Question

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Solve the equation on the interval 10,360"). (3 each) 1) 5 sin 0.1-sino 2) 2 cos20.3 cos 0 1 - 0 3) 4 tan2o. 7 tan 0-2 - 0 4] 3 sin x + sin = 0 Solve the right triangle shown in the figure. Round lengths to one decimal place and express angles to the nearest tenth of a degree. (+5) B с 5) B-35, b 572

Answer 1

1) 5 Sino +1 = 3 Sine OM, 2 Sino =-1 or, Sino Sino --Y2 The general solution of sing is, 7 TT +2975, O TITT +2771 6 6 77T +2017, then The solution is o +2719 6 u 2) 2 cos? 0 + 3COSO + 1 = 0 Let caso 202 + 30+1 = 0 On, 242 +24+4 + 1 = 0 Or, 2u ca+1) + 1 (u+1)=0 on, (24+1) Cu+1) = 0 ... u=-1/ caso = -1/2 +2071, 0 - 475 +2379 275 3

U = -1 Cos o = -1 :0 = 7T+277 The solution iso-2" +2771, 47T + 2TT), MT + 271 ni 3) 4 tan? 0 + 7 tano - 2 = 0 Let tanou : 4u? +70-2=0 -btſ b²_4ac i u = 2a -7† 172-4.4-2) 2.4 -7+81 -7 + 9 :. u= 8 8 4 iu -781 8 -7-9 8 -2 u = 4 tano = 4 The general solution of tano is ::8 = 'tan 67)+7 u= -2 tano =-2 r. 0 = tan! (-2) +77 n .". The solution is 8- tani' (4)+7m, tan' (-2)+7in CAno)

4) 3 Sim? x + Sinx = 0 let Sinx = le :. 3 u²+ U=0 on, u(34+1) = 0 U = 0, 34+1 = 0 ... - i. U = 3 ::ų = 0 OM, Sin Cx) = 0 The general solution of Sinx is, x = 27t n X= 7T +2717 u 3 - ㅎ Sinx= :: x = Sin' (-43) + 2710 The poolution X=27TN, 7T+27n, sin '613) +27Tn. 3 B Зго C-99.7 a = 81: 7 900 C А b=57.2 B = 35° , b= 57:2 we know that formula law of sines, a b Sim A Sin B Sin C b Sonne SinB 57.2 Sin 350 С Sinc С Sin 90° = 99.7 Es Sin 900 x 572 Sin 350 C A = 180°- 90°-350 = 55

а b Sin A Sin B а 51.2 π Sin (559) Sim 35° -> а 57.2 x Sin (559) Sin 35° =81:7 a = 81-7 LA = 550 < c = 90°, c=99. CAns)