The terminal ray of ∠A passes through the point (−6,3) .
∠A is drawn in standard position.
What is the value of secA ?
Enter your answer as an exact answer, in simplified form, in the box.
sec A = r / x
r = sqrt [ (-6)^2 + 3^2 ] = sqrt [ 36 + 9 ] = sqrt [45 ] = 3sqrt (5)
So
sec A = 3sqrt (5) / -6 = - sqrt (5) / 2