Question

In $\Delta ABC\angle A={50}^{\circ }$and$\angle B={60}^{\circ }$. By arranging the sides of the triangle in ascending order, we get :

• A. $AB<BC<CA$
• B. $CA<AB<BC$
• C. $BC<CA<AB$
• D. $BC<AB<CA$

Answer 1

The correct option is C $BC<CA<AB$

Using angle sum property in $△ABC$

$\angle A+\angle B+\angle C={180}^{\circ }\Rightarrow {50}^{\circ }+{60}^{\circ }+\angle C={180}^{\circ }\Rightarrow \angle C={70}^{\circ }$

In any triangle side opposite to the largest angle is the longest side.

Here $\angle C$ is largest and side opposite to it is $AB$

$\therefore AB$ is the longest side.

Then comes $\angle B$ and side opposite to it is $AC$

$\therefore AC$ is the second longest side.

Then comes $\angle A$ and side opposite to it is $AC$

$\therefore BC$ is the shortest side.

So the increasing order of sides is

$BC<AC<AB$