tringle (6+578) = 19 unit is semi perimeter 8- 12=9.5 Area = √5(S-a) (5-6) (s.cJ unit? 10.5) *(1.5)*(3.5)*(4:5) 14.9812 = 14.98 (two decimal Places? 2 Ans: A 14.98 / whit?