1. **Problem Statement:** Given the trigonometric values for angles A, B, C, D, E, and F, find the related angles in degrees and determine the quadrants where these angles lie. Also, find the two possible angles (Angle One and Angle Two) for each trigonometric value. 2. **Formulas and Rules:** - Use inverse trigonometric functions to find the reference angle: $\theta = \cos^{-1}(x)$, $\sin^{-1}(x)$, $\tan^{-1}(x)$, etc. - Quadrants for each function: - $\cos$ is positive in Quadrants I and IV. - $\sin$ is positive in Quadrants I and II. - $\tan$ is positive in Quadrants I and III. - $\cot$ has the same sign as $\tan$. - $\csc$ and $\sec$ have the same sign as $\sin$ and $\cos$ respectively. - Angles in degrees are found by converting radians to degrees: $\text{degrees} = \theta \times \frac{180}{\pi}$. - For each positive value, angles are $\theta$ and $360^\circ - \theta$ or $180^\circ - \theta$ depending on the function. - For negative values, angles are $180^\circ + \theta$ and $360^\circ - \theta$ or similar based on quadrant. 3. **Calculations:** **A. cos A = 0.579** - Reference angle: $\theta = \cos^{-1}(0.579) \approx 54.59^\circ$ - Cosine positive in Quadrants I and IV - Angle One: $54.59^\circ$ - Angle Two: $360^\circ - 54.59^\circ = 305.41^\circ$ - Quadrants: I and IV **B. sin B = -0.971** - Reference angle: $\theta = \sin^{-1}(-0.971) \approx -76.44^\circ$ - Take absolute value for reference: $76.44^\circ$ - Sine negative in Quadrants III and IV - Angle One: $180^\circ + 76.44^\circ = 256.44^\circ$ - Angle Two: $360^\circ - 76.44^\circ = 283.56^\circ$ - Quadrants: III and IV **C. tan C = 2.35** - Reference angle: $\theta = \tan^{-1}(2.35) \approx 67.38^\circ$ - Tangent positive in Quadrants I and III - Angle One: $67.38^\circ$ - Angle Two: $180^\circ + 67.38^\circ = 247.38^\circ$ - Quadrants: I and III **D. cot D = -0.123** - $\cot D = -0.123$ implies $\tan D = -\frac{1}{0.123} \approx -8.13$ - Reference angle: $\theta = \tan^{-1}(8.13) \approx 82.96^\circ$ - Tangent negative in Quadrants II and IV - Angle One: $180^\circ - 82.96^\circ = 97.04^\circ$ - Angle Two: $360^\circ - 82.96^\circ = 277.04^\circ$ - Quadrants: II and IV **E. csc E = \frac{11}{3} \approx 3.6667$** - $\sin E = \frac{1}{csc E} = \frac{3}{11} \approx 0.2727$ - Reference angle: $\theta = \sin^{-1}(0.2727) \approx 15.82^\circ$ - Sine positive in Quadrants I and II - Angle One: $15.82^\circ$ - Angle Two: $180^\circ - 15.82^\circ = 164.18^\circ$ - Quadrants: I and II **F. sec F = -\frac{19}{5} = -3.8$** - $\cos F = \frac{1}{sec F} = -\frac{5}{19} \approx -0.2632$ - Reference angle: $\theta = \cos^{-1}(-0.2632) \approx 105.25^\circ$ - Cosine negative in Quadrants II and III - Angle One: $105.25^\circ$ - Angle Two: $360^\circ - 105.25^\circ = 254.75^\circ$ - Quadrants: II and III 4. **Summary Table:** | Question | Related Angle (degrees) | Quadrants | Angle One (degrees) | Angle Two (degrees) | |---|---|---|---|---| | cos A = 0.579 | 54.59 | I, IV | 54.59 | 305.41 | | sin B = -0.971 | 76.44 | III, IV | 256.44 | 283.56 | | tan C = 2.35 | 67.38 | I, III | 67.38 | 247.38 | | cot D = -0.123 | 82.96 | II, IV | 97.04 | 277.04 | | csc E = 11/3 | 15.82 | I, II | 15.82 | 164.18 | | sec F = -19/5 | 105.25 | II, III | 105.25 | 254.75 | This completes the solution for all given trigonometric values.