Question

Help with trigonometry​

Answer 1

Answer:

○B. $\frac{\sqrt{6}}{2}$

Step-by-step explanation:

Trigonometric Identities

$\frac{y}{r} = sin\:θ$

$\frac{x}{r} = cos\:θ$

$\frac{y}{x} = tan\:θ$

$\frac{r}{x} = sec\:θ$

$\frac{r}{y} = csc\:θ$

$\frac{x}{y} = cot\:θ$

Radius Formula

${y}^{2} + {x}^{2} = {r}^{2}$

${[-\sqrt{15}]}^{2} + {[-\sqrt{10}]}^{2} = {r}^{2} → 15 + 10 = {r}^{2} → 25 = {r}^{2}\\ \\ 5 = r$

* Since we are talking about radii, we only want the NON-NEGATIVE root.

In this case, we will not be using the radius in our ratio, according to the trigonometric identity above because we are using the tangent ratio:

$\frac{\sqrt{6}}{2} = \frac{-\sqrt{15}}{-\sqrt{10}} = \sqrt{1\frac{1}{2}}$

I am joyous to assist you anytime.