If `sintheta=1/2, cosphi=1/3` then `theta+phi` belongs to, where `0

Options

A. `(pi/3, pi/2)`
B. `(pi/2, (2pi)/3)`
C. `(2pi/3, (5pi)/6)`
D. `((5pi)/6,pi)`

Answer

To solve the problem, we need to find the interval in which \( \theta + \phi \) belongs given \( \sin \theta = \frac{1}{2} \) and \( \cos \phi = \frac{1}{3} \) with the constraints \( 0 < \theta, \phi < \frac{\pi}{2} \).

### Step-by-Step Solution:

1. **Finding \( \theta \)**:
- We know that \( \sin \theta = \frac{1}{2} \).
- The angle \( \theta \) that satisfies this equation in the interval \( (0, \frac{\pi}{2}) \) is:
\[ \theta = \frac{\pi}{6} \]

2. **Finding \( \phi \)**:
- We have \( \cos \phi = \frac{1}{3} \).
- To find \( \phi \), we can use the inverse cosine function:
\[ \phi = \cos^{-1}\left(\frac{1}{3}\right) \]
- We know that \( \phi \) must lie between \( 0 \) and \( \frac{\pi}{2} \) since \( \cos \phi \) is positive.

3. **Determining the range for \( \phi \)**:
- Since \( \cos \phi = \frac{1}{3} \) is less than \( \cos \frac{\pi}{3} = \frac{1}{2} \), we can conclude:
\[ \phi > \frac{\pi}{3} \]
- Additionally, since \( \phi < \frac{\pi}{2} \), we have:
\[ \frac{\pi}{3} < \phi < \frac{\pi}{2} \]

4. **Finding the range for \( \theta + \phi \)**:
- Now we need to find the range for \( \theta + \phi \):
- The minimum value occurs when \( \phi \) is at its minimum:
\[ \theta + \phi > \frac{\pi}{6} + \frac{\pi}{3} = \frac{\pi}{6} + \frac{2\pi}{6} = \frac{3\pi}{6} = \frac{\pi}{2} \]
- The maximum value occurs when \( \phi \) is at its maximum:
\[ \theta + \phi < \frac{\pi}{6} + \frac{\pi}{2} = \frac{\pi}{6} + \frac{3\pi}{6} = \frac{4\pi}{6} = \frac{2\pi}{3} \]

5. **Final Result**:
- Therefore, the interval for \( \theta + \phi \) is:
\[ \frac{\pi}{2} < \theta + \phi < \frac{2\pi}{3} \]

### Conclusion:
The final answer is that \( \theta + \phi \) belongs to the interval \( \left( \frac{\pi}{2}, \frac{2\pi}{3} \right) \).