If in a `Delta ABC, angle A = 45^(@) , angle B = 75^(@),` then

Options

A. `a + c sqrt3 = 2b`
B. `a + c = 2b`
C. `a + c sqrt2 = 2b`
D. `a - c = 2b `

Answer

To solve the problem step by step, we will analyze the given angles in triangle ABC and find the necessary relationships.

### Step 1: Identify the Angles
Given:
- Angle A = 45°
- Angle B = 75°

### Step 2: Find Angle C
We know that the sum of angles in a triangle is 180°.
\[ \text{Angle C} = 180° - \text{Angle A} - \text{Angle B} \]
Substituting the values:
\[ \text{Angle C} = 180° - 45° - 75° = 60° \]

### Step 3: Summarize the Angles
Now we have:
- Angle A = 45°
- Angle B = 75°
- Angle C = 60°

### Step 4: Use the Law of Sines
We can apply the Law of Sines, which states:
\[ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} \]
Let’s denote the sides opposite to angles A, B, and C as a, b, and c respectively.

### Step 5: Express the Sides in Terms of One Side
Using the Law of Sines:
\[ \frac{a}{\sin 45°} = \frac{b}{\sin 75°} = \frac{c}{\sin 60°} \]
We can express the sides in terms of side a:
\[ b = a \cdot \frac{\sin 75°}{\sin 45°} \]
\[ c = a \cdot \frac{\sin 60°}{\sin 45°} \]

### Step 6: Calculate the Values of Sines
Using the known values of sine:
- \(\sin 45° = \frac{\sqrt{2}}{2}\)
- \(\sin 60° = \frac{\sqrt{3}}{2}\)
- \(\sin 75° = \sin(45° + 30°) = \sin 45° \cos 30° + \cos 45° \sin 30° = \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2} \cdot \frac{1}{2} = \frac{\sqrt{6} + \sqrt{2}}{4}\)

### Step 7: Substitute the Values
Now substituting these values into the equations for b and c:
\[ b = a \cdot \frac{\frac{\sqrt{6} + \sqrt{2}}{4}}{\frac{\sqrt{2}}{2}} = a \cdot \frac{\sqrt{6} + \sqrt{2}}{2\sqrt{2}} \]
\[ c = a \cdot \frac{\frac{\sqrt{3}}{2}}{\frac{\sqrt{2}}{2}} = a \cdot \frac{\sqrt{3}}{\sqrt{2}} \]

### Step 8: Final Relationships
We can summarize the relationships between the sides:
- \(b = a \cdot \frac{\sqrt{6} + \sqrt{2}}{2\sqrt{2}}\)
- \(c = a \cdot \frac{\sqrt{3}}{\sqrt{2}}\)

### Conclusion
We have found the angles and expressed the sides in terms of one side (a).