Find the value of cos(θ) for an angle θ in standard position with a terminal Ray that passes through the point (-5,-12)

D

Step-by-step explanation:

Let's draw a picture (see attachment).

We know that since our point is (-5, -12), it must pass in the third quadrant. Now, if we drop a perpendicular line from the point (-5, -12) to the x-axis, we have a right triangle. Notice that we already know the two leg lengths of this: -5 and -12.

Cosine is adjacent / hypotenuse. Here, the adjacent side to the angle θ is -5. We need to find the hypotenuse, which we can do so by using the Pythagorean Theorem:

c² = a² + b²

c² = (-5)² + (-12)²

c² = 25 + 144 = 169

c = 13

So, the hypotenuse is 13. Now, plug these values in:

cos(θ) = adjacent / hypotenuse = -5/13

The answer is D.

d. -5/13

Step-by-step explanation:

Hypotenuse = sqrt(5² + 12²) = 13

|cos(theta)| = 5/13

Since the angle is in quadrant 3, cos will be negative

Therefore, cos(theta) = -5/13

sss (side, side, side)

sas (side, angle, side)

asa (angle, side, angle)

aas (angle, angle, side)

hl (hypotenuse, leg)