Find the fifth roots of 243(cos 300° + i sin 300°).

In looking for the fifth roots, we will use De Moivre's theorem.
The formula for this problem is z^5 = 243 (cos 300 degress + i sin 300 degrees)

Where you'll also need the following data:
300/5 = 60
360/5 = 72
- you'll input these two after the cos and i sin (60+k*72) where k = 0,1,2,3,4

solution:

z^5 = 243 (cos 300 degrees + i sin 300 degrees)
z= 243^1/5 (cos 300 degrees + i sin 300 degrees)
z= 3 (cos (60 + k*72) degrees) + (i sin (60 + k*72) degrees)

so the following are the roots:

3 (cos 60 degrees + i sin 60 degrees)
3 (cos 132 degrees + i sin 132 degrees)
3 (cos 204 degrees + i sin 204 degrees)
3 (cos 276 degrees + i sin 276 degrees)
3 (cos 348 degrees + i sin 348 degrees)

Intersecting secant-tangent theorem states that if a tangent segment and a secant segment are drawn to a circle from an exterior point, then the square of the measure of the tangent segment is equal to the product of the measures of the secant segment and its external secant segment.so to break it down: 1. pq acts as the tangent and ps as the secant with an intersection at p.sr = 21rp = 3x + 3pq = 4x + 4following to equation: tangent squared = the two segments multiplied by each otherso (4x + 4) ^ 2 = (3x + 3) • (21) simplifies to 16x^2 - 31x - 47 = 0 or (16x - 47)(x + 1) x = -1 or x = 47/162. pq acts as the tangent and ps as the secant with an intersection at p (again! )sr = 16rp = xpq = 15(15) ^ 2 = 16 • x225 = 16xx = 225/163. the interesting chords are proportional to each other so a ratio is possible to set up: ap dp 10 3 + x = = —— = pc pb 8 xcross multiple to get 10x = (3 + x)(8)x = 12

You can find it by multiplying the areas it show u