(a) Given any \(\Delta\) ABC, Required: That < A + < B + < C = 180^\circ. Construction : Draw CP // BA and produce BC to Q. Proof: Using the above figure, \(a_{1} = a_{2}\) (alternate angles) \(b_{1} = b_{2}\) (corresponding angles) \(a_{1} + b_{1} + c = 180^\circ\) (angles on a straight line) \(\therefore a_{2} + b_{2} + c = 180^\circ\) \(< A + < B + < C = 180^\circ\) (proved) (b) < LNP = 46^\circ + 80^\circ = 126^\circ (exterior angles of triangle LMN) \(\theta = \frac{126^\circ}{2} = 63^\circ\) (bisector divides into 2 equal halves) \(< MLN = 180^\circ - 80^\circ = 100^\circ\) (angles on a straight line) \(< LQN = 180^\circ - (63^\circ + 100^\circ) = 17^\circ\)