15. angle f has a reference angle of 75°. if the terminal side of ∠f is in quadrant iii, what is a possible measure of ∠f?
16. the angle shown is in standard position and the terminal side passes through the point (-4,2).
a. find the sine value of the angle.
b. find the cosine value of the angle.
c. find the tangent value of the angle.
17. find the exact values of each of the following:
a. cos270°
b. tan150°
c. sin45°
18. for what values of θ, where 0° ≤ θ ≤ 360°, is cosθ = 0? explain.

Question

15. angle f has a reference angle of 75°. if the terminal side of ∠f is in quadrant iii, what is a possible measure of ∠f?
16. the angle shown is in standard position and the terminal side passes through the point (-4,2).
   a. find the sine value of the angle.
   b. find the cosine value of the angle.
   c. find the tangent value of the angle.
17. find the exact values of each of the following:
   a. cos270°
   b. tan150°
   c. sin45°
18. for what values of θ, where 0° ≤ θ ≤ 360°, is cosθ = 0? explain.

Accepted Answer

### Question 15
# Explanation:
## Step1: Recall reference angle in Quadrant III
In Quadrant III, the measure of an angle $ \theta $ is given by $ \theta = 180^\circ + \text{reference angle} $.
## Step2: Calculate the angle
The reference angle is $ 75^\circ $, so $ \angle F = 180^\circ + 75^\circ = 255^\circ $.
# Answer:
$ 255^\circ $

### Question 16
#### Part a: Sine value
# Explanation:
## Step1: Find the radius $ r $
For a point $ (x, y) $, $ r = \sqrt{x^2 + y^2} $. Here, $ x = -4 $, $ y = 2 $, so $ r = \sqrt{(-4)^2 + 2^2} = \sqrt{16 + 4} = \sqrt{20} = 2\sqrt{5} $.
## Step2: Use sine formula $ \sin\theta = \frac{y}{r} $
Substitute $ y = 2 $ and $ r = 2\sqrt{5} $, so $ \sin\theta = \frac{2}{2\sqrt{5}} = \frac{1}{\sqrt{5}} = \frac{\sqrt{5}}{5} $.
# Answer:
$ \frac{\sqrt{5}}{5} $

#### Part b: Cosine value
# Explanation:
## Step1: Use cosine formula $ \cos\theta = \frac{x}{r} $
Substitute $ x = -4 $ and $ r = 2\sqrt{5} $, so $ \cos\theta = \frac{-4}{2\sqrt{5}} = \frac{-2}{\sqrt{5}} = \frac{-2\sqrt{5}}{5} $.
# Answer:
$ \frac{-2\sqrt{5}}{5} $

#### Part c: Tangent value
# Explanation:
## Step1: Use tangent formula $ \tan\theta = \frac{y}{x} $
Substitute $ y = 2 $ and $ x = -4 $, so $ \tan\theta = \frac{2}{-4} = -\frac{1}{2} $.
# Answer:
$ -\frac{1}{2} $

### Question 17
#### Part a: $ \cos 270^\circ $
# Explanation:
## Step1: Recall the unit circle
$ 270^\circ $ is on the negative y - axis, and $ \cos\theta = x $-coordinate. At $ 270^\circ $, $ x = 0 $, so $ \cos 270^\circ = 0 $.
# Answer:
$ 0 $

#### Part b: $ \tan 150^\circ $
# Explanation:
## Step1: Find reference angle
$ 150^\circ $ is in Quadrant II, reference angle $ = 180^\circ - 150^\circ = 30^\circ $.
## Step2: Use tangent formula
$ \tan 150^\circ = -\tan 30^\circ $ (since tangent is negative in Quadrant II), and $ \tan 30^\circ = \frac{1}{\sqrt{3}} $, so $ \tan 150^\circ = -\frac{1}{\sqrt{3}} = -\frac{\sqrt{3}}{3} $.
# Answer:
$ -\frac{\sqrt{3}}{3} $

#### Part c: $ \sin 45^\circ $
# Explanation:
## Step1: Recall special angle values
For $ 45^\circ $, $ \sin 45^\circ = \frac{\sqrt{2}}{2} $.
# Answer:
$ \frac{\sqrt{2}}{2} $

### Question 18
# Explanation:
## Step1: Recall the unit circle
$ \cos\theta = 0 $ when the angle is on the y - axis (since $ \cos\theta = x $-coordinate, and on y - axis $ x = 0 $).
## Step2: Find angles in $ 0^\circ \leq \theta \leq 360^\circ $
The angles are $ 90^\circ $ and $ 270^\circ $ because at $ 90^\circ $ (positive y - axis) and $ 270^\circ $ (negative y - axis), the x - coordinate (cosine) is 0.
# Answer:
$ \theta = 90^\circ $ and $ \theta = 270^\circ $

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QUESTION IMAGE

Question

Explanation:

Question 15

Step1: Recall reference angle in Quadrant III

Step2: Calculate the angle

Step1: Find the radius \( r \)

Step2: Use sine formula \( \sin\theta = \frac{y}{r} \)

Step1: Use cosine formula \( \cos\theta = \frac{x}{r} \)

Answer:

Question 16

Part a: Sine value