MCQ
In the given figure, $AB \| DC$, $\angle\text{BAD}=90^\circ,\angle\text{CBD}=28^\circ$ and $\angle\text{BCE}=65^\circ.$ Then $\angle\text{ABD}=?$
- A$32^\circ$
- ✓$37^\circ$
- C$43^\circ$
- D$53^\circ$
✓
Answer
Correct option: B.
$37^\circ$
In $\triangle\text{DBC}$
$\angle\text{BCE}=\angle\text{DBC}+\angle\text{BDC}$ (Exterior angle property)
$65^\circ=28^\circ+\angle\text{BDC}$
$\text{BDC}=37^\circ$
As, $AB$ is parallel to $CD$
$\angle\text{ABD}=\angle\text{BDC}=37^\circ$ (Alternate interior angle).
$\angle\text{BCE}=\angle\text{DBC}+\angle\text{BDC}$ (Exterior angle property)
$65^\circ=28^\circ+\angle\text{BDC}$
$\text{BDC}=37^\circ$
As, $AB$ is parallel to $CD$
$\angle\text{ABD}=\angle\text{BDC}=37^\circ$ (Alternate interior angle).
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