PQRS is rhombus with `angle QPS=50^(@)`. Find `anglePQS`.

Answer

To find the angle \( \angle PQS \) in the rhombus \( PQRS \) where \( \angle QPS = 50^\circ \), we can follow these steps:

### Step 1: Understand the properties of a rhombus
In a rhombus, opposite angles are equal, and the diagonals bisect each other at right angles (90 degrees).

### Step 2: Identify the angles
Given that \( \angle QPS = 50^\circ \), we can denote:
- \( \angle QPS = 50^\circ \)
- Since opposite angles are equal in a rhombus, \( \angle PSQ = 50^\circ \) as well.

### Step 3: Use the property of diagonals
Let the diagonals intersect at point \( O \). The diagonals bisect the angles of the rhombus. Thus:
- \( \angle POQ = \frac{1}{2} \angle QPS = \frac{1}{2} \times 50^\circ = 25^\circ \)

### Step 4: Analyze triangle \( POQ \)
In triangle \( POQ \), we know:
- \( \angle POQ = 25^\circ \)
- \( \angle OPQ = \angle PQS \) (since \( PQ = PS \) in a rhombus)

### Step 5: Apply the triangle sum property
The sum of angles in triangle \( POQ \) is:
\[ \angle POQ + \angle OPQ + \angle PQO = 180^\circ \]
Substituting the known values:
\[ 25^\circ + \angle PQS + \angle PQO = 180^\circ \]
Since \( \angle PQO = \angle PQS \), we can denote \( \angle PQS \) as \( x \):
\[ 25^\circ + x + x = 180^\circ \]
This simplifies to:
\[ 25^\circ + 2x = 180^\circ \]

### Step 6: Solve for \( x \)
Subtract \( 25^\circ \) from both sides:
\[ 2x = 180^\circ - 25^\circ \]
\[ 2x = 155^\circ \]
Now divide by 2:
\[ x = \frac{155^\circ}{2} = 77.5^\circ \]

### Conclusion
Thus, \( \angle PQS = 77.5^\circ \).