`DeltaABC` , `angleA+angleB=65^(@)`, `angleB+angleC=140^(@)`,then `angleB` is equal to

Answer

To solve for angle B in triangle ABC given the equations:

1. \( \angle A + \angle B = 65^\circ \)
2. \( \angle B + \angle C = 140^\circ \)

We can follow these steps:

### Step 1: Use the angle sum property of a triangle.
The sum of the angles in a triangle is always \( 180^\circ \). Therefore, we can write:

\[ \angle A + \angle B + \angle C = 180^\circ \]

### Step 2: Substitute the known values.
From the first equation, we can express angle A in terms of angle B:

\[ \angle A = 65^\circ - \angle B \]

From the second equation, we can express angle C in terms of angle B:

\[ \angle C = 140^\circ - \angle B \]

### Step 3: Substitute these expressions into the angle sum property.
Now we substitute the expressions for angle A and angle C into the angle sum property:

\[ (65^\circ - \angle B) + \angle B + (140^\circ - \angle B) = 180^\circ \]

### Step 4: Simplify the equation.
Combine like terms:

\[ 65^\circ + 140^\circ - \angle B = 180^\circ \]

This simplifies to:

\[ 205^\circ - \angle B = 180^\circ \]

### Step 5: Solve for angle B.
Now, isolate angle B:

\[ 205^\circ - 180^\circ = \angle B \]

\[ \angle B = 25^\circ \]

### Conclusion:
Thus, the value of angle B is \( 25^\circ \).

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