Question

Please help me. I need help

Answer 1

Look at the first picture.

In Quadrant II tangent is negative and cosine is negative too.

We have:

$\tan\theta=-\dfrac{2}{3}=\dfrac{2}{-3}$

therefore we have the point (-3, 2) → x = -3 and y = 2.

Calculate r:

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

Calculate cosine:

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

Your answer is $\boxed{a.\ -\dfrac{3\sqrt{13}}{13}}$

Answer 2

Answer:

  a.  -(3√13)/13

Step-by-step explanation:

The cosine can be found from the tangent by way of the secant.

  tan(θ)² +1 = sec(θ)² = 1/cos(θ)²

Then ...

  cos(θ) = ±1/√(tan(θ)² +1)

The cosine is negative in the second quadrant, so we will choose that sign.

  cos(θ) = -1/√((-2/3)² +1) = -1/√(4/9 +1) = -1/√(13/9)

  cos(θ) = -3/√13 = -(3√13)/13 . . . . . matches your selection A