Question

$A,B,C$ are the three angles of a triangle. If $A-B={15}^{\circ }$, $B-C={30}^{\circ }$, find $\angle$A, $\angle$B and $\angle$C .

Answer 1

We have, $A-B={15}^{\circ }$ and $B-C={30}^{\circ }$

$\Rightarrow$ $A={15}^{\circ }+B$ and $C=B-{30}^{\circ }$

Since, In a traingle, $\angle A+\angle B+\angle C={180}^{\circ }$
Substitute values of $A$ and $B$ in above equation, we get
$\Rightarrow 15+B+B+B-30=180$
$\Rightarrow 3B-15=180$
$\Rightarrow 3B=180+15$
$\Rightarrow 3B=195$
$\Rightarrow B=\frac{195}{3}$
$\therefore \angle B={65}^{\circ }$

Thus, $\angle A={15}^{\circ }+{65}^{\circ }={80}^{\circ }$ and $\angle C={65}^{\circ }-{30}^{\circ }$
$={35}^{\circ }$

Hence, $\angle A={80}^{\circ },\angle B={65}^{\circ }$ and $\angle C={35}^{\circ }$