Question

$\text{In the given figure,}BAD||EF,\angle AEF={55}^{\circ }and\angle ACB={25}^{\circ },find\angle ABC.$

Answer 1

ANSWER:
In the given figure, EF || BD and CE is the transversal.
∴ $\angle CAD=\angle AEF$
(Pair of corresponding angles)
⇒$\angle CAD={55}^{\circ }$
In $△ABC$,
$\angle CAD=\angle ABC+\angle ACB$
(Exterior angle of a triangle is equal to the sum of the two interior opposite angles)
⇒ ${55}^{\circ }=\angle ABC+{25}^{\circ }$
⇒ $\angle ABC=55°-25°={30}^{\circ }$
Thus, the measure of $\angle ABCis{30}^{\circ }$