To determine in which of the given cases a triangle ABC can exist, we will analyze each option based on the sine rule and the conditions provided. The sine rule states that:
\[
\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}
\]
This implies that \( b \sin A = a \sin B \).
Let's evaluate each option step by step:
### Option (a): \( b \sin A = a, A < \frac{\pi}{2} \)
1. Given \( b \sin A = a \) and \( A < \frac{\pi}{2} \).
2. Since \( \sin A \) is positive and less than 1 for \( A < \frac{\pi}{2} \), we can conclude that \( b \) must be greater than \( a \) for the equation to hold.
3. Therefore, a triangle can exist in this case.
**Conclusion**: **Possible**.
### Option (b): \( b \sin A > a, A > \frac{\pi}{2} \)
1. Given \( b \sin A > a \) and \( A > \frac{\pi}{2} \).
2. For \( A > \frac{\pi}{2} \), \( \sin A \) is still positive but greater than 1, which means \( b \sin A \) could potentially exceed \( a \).
3. However, this condition does not guarantee that a triangle can exist since \( A \) being obtuse does not ensure the existence of sides \( a \) and \( b \) that satisfy the triangle inequality.
**Conclusion**: **Not Possible**.
### Option (c): \( b \sin A > a, A < \frac{\pi}{2} \)
1. Given \( b \sin A > a \) and \( A < \frac{\pi}{2} \).
2. Here, since \( \sin A \) is positive and less than 1, \( b \) must be sufficiently larger than \( a \) for the inequality to hold.
3. Therefore, this condition can also allow for the existence of a triangle.
**Conclusion**: **Possible**.
### Option (d): \( b \sin A < a, A < \frac{\pi}{2}, b > a \)
1. Given \( b \sin A < a \) and \( A < \frac{\pi}{2} \), with \( b > a \).
2. Since \( \sin A \) is positive and less than 1, \( b \sin A \) can be less than \( a \) if \( b \) is not significantly larger than \( a \).
3. Thus, this condition can also allow for the existence of a triangle.
**Conclusion**: **Possible**.
### Option (e): \( b \sin A < a, A > \frac{\pi}{2}, b = a \)
1. Given \( b \sin A < a \) and \( A > \frac{\pi}{2} \) with \( b = a \).
2. Here, \( b \sin A \) would be equal to \( a \sin A \) for \( b = a \), and since \( \sin A \) is positive but less than 1, \( b \sin A \) cannot be less than \( a \).
3. Therefore, this condition does not allow for the existence of a triangle.
**Conclusion**: **Not Possible**.
### Final Summary of Options:
- **(a)** Possible
- **(b)** Not Possible
- **(c)** Possible
- **(d)** Possible
- **(e)** Not Possible
### Correct Answers:
The cases in which a triangle ABC can exist are **(a), (c), and (d)**.
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