MCQ
In the adjoining figure, $BC = AC$. If $\angle\text{ACD} = 115^\circ,$ the $\angle\text{A}$ is: 
- A$50^\circ$
- B$65^\circ$
- ✓$57.5^\circ$
- D$70^\circ$
✓
Answer
Correct option: C.
$57.5^\circ$
As $BC = AC$, therefore triangle ABC is an isoscelestriangle.
Given $\angle\text{ACD} = 115^\circ,\ \angle\text{ACB} = 180 - 115 = 65^\circ$ (Linear Pair)
As $AC = BC$, therefore $\angle\text{A} =\angle\text{B}$
As sum of all the three angles of atriangle is $180^\circ $
Therefore. $\angle\text{A}+\angle\text{B} +\angle\text{ACB} = 180^\circ$
$\angle\text{A} =\angle\text{B}=57.5^\circ$
Given $\angle\text{ACD} = 115^\circ,\ \angle\text{ACB} = 180 - 115 = 65^\circ$ (Linear Pair)
As $AC = BC$, therefore $\angle\text{A} =\angle\text{B}$
As sum of all the three angles of atriangle is $180^\circ $
Therefore. $\angle\text{A}+\angle\text{B} +\angle\text{ACB} = 180^\circ$
$\angle\text{A} =\angle\text{B}=57.5^\circ$
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