Pregunta
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3. The meavure of \( \angle P \) in betwons \( O \) and 1309. Determine all pouble values of \( \angle P \) \( \begin{array}{ll}\text { a) } \sin P=0.35 & \text { b) } \sin P=0.51 \\ \text { a } \cos P=-0.92 & \text { d) } \tan P=-0.87\end{array} \)

Ask by Hobbs Higgins. in Canada
Jan 23,2025

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The possible values of angle \( P \) are: - \( P \approx 20.49^\circ \) and \( P \approx 159.51^\circ \) when \( \sin P = 0.35 \) - \( P \approx 30.71^\circ \) and \( P \approx 149.29^\circ \) when \( \sin P = 0.51 \) - \( P \approx 156.93^\circ \) and \( P \approx 203.07^\circ \) when \( \cos P = -0.92 \) - \( P \approx 139.30^\circ \) and \( P \approx 319.30^\circ \) when \( \tan P = -0.87 \)

Solución

Certainly! Let's determine all possible values of angle \( P \) for each given trigonometric equation within the range \( 0^\circ \leq P < 360^\circ \). ### General Approach: For each trigonometric function: - **Sine (\( \sin \theta \))**: Positive in Quadrants I and II. - **Cosine (\( \cos \theta \))**: Positive in Quadrants I and IV. - **Tangent (\( \tan \theta \))**: Positive in Quadrants I and III. We'll find the reference angle using the inverse trigonometric function and then determine the actual angles based on the quadrant(s) where the function is positive or negative. --- ### a) \( \sin P = 0.35 \) **Step 1: Find the reference angle** \[ P_{\text{ref}} = \sin^{-1}(0.35) \approx 20.49^\circ \] **Step 2: Determine the quadrants where sine is positive** Sine is positive in Quadrants I and II. **Step 3: Find all possible angles** \[ P_1 = P_{\text{ref}} \approx 20.49^\circ \] \[ P_2 = 180^\circ - P_{\text{ref}} \approx 159.51^\circ \] **Solutions:** \[ P \approx 20.49^\circ \quad \text{and} \quad P \approx 159.51^\circ \] --- ### b) \( \sin P = 0.51 \) **Step 1: Find the reference angle** \[ P_{\text{ref}} = \sin^{-1}(0.51) \approx 30.71^\circ \] **Step 2: Determine the quadrants where sine is positive** Sine is positive in Quadrants I and II. **Step 3: Find all possible angles** \[ P_1 = P_{\text{ref}} \approx 30.71^\circ \] \[ P_2 = 180^\circ - P_{\text{ref}} \approx 149.29^\circ \] **Solutions:** \[ P \approx 30.71^\circ \quad \text{and} \quad P \approx 149.29^\circ \] --- ### c) \( \cos P = -0.92 \) **Step 1: Find the reference angle** \[ P_{\text{ref}} = \cos^{-1}(0.92) \approx 23.07^\circ \] **Step 2: Determine the quadrants where cosine is negative** Cosine is negative in Quadrants II and III. **Step 3: Find all possible angles** \[ P_1 = 180^\circ - P_{\text{ref}} \approx 156.93^\circ \] \[ P_2 = 180^\circ + P_{\text{ref}} \approx 203.07^\circ \] **Solutions:** \[ P \approx 156.93^\circ \quad \text{and} \quad P \approx 203.07^\circ \] --- ### d) \( \tan P = -0.87 \) **Step 1: Find the reference angle** \[ P_{\text{ref}} = \tan^{-1}(0.87) \approx 40.70^\circ \] **Step 2: Determine the quadrants where tangent is negative** Tangent is negative in Quadrants II and IV. **Step 3: Find all possible angles** \[ P_1 = 180^\circ - P_{\text{ref}} \approx 139.30^\circ \] \[ P_2 = 360^\circ - P_{\text{ref}} \approx 319.30^\circ \] **Solutions:** \[ P \approx 139.30^\circ \quad \text{and} \quad P \approx 319.30^\circ \] --- **Summary of Solutions:** - **a) \( \sin P = 0.35 \)** - \( P \approx 20.49^\circ \) - \( P \approx 159.51^\circ \) - **b) \( \sin P = 0.51 \)** - \( P \approx 30.71^\circ \) - \( P \approx 149.29^\circ \) - **c) \( \cos P = -0.92 \)** - \( P \approx 156.93^\circ \) - \( P \approx 203.07^\circ \) - **d) \( \tan P = -0.87 \)** - \( P \approx 139.30^\circ \) - \( P \approx 319.30^\circ \) **Note:** All angle measurements are rounded to two decimal places.

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The Deep Dive

For the sine values, \( \sin P = 0.35 \) gives us two solutions in one full rotation of a circle (0° to 360°), which are approximately \( 20.5° \) and \( 159.5° \). This means that \( P \) can be in the first quadrant or the second quadrant. For \( \sin P = 0.51 \), the approximate angles are \( 30.4° \) and \( 149.6° \). When it comes to cosine, \( \cos P = -0.92 \) indicates that the angle \( P \) is in the second or third quadrants. The angles give solutions around \( 143.1° \) and \( 216.9° \). Lastly, for \( \tan P = -0.87 \), \( P \) is in the second or fourth quadrant, resulting in angles approximately \( 143.2° \) and \( 323.2° \). So, there you have it! Multiple options to explore those angle measurements!

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