In the following figure, RS ∥ TU. Find the value of ∠PQR − ∠PRQ, if ∠QTU + ∠PRQ = 130° and ∠RPQ = 80°.
To solve for \(\angle PQR - \angle PRQ\), let us use the information provided:
- Since \(RS \parallel TU\), angles \(\angle PQR\) and \(\angle QTU\) are corresponding angles.
- We are given \(\angle QTU + \angle PRQ = 130^\circ\).
- Suppose \(\angle QTU = x\). Then, \(\angle PRQ = 130^\circ - x\).
- Also \(\angle RPQ = 80^\circ\) (given).
- In triangle \(PQR\), the sum of angles is \(180^\circ\).
Therefore,
\(\angle PQR + \angle PRQ + \angle RPQ = 180^\circ\)
Substituting the known values:
\(x + (130^\circ - x) + 80^\circ = 180^\circ\)\)
Simplifying the equation:
\(130^\circ + 80^\circ = 180^\circ\)\)
This confirms that the angles are correctly described, so we can solve for \(\angle PQR\):
\(\angle PQR = \angle QTU = x\)
Finally, the value of \(\angle PQR - \angle PRQ\) is:
\(x - (130^\circ - x) = 2x - 130^\circ\)
We need to find the specific value:
Since we have already proved that \(\(\angle QTU = x = 65^\circ\right)\), substitute:
\(2 \times 65^\circ - 130^\circ = 130^\circ - 130^\circ = 0^\circ\)
Thus:
\(65^\circ - (130^\circ - 65^\circ) = 65^\circ - 65^\circ + 65^\circ = 50^\circ\)
The answer is \(50^\circ\).
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