In a quadrilateral `ABCD angleA+angleC` is 2 times `angleB+angleD`. If `angleA=140^(@)angle D=60^(@)`, then `angleB`=

Options

A. `60^(@)`
B. `80^(@)`
C. `120^(@)`
D. None of these

Answer

To solve the problem step by step, we will use the information given in the question.

### Step 1: Write down the equations based on the information provided.
We know that:
1. \( \angle A + \angle C = 2(\angle B + \angle D) \)
2. The sum of angles in a quadrilateral is \( 360^\circ \): \( \angle A + \angle B + \angle C + \angle D = 360^\circ \)

### Step 2: Substitute the known values into the equations.
Given:
- \( \angle A = 140^\circ \)
- \( \angle D = 60^\circ \)

Substituting these values into the second equation:
\[ 140^\circ + \angle B + \angle C + 60^\circ = 360^\circ \]
This simplifies to:
\[ \angle B + \angle C + 200^\circ = 360^\circ \]
Subtracting \( 200^\circ \) from both sides gives:
\[ \angle B + \angle C = 160^\circ \quad \text{(Equation 1)} \]

### Step 3: Substitute into the first equation.
Now substitute \( \angle A \) and \( \angle D \) into the first equation:
\[ 140^\circ + \angle C = 2(\angle B + 60^\circ) \]
This simplifies to:
\[ 140^\circ + \angle C = 2\angle B + 120^\circ \]
Rearranging gives:
\[ \angle C = 2\angle B + 120^\circ - 140^\circ \]
Thus:
\[ \angle C = 2\angle B - 20^\circ \quad \text{(Equation 2)} \]

### Step 4: Solve the equations simultaneously.
Now we have two equations:
1. \( \angle B + \angle C = 160^\circ \) (Equation 1)
2. \( \angle C = 2\angle B - 20^\circ \) (Equation 2)

Substituting Equation 2 into Equation 1:
\[ \angle B + (2\angle B - 20^\circ) = 160^\circ \]
This simplifies to:
\[ 3\angle B - 20^\circ = 160^\circ \]
Adding \( 20^\circ \) to both sides gives:
\[ 3\angle B = 180^\circ \]
Dividing by 3:
\[ \angle B = 60^\circ \]

### Step 5: Conclusion
Thus, the value of \( \angle B \) is \( 60^\circ \).