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 start with the given equation in triangle \( ABC \):

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

### Step 1: Rearranging the Equation
We can rearrange the equation to isolate terms involving \( a^2 \):

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

This simplifies to:

\[ a^2 (\sin B - 1) = b^2 + c^2 - a^2 \]

### Step 2: Using the Cosine Rule
From the cosine rule, we know that:

\[ a^2 = b^2 + c^2 - 2bc \cos A \]

We can substitute this into our rearranged equation:

\[ a^2 (\sin B - 1) = b^2 + c^2 - (b^2 + c^2 - 2bc \cos A) \]

This simplifies to:

\[ a^2 (\sin B - 1) = 2bc \cos A \]

### Step 3: Analyzing the Equation
Now, we have:

\[ a^2 (\sin B - 1) = 2bc \cos A \]

### Step 4: Considering the Values of Sine
Since \( \sin B \) can only take values between -1 and 1, we can analyze the implications:

- If \( \sin B = 1 \), then \( B = 90^\circ \) and \( a^2 (1 - 1) = 0 \), which is valid.
- If \( \sin B < 1 \), then \( \sin B - 1 < 0 \), which implies \( a^2 (\sin B - 1) < 0 \).

### Step 5: Conclusion About Angle A
Since \( 2bc \cos A \) must also be less than or equal to zero, we conclude that:

- If \( \cos A < 0 \), then \( A \) must be an obtuse angle (i.e., \( 90^\circ < A < 180^\circ \)).

Thus, we can conclude that:

**The angle \( A \) is obtuse.**

### Final Answer
The correct option is that angle \( A \) is obtuse.

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