MCQ
In an isosceles $\triangle\text{ABC},$ if AB = AC and $\angle\text{A} = 90^\circ,$ Find $\angle\text{B}.$
- A$60^\circ$
- B$80^\circ$
- C$95^\circ$
- ✓$ 45^\circ$
✓
Answer
Correct option: D.
$ 45^\circ$
We know that sum of all angle of a triangle is $180^\circ$
So, $\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$
$\angle\text{A} = 90^\circ$
$AB = AC$
$\angle\text{B} = \angle\text{C}$ (The angle opposite to equal side is also equal)
$90^\circ + \angle\text{B} + \angle\text{C} = 180^\circ$
$\angle\text{B} + \angle\text{C} = 180^\circ - 90^\circ$
$\angle\text{B} + \angle\text{C} = 90^\circ ( \angle\text{B} = \angle\text{C})$
$2\angle\text{B} = 90^\circ$
So, $\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$
$\angle\text{A} = 90^\circ$
$AB = AC$
$\angle\text{B} = \angle\text{C}$ (The angle opposite to equal side is also equal)
$90^\circ + \angle\text{B} + \angle\text{C} = 180^\circ$
$\angle\text{B} + \angle\text{C} = 180^\circ - 90^\circ$
$\angle\text{B} + \angle\text{C} = 90^\circ ( \angle\text{B} = \angle\text{C})$
$2\angle\text{B} = 90^\circ$
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