Question

What is 6[cos(70°) i sin(70°)] ÷ 2[cos(5°) i sin(5°)]? 3[cos(75°) i sin(75°)] 4[cos(75°) i sin(75°)] 3[cos(65°) i sin(65°)] 4[cos(65°) i sin(65°)].

Answer 1

Answer:c.

Explanation:

Answer 2

Explanation:

If we have

`(r( \cos(x) + i \sin(x) )/(s( \cos(v) + i \: sin(w)) = (r)/(s) ( \cos(x - v) + i \: sin(x - w)`

So first we know the modulus is 6 and 2 for the second one so we get

`(6)/(2) = 3`

First degree is 70 , and second is 5 so our degree will be 65 in the second because

70-5=65

Our answer is

`3( \cos(65) + i \: \sin(65) )`