MCQ
In $\triangle\text{ABC},$ $BC = AB$ and $\angle\text{B} = 80^\circ.$ Then $\angle\text{A}$ is equal to:
- A$100^\circ$
- B$80^\circ$
- C$40^\circ$
- ✓$50^\circ$
✓
Answer
Correct option: D.
$50^\circ$
Given: $\triangle\text{ABC},\ \text{(BC=AB)}$ and $\angle\text{B} = 80^\circ$
As $BC = AB$
So it is an isosceles triangle.
let $\angle\text{C} = \angle\text{A} = \text{x}$
$\angle B = \angle\text{B} = 80^\circ$ (given)
As we know $\angle\text{A} + \angle\text{B} + \angle\text{C} = 180^\circ$
$\Rightarrow x + 80^\circ + x = 180^\circ$
$\Rightarrow 2x = 180^\circ - 80^\circ$
$\Rightarrow 2x = 100^\circ$
$\Rightarrow x = 50^\circ$
So, $\angle\text{C} = \angle\text{A} = 50^\circ$
As $BC = AB$
So it is an isosceles triangle.
let $\angle\text{C} = \angle\text{A} = \text{x}$
$\angle B = \angle\text{B} = 80^\circ$ (given)
As we know $\angle\text{A} + \angle\text{B} + \angle\text{C} = 180^\circ$
$\Rightarrow x + 80^\circ + x = 180^\circ$
$\Rightarrow 2x = 180^\circ - 80^\circ$
$\Rightarrow 2x = 100^\circ$
$\Rightarrow x = 50^\circ$
So, $\angle\text{C} = \angle\text{A} = 50^\circ$
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