Question

In a $△ABC$, $\angle A-\angle B={30}^{\circ }$ and $\angle B-\angle C={42}^{\circ }$; find $\angle A$.

• A. ${84}^o$
• B. ${94}^o$
• C. ${32}^o$
• D. none of the above

Answer 1

The correct option is B ${94}^o$

In $△ABC$,
given, $\angle A-\angle B={30}^o$ ....$\left(I\right)$
and $\angle B-\angle C={42}^o$ .....$\left(II\right)$.

Also, by angle sum property, the sum of angles of the triangle $={180}^o$.
$\angle A+\angle B+\angle C={180}^o$ ....$\left(III\right)$.

On subtracting $\left(I\right)$ and $\left(II\right)$, we get,

$\angle A-\angle B-\angle B+\angle C={30}^o-{42}^o$

$⟹$ $\angle A-2\angle B+\angle C=-{12}^o$ ....$\left(IV\right)$.

On subtracting $\left(III\right)$ and $\left(IV\right)$, we have,

$\angle A+\angle B+\angle C$ $-\angle A+2\angle B-\angle C={180}^o+{12}^o$

$⟹$$3\angle B={192}^o$

$⟹$$\angle B={64}^o$ ....$\left(V\right)$.

Substitute $\left(V\right)$ in $\left(I\right)$, we get,

$\angle A-\angle B={30}^o$

$⟹$$\angle A-{64}^o={30}^o$

$⟹$$\angle A={30}^o+{64}^o$

$⟹$$\angle A={94}^o$.

Therefore, option $B$ is correct.