In a `DeltaABC, bar(AB)^2 + bar(AC)^2 = bar(BC)^2` and `bar(BC) = sqrt(2) bar(AB)` , then `/_ABC `is :

Options

A. A)`30^@`
B. B)`45^@`
C. C)`60^@`
D. D)`90^@`

Answer

To solve the problem, we need to analyze the given information about triangle ABC.

### Step-by-Step Solution:

1. **Understanding the Given Information:**
We are given that in triangle ABC, \( AB^2 + AC^2 = BC^2 \) and \( BC = \sqrt{2} \cdot AB \).

2. **Identifying the Type of Triangle:**
The equation \( AB^2 + AC^2 = BC^2 \) indicates that triangle ABC is a right triangle (by the Pythagorean theorem). The right angle is at vertex A.

3. **Substituting the Lengths:**
Let's denote \( AB = x \). Then, from the second equation, we have:
\[ BC = \sqrt{2} \cdot x \]

4. **Using the Pythagorean Theorem:**
Since \( AB^2 + AC^2 = BC^2 \), substituting the lengths we have:
\[ x^2 + AC^2 = (\sqrt{2} \cdot x)^2 \]
This simplifies to:
\[ x^2 + AC^2 = 2x^2 \]

5. **Solving for AC:**
Rearranging the equation gives:
\[ AC^2 = 2x^2 - x^2 = x^2 \]
Taking the square root of both sides, we find:
\[ AC = x \]

6. **Conclusion about the Triangle:**
Since \( AB = AC = x \), triangle ABC is an isosceles right triangle (with angles 45°-45°-90°).

7. **Finding Angle ABC:**
In an isosceles right triangle, the angles opposite the equal sides are equal. Therefore:
\[ \angle ABC = \angle ACB = 45^\circ \]
And since the angles in a triangle sum up to 180°, we have:
\[ \angle A + \angle ABC + \angle ACB = 180^\circ \]
\[ 90^\circ + 45^\circ + 45^\circ = 180^\circ \]

8. **Final Answer:**
Therefore, the angle \( \angle ABC \) is \( 45^\circ \).