MCQ
In the given figure, $\angle OAB =110^{\circ}$ and $\angle BCD =130^{\circ}$ then $\angle ABC$ is equal to
- A$50^{\circ}$
- ✓$60^{\circ}$
- C$40^{\circ}$
- D$70^{\circ}$
✓
Answer
Correct option: B.
$60^{\circ}$
$ 60^{\circ}$
In the given figure, $OA \| CD$.
Construction: Extend $OA$ such that it intersects $BC$ at $E$.
Now, $OE \| CD$ and $BC$ is a transversal.
$\therefore \angle AEC =\angle BCD =130^{\circ}\ ($Pair of corresponding angles$)$
Also, $\angle OAB +\angle BAE =180^{\circ} \ ($Linear pair$)$
$\therefore 110^{\circ}+\angle BAE =180^{\circ}$
$\Rightarrow \angle BAE =180^{\circ}-110^{\circ}=70^{\circ}$
In $\triangle ABE$
$\angle AEC =\angle BAE +\angle ABE \ldots \ ($In a triangle, exterior angle is equal to the sum of two opposite interior angles$)$
$\therefore 130^{\circ}=70^{\circ}+x^0$
$\Rightarrow x^0=130^{\circ}-70^{\circ}=60^{\circ}$
Thus, the measure of angle $\angle ABC$ is $60^{\circ}$
In the given figure, $OA \| CD$.
Construction: Extend $OA$ such that it intersects $BC$ at $E$.
Now, $OE \| CD$ and $BC$ is a transversal.
$\therefore \angle AEC =\angle BCD =130^{\circ}\ ($Pair of corresponding angles$)$
Also, $\angle OAB +\angle BAE =180^{\circ} \ ($Linear pair$)$
$\therefore 110^{\circ}+\angle BAE =180^{\circ}$
$\Rightarrow \angle BAE =180^{\circ}-110^{\circ}=70^{\circ}$
In $\triangle ABE$
$\angle AEC =\angle BAE +\angle ABE \ldots \ ($In a triangle, exterior angle is equal to the sum of two opposite interior angles$)$
$\therefore 130^{\circ}=70^{\circ}+x^0$
$\Rightarrow x^0=130^{\circ}-70^{\circ}=60^{\circ}$
Thus, the measure of angle $\angle ABC$ is $60^{\circ}$
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