Question
In $\triangle\text{ABC},$ if $\angle\text{A}=40^\circ$ and $\angle\text{B}=60^\circ.$ Determine the longest and shortest sides of the triangle.
✓
Answer
Given that in $\triangle\text{ABC},\angle\text{A}=40^\circ$ and $\angle\text{B}=60^\circ$
We have to find longest and shortest side
We know that,
Sum of angles in a triangle 180°
$\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$
$40^\circ+60^\circ-\big((10)0^\circ\big)=180^\circ$
$\angle\text{C}=180^\circ$
Now,
$\Rightarrow40^\circ<60^\circ<80^\circ=\angle\text{A}<\angle\text{B}<\angle\text{C}$
$\Rightarrow\angle\text{C}$ is greater angle and $\angle\text{A}$ is smaller angle.
Now, $\angle\text{A}<\angle\text{B}<\angle\text{C}$
$\Rightarrow\text{BC}<\text{AAC}<\text{AB}$ [Side opposite to greater angle is larger and side opposite to smaller angle is smaller]
AB is longest and BC is smallest or shortest side.
We have to find longest and shortest side
We know that,
Sum of angles in a triangle 180°
$\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$
$40^\circ+60^\circ-\big((10)0^\circ\big)=180^\circ$
$\angle\text{C}=180^\circ$
Now,
$\Rightarrow40^\circ<60^\circ<80^\circ=\angle\text{A}<\angle\text{B}<\angle\text{C}$
$\Rightarrow\angle\text{C}$ is greater angle and $\angle\text{A}$ is smaller angle.
Now, $\angle\text{A}<\angle\text{B}<\angle\text{C}$
$\Rightarrow\text{BC}<\text{AAC}<\text{AB}$ [Side opposite to greater angle is larger and side opposite to smaller angle is smaller]
AB is longest and BC is smallest or shortest side.
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