In a `DeltaABC," if " a^(2)sinB=b^(2)+c^(2)`, then :

Options

A. `angleA` is obtuse
B. `angleA` is acute
C. `angleB` is abtuse
D. `angleA` is right angle

Answer

To solve the problem, we need to analyze the given equation in the context of triangle ABC. The equation provided is:

\[ a^2 \sin B = b^2 + c^2 \]

We will use the cosine rule and properties of triangles to deduce the nature of angle A.

### Step-by-step Solution:

1. **Start with the given equation:**
\[ a^2 \sin B = b^2 + c^2 \]

2. **Rearrange the equation:**
Subtract \( a^2 \) from both sides:
\[ a^2 \sin B - a^2 = b^2 + c^2 - a^2 \]
This can be rewritten as:
\[ a^2 (\sin B - 1) = b^2 + c^2 - a^2 \]

3. **Use the cosine rule:**
According to the cosine rule, we have:
\[ \cos A = \frac{b^2 + c^2 - a^2}{2bc} \]
Therefore, we can express \( b^2 + c^2 - a^2 \) as:
\[ b^2 + c^2 - a^2 = 2bc \cos A \]

4. **Substitute into the equation:**
Substitute this back into our rearranged equation:
\[ a^2 (\sin B - 1) = 2bc \cos A \]

5. **Analyze the signs:**
- Since \( a, b, c \) are lengths of the sides of the triangle, they are positive.
- The term \( \sin B - 1 \) must be negative because \( \sin B \) can only reach a maximum of 1, and it cannot equal 1 unless \( B = 90^\circ \). Thus, \( \sin B - 1 < 0 \).

6. **Conclude about \( \cos A \):**
Since \( a^2 (\sin B - 1) < 0 \) and \( 2bc > 0 \), we have:
\[ \cos A < 0 \]
This implies that angle A must be obtuse (greater than \( 90^\circ \)).

7. **Final conclusion:**
Therefore, we conclude that:
\[ \text{Angle A is obtuse.} \]