.
12.
13.
ID: 2.2-44
Find all values of θ, if θ is in the interval [0,360°) and has the given function value.
cos (θ) = 1
2
A. 150° and 210°
B. 60° and 120°
C. 210° and 330°
D. 60° and 300°
Answer: D. 60° and 300°
ID: 2.3-19
Find a value of in [0 , 90 ] that satisfies the statement. Leave the answer in decimal degrees rounded to seven decimal places, if 
necessary.
θ ° °
sec θ = 2.165
A. 24.7919207°
B. 30.8660444°
C. 62.4906237°
D. 27.5093763°
Answer: C. 62.4906237°
ID: 2.4-11
Solve the right triangle. Round to three decimal places.
a = 3.4 cm, b = 2.5 cm, C = 90°
A. A = 53.673°, B = 36.327°, c = 4.220 cm
B. A = 47.332°, B = 42.668°, c = 5.900 cm
C. A = 49.173°, B = 40.827°, c = 4.220 cm
D. A = 36.327°, B = 53.673°, c = 4.220 cm
Answer: A. A = 53.673°, B = 36.327°, c = 4.220 cm
T1 (DO NOT ASSIGN) https://xlitemprod.pearsoncmg.com/api/v1/print/highered
4 of 8 9/25/23, 9:04 AM
cos(theta) = 1/2 > 0 => theta in Quadrant I or IV
cos(60) = 1/2, so what angle in Quadrant IV has
a reference angle of 60? 
For Quadrant IV, subtract reference angle from 360 => 300

Let's break down the key elements in your text and go through each mathematical problem presented:

### Problem 1:
You are asked to find all values of \(\theta\) in the interval \([0^\circ, 360^\circ)\) such that \(\cos(\theta) = \frac{1}{2}\). The answer provided is D: \(60^\circ\) and \(300^\circ\).

**Explanation**:
- The cosine function is positive in Quadrants I and IV.
- \(\cos(60^\circ) = \frac{1}{2}\), and in Quadrant IV, the angle with the same reference angle of \(60^\circ\) is \(360^\circ - 60^\circ = 300^\circ\).
Thus, the angles are \(60^\circ\) and \(300^\circ\).

### Problem 2:
You're asked to find the value of \(\theta\) in the interval \([0^\circ, 90^\circ]\) such that \(\sec(\theta) = 2.165\), and the answer is C: \(62.4906237^\circ\).

**Explanation**:
- \(\sec(\theta) = 2.165\) means that \(\cos(\theta) = \frac{1}{2.165}\).
- Using a calculator, \(\theta = \cos^{-1}\left(\frac{1}{2.165}\right) \approx 62.4906237^\circ\).

### Problem 3:
This problem involves solving a right triangle where \(a = 3.4 \, \text{cm}\), \(b = 2.5 \, \text{cm}\), and \(\angle C = 90^\circ\). The correct solution is A: \(\angle A = 53.673^\circ\), \(\angle B = 36.327^\circ\), and \(c = 4.220 \, \text{cm}\).

**Explanation**:
- Use the Pythagorean theorem to find the hypotenuse:
  \[
  c = \sqrt{a^2 + b^2} = \sqrt{3.4^2 + 2.5^2} = \sqrt{11.56 + 6.25} = \sqrt{17.81} \approx 4.220 \, \text{cm}.
  \]
- Find angles using trigonometry:
  \[
  \tan(A) = \frac{a}{b} = \frac{3.4}{2.5}, \quad A = \tan^{-1}\left(\frac{3.4}{2.5}\right) \approx 53.673^\circ.
  \]
  Then, \(B = 90^\circ - A = 90^\circ - 53.673^\circ = 36.327^\circ\).

### Extra Steps for Reference Angle in Problem 1:
- **Cosine** is positive in Quadrant I and IV.
- In Quadrant I, \(\theta = 60^\circ\).
- In Quadrant IV, subtract the reference angle from \(360^\circ\): \(360^\circ - 60^\circ = 300^\circ\).

Would you like more details on any of these steps?

### Here are 5 related questions to further your understanding:
1. What are the general solutions for \(\cos(\theta) = \frac{1}{2}\) in radians?
2. How do you find the reference angle for a given trigonometric value?
3. How can you derive the secant function's inverse on a calculator?
4. What is the general rule for solving a right triangle given two sides?
5. How can you find angles using sine or cosine in a non-right triangle?

### Tip:
For trigonometric problems, always remember which quadrants the sine, cosine, and tangent functions are positive in: All Students Take Calculus (ASTC) for Quadrants I, II, III, and IV respectively.

12.13. ID: 2.2-44 Find all values of θ, if θ is in the interval [0,360°) and has the given function value. cos(θ) = 1/2. A. 150° and 210°, B. 60° and 120°, C. 210° and 330°, D. 60° and 300°. Answer: D. 60° and 300°. ID: 2.3-19 Find a value of θ in [0°, 90°] that satisfies the statement sec(θ) = 2.165. A. 24.7919207°, B. 30.8660444°, C. 62.4906237°, D. 27.5093763°. Answer: C. 62.4906237°. ID: 2.4-11 Solve the right triangle. a = 3.4 cm, b = 2.5 cm, C = 90°. A. A = 53.673°, B = 36.327°, c = 4.220 cm, B. A = 47.332°, B = 42.668°, c = 5.900 cm, C. A = 49.173°, B = 40.827°, c = 4.220 cm, D. A = 36.327°, B = 53.673°, c = 4.220 cm. Answer: A. A = 53.673°, B = 36.327°, c = 4.220 cm.

Learn how to solve trigonometric equations like cos(θ) = 1/2 and sec(θ) = 2.165. This solution includes using inverse trigonometric functions and applying the Pythagorean theorem to solve right triangles. Ideal for students in Grades 10-12.

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