MCQ
In a $\triangle\text{ABC},$ If $\angle\text{A}= 45^\circ$ and $\angle\text{B}= 70^\circ.$ Determine the shortest sides of the triangles.
- A$AC$
- ✓$BC$
- C$CA$
- DAll are equal.
✓
Answer
Correct option: B.
$BC$
$\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$
$\angle\text{A}=45^\circ$
$\angle\text{B}=70^\circ$
$\angle\text{C}+45^\circ+70^\circ=180^\circ$
$\angle\text{C}+115^\circ=180^\circ$
$\angle\text{C}=180^\circ-115^\circ$
$\angle\text{C}=65^\circ$
$\angle\text{A}$ is shortest angle and the side opp to shortest angle is shortest.
So, $BC$ is the shotest side.
$\angle\text{A}=45^\circ$
$\angle\text{B}=70^\circ$
$\angle\text{C}+45^\circ+70^\circ=180^\circ$
$\angle\text{C}+115^\circ=180^\circ$
$\angle\text{C}=180^\circ-115^\circ$
$\angle\text{C}=65^\circ$
$\angle\text{A}$ is shortest angle and the side opp to shortest angle is shortest.
So, $BC$ is the shotest side.
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