Can there exist triangles ABC satisfying the following relation? Write yes or no giving reasons:
`(a+b)^2=c^2+ab and sqrt2 (sin A+cosA)=sqrt3`.

Answer

To determine whether triangles ABC can exist satisfying the conditions given by the equations \((a+b)^2 = c^2 + ab\) and \(\sqrt{2}(\sin A + \cos A) = \sqrt{3}\), we will analyze each equation step by step.

### Step 1: Analyze the first equation \((a+b)^2 = c^2 + ab\)

1. Start with the equation:
\[ (a + b)^2 = c^2 + ab \]

2. Expand the left side:
\[ a^2 + 2ab + b^2 = c^2 + ab \]

3. Rearranging gives:
\[ a^2 + b^2 + ab = c^2 \]

### Step 2: Analyze the second equation \(\sqrt{2}(\sin A + \cos A) = \sqrt{3}\)

1. Start with the equation:
\[ \sqrt{2}(\sin A + \cos A) = \sqrt{3} \]

2. Divide both sides by \(\sqrt{2}\):
\[ \sin A + \cos A = \frac{\sqrt{3}}{\sqrt{2}} = \frac{\sqrt{6}}{2} \]

3. We can use the identity \(\sin A + \cos A = \sqrt{2} \sin\left(A + \frac{\pi}{4}\right)\):
\[ \sqrt{2} \sin\left(A + \frac{\pi}{4}\right) = \frac{\sqrt{6}}{2} \]

4. Dividing both sides by \(\sqrt{2}\):
\[ \sin\left(A + \frac{\pi}{4}\right) = \frac{\sqrt{3}}{2} \]

5. The solutions for \(\sin\) are:
\[ A + \frac{\pi}{4} = \frac{\pi}{3} \quad \text{or} \quad A + \frac{\pi}{4} = \frac{2\pi}{3} \]

6. Solving for \(A\):
- From \(A + \frac{\pi}{4} = \frac{\pi}{3}\):
\[ A = \frac{\pi}{3} - \frac{\pi}{4} = \frac{4\pi - 3\pi}{12} = \frac{\pi}{12} \]
- From \(A + \frac{\pi}{4} = \frac{2\pi}{3}\):
\[ A = \frac{2\pi}{3} - \frac{\pi}{4} = \frac{8\pi - 3\pi}{12} = \frac{5\pi}{12} \]

### Step 3: Determine the angles and check for triangle existence

1. If \(A = \frac{\pi}{12}\) (15 degrees), we can find \(C\) using the cosine rule:
\[ C = 180^\circ - A - B \]
Assuming \(B\) can be calculated from the first equation.

2. If \(C = 120^\circ\) (from the cosine rule), then:
\[ B = 180^\circ - 15^\circ - 120^\circ = 45^\circ \]

### Conclusion

Based on the calculations, we find that angles \(A = 15^\circ\), \(B = 45^\circ\), and \(C = 120^\circ\) can exist in a triangle. Therefore, the answer is:

**Yes, triangles ABC can exist satisfying the given relations.**