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Construct a \(\triangle{ABC}\) in which QR = 6 cm, \(\angle{Q}\) = \(60^\circ\) and PR - PQ = 2cm.
Answer :
Given: In \(\triangle{ABC}\) in which QR = 6 cm, \(\angle{Q}\) = \(60^\circ\) and PR - PQ = 2cm.
Steps of construction: 
1. Draw the base QR = 6 cm.
2. At the point Q make an \(\angle{XQR}\) = \(60^\circ\).
3. Cut line segment QS = PR - PQ (= 2 cm) from the line QX extended on opposite side of line segment QR.
4. Let LM intersect QX at P.
5. Join PR.
Hence, ABC is the required triangle.