Question

$Ina\Delta ABC,if\angle A-\angle B={42}^{\circ }and\angle B-\angle C={21}^{\circ }then\angle B=?\left(a\right){32}^{\circ }\left(b\right){63}^{\circ }\left(c\right){53}^{\circ }\left(d\right){95}^{\circ }$

Answer 1

ANSWER:
(c) ${53}^{\circ }$
Let
$\angle A-\angle B={42}^{\circ }...iand\angle B-\angle C={21}^{\circ }$ ...ii

Adding (i) and (ii), we get:$\angle A-\angle C=63°\angle B=\angle A-{42}^{\circ }$

Using i $\angle C=\angle A-{63}^{\circ }$

Using iii
∴ $\angle A+\angle B+\angle C={180}^{\circ }$

Sum of the angles of a triangle⇒$\angle A+\angle A-{42}^{\circ }+\angle A-{63}^{\circ }={180}^{\circ }\Rightarrow 3\angle A-{105}^{\circ }={180}^{\circ }\Rightarrow 3\angle A={285}^{\circ }$
⇒$\angle A={95}^{\circ }\therefore \angle B=95-{42}^{\circ }\Rightarrow \angle B={53}^{\circ }$