Question

Question 1 of 7, Step 1 of 1 0/7 Correct 1 Solve for the remaining angles and side of the two triangles that can be created. Round to the nearest hundredth: B = 30°, b = 4, a = 6 Answer How to enter your answer Keypad Keyboard Shortcuts Triangle 1: (where angle A is acute): Triangle 2: (where angle A is obtuse): A = | 0 0 A= C= C= C=

Answer 1

Sulution:- Given B = 390 b=4, a=6 using sine law b. sinA sin B sine c sinc a b sinA sinA sin B sin B b. + sina - sinB A = sin) b in! (simes) T A = sinta sin 30 6 6 4 LA - 48.59 we know LA ELB ELLE 1800 - LC= 180°-LA-LB = 1800 - 48.59 - 30 LC 101.410 Again using sine law to find c. b. C sinc. b sinB sinc 11 с sin B T c = 4 · sin 101.41 sin 30 T 7.84 Triangle 1 (where angle A is acute) A = 48.599,5 = 101.410 c / = 7.84

we know that sin (180 - 0) = sino sin ( 180 - 48.59° ) sio 131.41 0 LA= 131:41 As we know LA + LB +L(= 180° + LC=180°- LA-LB L(= 180'e 131.410 - 300 LC 0 18.59 Again using sine law b sins since sinc c = b sin B sinc c = 4 .sin 18.59 SIN 30 2.55 Triangle 2 (where angle A is obtuse ) 131.410 C = 18.599 A C = 2.55 Answer :- A is acute A is obtuse A - 48.59° Ą = 131.40 C 101.419 c = 18.59 C 7.84 C - 2.55