Question

Write the complex number in the form a + bi. 5) Vocos 315º + i sin 3159) 5) Perform the indicated operation. Write the answer in the form a +bi. 8(cos i sin 6) 3(cos + i sin 7) Draw a picture and label all the parts. Solve the triangle for the 3 missing parts. 7) a-8.3 b - 13.8 c=15.7 Also ...Give me the area:

Answer 1

Z & Xt 71: V Coso 2 X= a 6 54 6 Cos 315'+ i sin 315) krig (or at ib [v=1To 10:315) X= Cos 3150 yars r sino r: 13 ilovs a. 13 ya Vī sín 315 y 13 (@r) bs-3 13 - 113 (which is in complex.form] b> 8 (costo 3 (cos o tišino 7-8,0-8 8 l cos ti sia ) => a = 8 Cos (0) = 0 b=rsino = ..; Z- T ti sin TT av Coso 8 Sin (3) 8

s(cost tisini 0+118) 0+81 173,03 3 (Cos D tisin El 6 m). -4 a = 3 Cosmo a tib= 3 + 13 3/3 TT 6 = 3 sin onlor ܪ ܀ i 4r - t 3 T 6 8( cos tisins 10+i8 ) 3 (cos y tisini 3[3 +i) 8( cos + i sin ) 3 (Cost isin los + horsi G

label all the parts. Solve 3 missing parts Draw а picture and the triangle for the 17) a = 8.3 13.8 с 15,7 b b= 13.8 a=8.3 from cosine Rule B C-157: a²= 6²+c²-2bc cosA measurements are not for full scale. cosA a²-16²+(²) -abc LA cos a ²-16²+c2 - abc .: LA = cosi! 19.3)2 – ((15-8)*+(15.772) = 31.86 -(23 (138)(15.7) = 31.859 O

LA = 31.86° from Sine Rule Sinct SinB b a LB = Ab Sin sin(31-86) x13.8 61.3569 LB 61.36° SinLA Since a с LC -sic.co si (sin tal to 150 .86.8144 LC =86.819 LA = 31.86° LB = 61.36° LC = 86.81°

Area of the triangle = JS(8-a)(s-bXS - ?) where 8 = atbtc 2. .:Si 8.3 +13.8+15.7 +15:1 189 10 is = 18.9 .. Area of the triangle is. A = 18.9 (1819-8-3)(18.9 - 13:8) (18-9-15-7) = 57:1798852 Area of the 571799 Sq units treangle