Q.2 Construct \(\triangle \)PQR if PQ = 5 cm, m?PQR = 105° and \(\angle \)QRP = 40°.

(Hint: Recall angle-sum property of a triangle)

(Hint: Recall angle-sum property of a triangle)

Given: \(\angle PQR = 40°, \angle QRP = 105°\)

\(\therefore \angle PQR + \angle QRP + \angle QPR = 180°\) (Angle sum property of a triangle)

105° + 40° +\( \angle \)QPR = 180°

145° + \(\angle \)QPR = 180°

\(\therefore \angle QPR = 180° – 145° = 35°\)

Step 1. Draw a line segment PQ = 5 cm.

Step 2. At point P, draw an angle \(\angle \) P = 35°

Step 3 At point Q draw an angle \(\angle \) Q = 105°

Step 4. Now the two rays intersect at the point R.

Then, \(\triangle \)PQR is the required triangle.