In a `DeltaABC`, If `/_A + B = 135^@ `and `/_C + 2 /_B = 180^@`, then the correct relation is :

Options

A. `CA gt AB`
B. `CA = AB`
C. `CA lt AB`
D. `CA + AB = CB `

Answer

To solve the problem, we need to find the angles of triangle ABC based on the given conditions:

1. **Given Conditions**:
- Angle A + Angle B = 135°
- Angle C + 2 * Angle B = 180°

2. **Step 1: Express Angle C in terms of Angle B**:
From the second condition, we can express Angle C as:
\[ \text{Angle C} = 180° - 2 \times \text{Angle B} \]

3. **Step 2: Substitute Angle C into the triangle angle sum property**:
We know that the sum of angles in a triangle is 180°:
\[ \text{Angle A} + \text{Angle B} + \text{Angle C} = 180° \]
Substituting the expression for Angle C:
\[ \text{Angle A} + \text{Angle B} + (180° - 2 \times \text{Angle B}) = 180° \]

4. **Step 3: Simplify the equation**:
Simplifying the equation gives:
\[ \text{Angle A} + \text{Angle B} + 180° - 2 \times \text{Angle B} = 180° \]
\[ \text{Angle A} - \text{Angle B} = 0 \]
Thus,
\[ \text{Angle A} = \text{Angle B} \]

5. **Step 4: Use the first condition to find the angles**:
Since we know that Angle A + Angle B = 135°, we can substitute Angle A with Angle B:
\[ \text{Angle B} + \text{Angle B} = 135° \]
\[ 2 \times \text{Angle B} = 135° \]
\[ \text{Angle B} = \frac{135°}{2} = 67.5° \]

6. **Step 5: Find Angle A**:
Since Angle A = Angle B:
\[ \text{Angle A} = 67.5° \]

7. **Step 6: Find Angle C**:
Now, substituting Angle B back to find Angle C:
\[ \text{Angle C} = 180° - 2 \times 67.5° = 180° - 135° = 45° \]

8. **Final Angles**:
- Angle A = 67.5°
- Angle B = 67.5°
- Angle C = 45°

9. **Step 7: Determine the relationship between the sides**:
Since Angle A = Angle B, the sides opposite to these angles (AC and BC) are equal. Therefore, we have:
\[ AC = BC \]
Since Angle C is less than Angle A and Angle B, the side opposite to Angle C (AB) is less than the sides opposite to Angle A and Angle B:
\[ AB < AC \quad \text{and} \quad AB < BC \]

Thus, the correct relation is:
\[ AB < AC \quad \text{or} \quad AB < BC \]