Question

Find the value of cot (theta) for an angle (theta) in standard position with a terminal ray that passes through the point (-9,-5).

a. 5/9

b.5/(sqrt 106)

c. 9/5

d. 9/(sqrt 106)

Answer 1

$P(x,\ y),\ r=\sqrt{x^2+y^2}\\\\\cos\theta=\dfrac{x}{r}$

$(-9,\ -5)\to x=-9,\ y=-5\\\\r=\sqrt{(-9)^2+(-5)^2}=\sqrt{81+25}=\sqrt{106}\\\\\cos\theta=\dfrac{-9}{\sqrt{106}}$