Question
In a $\triangle\text{ABC},$ if $\angle\text{A}-\angle\text{B}=42^\circ$ and $\angle\text{B}-\angle\text{C}=21^\circ$ then $\angle\text{B}=?$
✓
Answer
- 53º
Solution:
$\angle\text{A}-\angle\text{B}=42^\circ$
$\Rightarrow\angle\text{A}=\angle\text{B}+42^\circ$
$\angle\text{B}-\angle\text{C}=21^\circ$
$\Rightarrow\angle\text{C}=\angle\text{B}-21^\circ$
In $\triangle\text{ABC},$
$\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$
$\Rightarrow\angle\text{B}+42^\circ+\angle\text{B}+\angle\text{B}-21^\circ=180^\circ$
$\Rightarrow3\angle\text{B}=159$
$\Rightarrow\angle\text{B}=53^\circ$
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