MCQ
- ✓$50^{\circ}$
- B$40^{\circ}$
- C$60^{\circ}$
- D$70^{\circ}$
✓
Answer
Correct option: A.
$50^{\circ}$
(a) $50^{\circ}$
In $\triangle O A B$, we have
$O A=O B \Rightarrow \angle O B A=\angle O A B \Rightarrow \angle O B A=40^{\circ}$
Using angle sum property in $\triangle A O B$, we obtain
$\begin{array}{ll}& \angle O A B+\angle O B A+\angle A O B=180^{\circ} \\
\Rightarrow & 40^{\circ}+40^{\circ}+\angle A O B=180^{\circ} \Rightarrow \angle A O B=100^{\circ}\end{array}$
Thus, $\operatorname{arc} A B$ subtends $\angle A O B=100^{\circ}$ at the centre and $\angle A C B$ at a point on the circumference.
$\therefore \quad \angle A C B=\frac{1}{2} \angle A O B=50^{\circ}$
In $\triangle O A B$, we have
$O A=O B \Rightarrow \angle O B A=\angle O A B \Rightarrow \angle O B A=40^{\circ}$
Using angle sum property in $\triangle A O B$, we obtain
$\begin{array}{ll}& \angle O A B+\angle O B A+\angle A O B=180^{\circ} \\
\Rightarrow & 40^{\circ}+40^{\circ}+\angle A O B=180^{\circ} \Rightarrow \angle A O B=100^{\circ}\end{array}$
Thus, $\operatorname{arc} A B$ subtends $\angle A O B=100^{\circ}$ at the centre and $\angle A C B$ at a point on the circumference.
$\therefore \quad \angle A C B=\frac{1}{2} \angle A O B=50^{\circ}$
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