MCQ
In $\triangle\text{ABC},$ if $AB = AC$ and $\angle\text{ACD}=1200,$ find $\angle\text{A}.$ 
- A$70^\circ$
- BNone of these
- C$50^\circ$
- ✓$60^\circ$
✓
Answer
Correct option: D.
$60^\circ$
In $\triangle\text{ABC}, \ \text{AB} = \text{AC}$
$\Rightarrow \angle\text{ABC} = \angle\text{ACB}$
Also, $\angle\text{ACD}=120^\circ$
$\Rightarrow \angle\text{ACB} = 180^\circ- \angle\text{ACD}$ (Linear pair)
$\Rightarrow \angle\text{ACB} = 180^\circ- 120^\circ = 60^\circ$
$\Rightarrow \angle\text{ABC} = 60^\circ$
By using angle sum property, we have
$\angle\text{ABC} + \angle\text{ACB} + \angle\text{BAC} = 180^\circ$
$60^\circ + 60^\circ+ \angle\text{A} = 180^\circ$
or, $\angle\text{A} = 180^\circ - 120^\circ= 60^\circ$
$\Rightarrow \angle\text{ABC} = \angle\text{ACB}$
Also, $\angle\text{ACD}=120^\circ$
$\Rightarrow \angle\text{ACB} = 180^\circ- \angle\text{ACD}$ (Linear pair)
$\Rightarrow \angle\text{ACB} = 180^\circ- 120^\circ = 60^\circ$
$\Rightarrow \angle\text{ABC} = 60^\circ$
By using angle sum property, we have
$\angle\text{ABC} + \angle\text{ACB} + \angle\text{BAC} = 180^\circ$
$60^\circ + 60^\circ+ \angle\text{A} = 180^\circ$
or, $\angle\text{A} = 180^\circ - 120^\circ= 60^\circ$
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