Question

In fig 5.24, $\square$ABCD is a parallelogram. $\angle$ADB = 40$°$, $\angle$ABD = 30$°$, find the measure of $\angle$BAD, $\angle$DCB, $\angle$ADC and $\angle$ABC.

Answer 1

In ΔABD, we have:

$\angle$BAD + $\angle$ADB + $\angle$ABD = 180° ...[Angle addition property of triangle]
$\Rightarrow$ $\angle$BAD + 40° + 30° = 180°
$\Rightarrow$ $\angle$BAD + 70° = 180°
$\Rightarrow$ $\angle$BAD = 180° – 70°
$\Rightarrow$ $\angle$BAD = 110°

In parallelogram ABCD, $\angle$BAD and $\angle$DCB are opposite angles and we know that opposite angles in a parallelogram are congruent to each other.

Thus, we have:
$\angle$DCB = $\angle$BAD = 110°
$\angle$BAD + $\angle$ADC = 180° [In a parallelogram, adjacent angles are supplementary.]
∵ $\angle$BAD = 110°
We have:
110^o + $\angle$ADC = 180°
$\Rightarrow$$\angle$ADC = 180° – 110°
$\Rightarrow$$\angle$ADC = 70°
$\angle$ABC = $\angle$ADC = 70° [Opposite angles of a parallelogram are congruent to each other.]
Thus, we get:
$\angle$BAD = 110°, $\angle$DCB = 110°, $\angle$ADC = 70° and $\angle$ABC = 70°