Mathematics

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


ANSWER

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.
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