Precalculus - need help thanks!

Question

If cos(alpha) = negative square root of 3/4 and (alpha) is in the third quadrant, find the exact value for tan(alpha)

Answer 1

CosA = -sqr(3/4) = -(1/2)sqr3 = - sqr3/2

sqr3 over 2 is the 30-60-90 triangle, with A=30 degrees = pi/6 radians

But in the 3rd quadrant A=30 + 180 = 210 degrees = pi/6 + pi = 7pi/6

tan 210 = tan30 = tan pi/6 = 1/sqr3 = (sqr3)/3

which equals approximately 1.732/3 = 0.577

square root of 3 over 3 is as exact as it gets