Question

Please help me with this right away. Please show the work

Answer 1

Look at the picture.

Definitions:

$\sin\theta=\dfrac{y}{r}\\\\\cos\theta=\dfrac{x}{r}\\\\\tan\theta=\dfrac{y}{x}\\\\\cot\theta=\dfrac{x}{y}$

We have:

$\tan\theta=-\dfrac{2}{3}$ and Quadrant II (x < 0, y > 0)

Therefore x=-3 and y = 2. Calculate r:

$r=\sqrt{(-3)^2+2^2}=\sqrt{9+4}=\sqrt{13}$

Substitute to the formula of cosine:

$\cos\theta=\dfrac{-3}{\sqrt{13}}=-\dfrac{3}{\sqrt{13}}\cdot\dfrac{\sqrt{13}}{\sqrt{13}}=-\dfrac{3\sqrt{13}}{13}$

Answer: a.