Question

In Fig., AB and CD are parallel lines intersected by a transversal PQ at L and M respectively. If ∠CMQ = 60°, find all other angles in the figure.

Answer 1

$\angle$ALM = $\angle$CMQ = $60°$ (Corresponding angles)
$\angle$LMD = $\angle$CMQ = $60°$ (Vertically opposite angles)
$\angle$ALM = $\angle$PLB = $60°$ (Vertically opposite angles)
Since
$\angle$CMQ + $\angle$QMD = $180°$ (Linear pair)
$\therefore \angle$QMD = $180°-60°=120°$
$\angle$QMD = $\angle$MLB = $120°$ (Corresponding angles)
$\angle$QMD = $\angle$CML = $120°$ (Vertically opposite angles)
$\angle$MLB = $\angle$ALP = $120°$ (Vertically opposite angles)