ABCD is a rhombus in which $\angle \text{ADB}={25}^{\circ }$. Then, the value of (2 $\angle \text{BAD}$ − $\angle \text{ABC}$) is :

The correct answer is 210°.

Step-by-step Explanation:

Step 1: Understand the properties of rhombus ABCD.
In a rhombus ABCD:
1. Opposite sides are parallel (AB || CD and AD || BC).
2. All sides are equal in length (AB = BC = CD = DA).
3. The diagonals bisect the angles of the rhombus. Therefore, the diagonal BD bisects $\angle \text{ADC}$.
4. Adjacent angles are supplementary, which means:

$\angle \text{BAD}+\angle \text{ADC}={180}^{\circ }$

$\angle \text{BAD}+\angle \text{ABC}={180}^{\circ }$

Step 2: Find $\angle \text{ADC}$.
Since the diagonal BD bisects $\angle \text{ADC}$, we have:

$\angle \text{ADC}=2\times \angle \text{ADB}$

$\angle \text{ADC}=2\times {25}^{\circ }={50}^{\circ }$

Step 3: Find $\angle \text{BAD}$ and $\angle \text{ABC}$.
Since adjacent angles of a rhombus are supplementary:

$\angle \text{BAD}={180}^{\circ }-\angle \text{ADC}={180}^{\circ }-{50}^{\circ }={130}^{\circ }$

$\angle \text{ABC}=\angle \text{ADC}={50}^{\circ }$

Step 4: Calculate the required expression (2 $\angle \text{BAD}$ − $\angle \text{ABC}$).
Substitute the values of $\angle \text{BAD}$ and $\angle \text{ABC}$ into the expression:

$2\angle \text{BAD}-\angle \text{ABC}=2\left({130}^{\circ }\right)-{50}^{\circ }$

$={260}^{\circ }-{50}^{\circ }={210}^{\circ }$