Question

In Fig., AB || CD and a transversal PQ cuts them at L and M respectively. If ∠QMD = 100°, find all other angles.

Answer 1

In the given figure, AB || CD, PQ is a transversal line and $\angle$QMD = 100°.
Thus, we have:
$\angle$DMQ + $\angle$QMC = 180° (Linear pair)
$\therefore \angle \mathrm{QMC}=180°-\angle \mathrm{DMQ}=180°-100°=80°$
Thus,
$\angle$DMQ = $\angle$BLM = 100° (Corresponding angles)
$\angle$DMQ = $\angle$CML = 100° (Vertically opposite angles)
$\angle$BLM = $\angle$PLA = 100° (Vertically opposite angles)
Also,
$\angle$CMQ = $\angle$ALM = 80° (Corresponding angles)
$\angle$CMQ = $\angle$DML = 80° (Vertically opposite angles)
$\angle$ALM = $\angle$PLB = 80° (Vertically opposite angles)