If 5cosA+3=0 and 180°<A<360° determine by the means of a diagram:
•tan^2 A
•sin A over cos A
5cosA+3=0 and 180° < A < 360°
cosA = -3/5, from your domain, A is in quad II
construct a right-angled triangle in quad II with a base of 3 and
hypotenuse of 5
x^2 + y^2 = r^2
9 + y^2 = 25
y = ± √16 = ±4
but we are in II, so y = +4
so you have: x = -3, y = 4, r = 5 and you can find any of the 6 trig ratios
tan^2 A = (- 4/3)^2 = 16/9
sinA over cosA
= sinA/cosA
= tanA, by definition
= -4/3