Question

In. triangle $ABC$, $\angle B$ $={144}^o$ and $\angle C$ $={124}^o$.
Calculate smallest angle of the triangle.

• A. ${36}^o$
• B. ${56}^o$
• C. ${46}^o$
• D. none of these

Answer 1

The correct option is A ${36}^o$

Given, $\angle A+\angle B=144$ ...$\left(I\right)$
Also, $\angle A+\angle C=124$ ...$\left(II\right)$.

In $△ABC$,
$\angle A+\angle B+\angle C={180}^o$ ..... [By angle sum property].

Add $\left(I\right)$ and $\left(II\right)$,
$\angle A+\angle B+\angle A+\angle C={144}^o+{124}^o$.

Substitute $\left(I\right)$,
${180}^o+\angle A={268}^o$
$⟹$ $\angle A={268}^o-{180}^o$
$⟹$ $\angle A={88}^o$ ...$\left(III\right)$.

Substitute $\left(III\right)$ in $\left(I\right)$,
$\angle A+\angle B={144}^o$
$⟹$ ${88}^o+\angle B={144}^o$
$⟹$ $\angle B={56}^o$ ...$\left(IV\right)$.

Substitute $\left(III\right)$ in $\left(II\right)$,
$\angle A+\angle C={124}^o$
$⟹$ ${88}^o+\angle C={124}^o$
$⟹$ $\angle C={36}^o$ ...$\left(V\right)$.

From $\left(III\right)$, $\left(IV\right)$ and $\left(V\right)$,

we get, the smallest angle of the $△ABC$ is $\angle C={36}^o$.

Hence, option $A$ is correct.