MCQ 11 Mark
Statement-1 (A): In Fig.(i), if AOB is a diameter and $\angle A D C=120^{\circ}$, then $\angle C A B=30^{\circ}$.
Statement-2 (R): In Fig.(ii), AOB is a diameter and $\angle A D C=120^{\circ}$, then $\angle B A C=30^{\circ}$.
Statement-2 (R): In Fig.(ii), AOB is a diameter and $\angle A D C=120^{\circ}$, then $\angle B A C=30^{\circ}$.
- ✓Statement-1 and Statement-2 are True; Statement-2 is a correct explanation for Statement-1.
- BStatement-1 and Statement-2 are True; Statement-2 is not a correct explanation for Statement-1.
- CStatement-1 is True, Statement-2 is False.
- DStatement-1 is False, Statement-2 is True.
Answer
View full question & answer→Correct option: A.
Statement-1 and Statement-2 are True; Statement-2 is a correct explanation for Statement-1.
(a)
In Fig. (ii), ABCD is a cyclic quadrilateral such that $\angle A D C=120^{\circ}$. Therefore,
$\angle A D C+\angle A B C=180^{\circ} \Rightarrow \angle A BC=180^{\circ}-120^{\circ}=60^{\circ}$
We find that $\angle A C B=90^{\circ}$ (angle in a semi-circle).
Using angle sum property in $\triangle A C B$, we obtain
$\angle A C B+\angle B A C+\angle A B C=180^{\circ} \Rightarrow 90^{\circ}+\angle B AC+60^{\circ}=180^{\circ} \Rightarrow \angle B A C=30^{\circ}$
So, statement-2 is true.
In Fig.(i), join BC. We observe that ABCD is a cyclic quadrilateral.
$\therefore \quad \angle A D C+\angle A B C=180^{\circ} \Rightarrow 120^{\circ}+\angle A BC=180^{\circ} \Rightarrow \angle A B C=60^{\circ}$
Since AOB is a diameter of the circle. Therefore, $\angle A C B=90^{\circ}$.
Using angle sum property in $\triangle A B C$, we obtain $\angle C A B=30^{\circ}$.
So, statement- 1 is also true. We also observe that statement- 2 is a correct explanation for statement-1. Hence, option (a) is correct.
In Fig. (ii), ABCD is a cyclic quadrilateral such that $\angle A D C=120^{\circ}$. Therefore,
$\angle A D C+\angle A B C=180^{\circ} \Rightarrow \angle A BC=180^{\circ}-120^{\circ}=60^{\circ}$
We find that $\angle A C B=90^{\circ}$ (angle in a semi-circle).
Using angle sum property in $\triangle A C B$, we obtain
$\angle A C B+\angle B A C+\angle A B C=180^{\circ} \Rightarrow 90^{\circ}+\angle B AC+60^{\circ}=180^{\circ} \Rightarrow \angle B A C=30^{\circ}$
So, statement-2 is true.
In Fig.(i), join BC. We observe that ABCD is a cyclic quadrilateral.
$\therefore \quad \angle A D C+\angle A B C=180^{\circ} \Rightarrow 120^{\circ}+\angle A BC=180^{\circ} \Rightarrow \angle A B C=60^{\circ}$
Since AOB is a diameter of the circle. Therefore, $\angle A C B=90^{\circ}$.
Using angle sum property in $\triangle A B C$, we obtain $\angle C A B=30^{\circ}$.
So, statement- 1 is also true. We also observe that statement- 2 is a correct explanation for statement-1. Hence, option (a) is correct.
