Mathematics

In $\Delta ABC,$ if $\angle A = 40^{\circ}$ and $\angle B = 60^{\circ}.$ Determine the longest and shortest sides of the triangle.

Longest = AB, shortest = BC

SOLUTION

In $\triangle ABC,$
$\angle A=40^o$ and $\angle B=60^o$
We know, that, sum of angles in a triangle is $180^o$
$\Rightarrow$  $\angle A+\angle B+\angle C=180^o$
$\Rightarrow$  $40^o+60^o+\angle C=180^o$
$\Rightarrow$  $\angle C=180^o-(40^o+60^o)$
$\Rightarrow$  $\angle C=180^o-100^o$
$\therefore$  $\angle C=80^o$
Now,
$40^o<60^o<80^o$
$\angle A<\angle B<\angle C$
$\Rightarrow$  $\angle C$ is greater angle and $\angle A$ is smaller angle.
Now,
$\angle A<\angle B<\angle C$
$BC<AC<AB$
Since, side opposite to greater angle is larger and side opposite to smaller angle is smaller.
$\therefore$  $AB$ is longest and $BC$ is shortest side. You're just one step away

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