Find the fifth roots of 243(cos 240° + i sin 240°).

See below.

Step-by-step explanation:

Fifth root of 243 = 3,

Suppose r( cos Ф + i sinФ) is the fifth root of 243(cos 240 + i sin 240),

then r^5( cos  Ф  + i sin  Ф )^5 = 243(cos 240 + i sin 240).

Equating equal parts and using de Moivre's theorem:

r^5 =243  and  cos  5Ф  + i sin  5Ф = cos 240 + i sin 240

r = 3 and  5Ф = 240 +360p so Ф =  48 + 72p

So Ф = 48, 120, 192, 264, 336  for   48 ≤ Ф < 360

So there are 5 distinct solutions given by:

3(cos 48 + i sin 48),

3(cos 120 + i sin 120),

3(cos 192 + i sin 192),

3(cos 264 + i sin 264),

3(cos 336 + i sin 336).. (Answer).

1. 16.64

a^2+b^2=c^2

pr^2=pq^2+qr^2

pr^2=14^2+ 9^2

pr^2=196+81

pr= sqr root of 277

16.64

2. 11.31

c^2-a^2=b^2

18^2- 14^2=b^2

b^2=324-196

b=sqr root of 126

b=11.31

hope it ! (btw im in connections academy to, good luck! )