Question

In a triangle ABC, $\angle A-\angle B={15}^o$ and $\angle B-\angle C={30}^o$, find $\angle A,\angle B$ and $\angle C$.

Answer 1

In the $△ABC$ according to the given condition we have $\angle B-\angle C={30}^o$

$\Rightarrow \angle B={30}^o+\angle C$

$\angle A-\angle B={15}^o$

$\Rightarrow \angle A={15}^o+\angle B={45}^o+\angle C$

Now, using the property that sum of all the interior angle of a triangle is ${180}^o$ we get

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

$\Rightarrow {45}^o+\angle C+{30}^o+\angle C+\angle C={180}^o$

$\Rightarrow 3\angle C={105}^o$

$\Rightarrow \angle C={35}^o$

$\therefore \angle A={80}^o,\angle B={65}^o,\angle C={35}^o$