In `DeltaABC`, if `sin A + sin B + sin C= 1 + sqrt2 and cos A+cos B+cosC =sqrt2` then the triangle is

Options

A. equilateral
B. isosceles
C. right angled
D. right angle isosceles

Explanation

To determine the type of triangle \( \Delta ABC \) given the conditions \( \sin A + \sin B + \sin C = 1 + \sqrt{2} \) and \( \cos A + \cos B + \cos C = \sqrt{2} \), we can analyze the given equations step by step.

### Step 1: Analyze the given conditions
We have two equations:
1. \( \sin A + \sin B + \sin C = 1 + \sqrt{2} \)
2. \( \cos A + \cos B + \cos C = \sqrt{2} \)

### Step 2: Consider the angles of the triangle
Since \( A + B + C = 180^\circ \), we can use the properties of sine and cosine for specific angles to check for possible triangle types.

### Step 3: Check if the triangle is equilateral
If the triangle is equilateral, then \( A = B = C = 60^\circ \).
- Calculate \( \sin A + \sin B + \sin C \):
\[ \sin 60^\circ + \sin 60^\circ + \sin 60^\circ = 3 \cdot \frac{\sqrt{3}}{2} = \frac{3\sqrt{3}}{2} \approx 2.598 \]
- This does not equal \( 1 + \sqrt{2} \) (approximately 2.414), so it is not equilateral.

### Step 4: Check if the triangle is isosceles
Assume \( A = B \) and \( C \) is different. Let’s try \( A = B = 30^\circ \) and \( C = 120^\circ \).
- Calculate \( \sin A + \sin B + \sin C \):
\[ \sin 30^\circ + \sin 30^\circ + \sin 120^\circ = \frac{1}{2} + \frac{1}{2} + \frac{\sqrt{3}}{2} = 1 + \frac{\sqrt{3}}{2} \approx 1.866 \]
- This does not equal \( 1 + \sqrt{2} \), so it is not isosceles.

### Step 5: Check if the triangle is right-angled
Assume \( A = 90^\circ \), \( B = 30^\circ \), and \( C = 60^\circ \).
- Calculate \( \sin A + \sin B + \sin C \):
\[ \sin 90^\circ + \sin 30^\circ + \sin 60^\circ = 1 + \frac{1}{2} + \frac{\sqrt{3}}{2} = 1 + 1 = 2 \]
- This does not equal \( 1 + \sqrt{2} \), so it is not a right triangle.

### Step 6: Check if the triangle is right-angled isosceles
Assume \( A = 90^\circ \), \( B = 45^\circ \), and \( C = 45^\circ \).
- Calculate \( \sin A + \sin B + \sin C \):
\[ \sin 90^\circ + \sin 45^\circ + \sin 45^\circ = 1 + \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} = 1 + \sqrt{2} \]
- This matches the first condition.

- Now check \( \cos A + \cos B + \cos C \):
\[ \cos 90^\circ + \cos 45^\circ + \cos 45^\circ = 0 + \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} = \sqrt{2} \]
- This matches the second condition.

### Conclusion
Since both conditions are satisfied with the angles \( A = 90^\circ \), \( B = 45^\circ \), and \( C = 45^\circ \), we conclude that the triangle is a **right-angled isosceles triangle**.