Question
In $\triangle\text{PQR},$ if $\angle\text{P}-\angle\text{Q}=42^\circ$ and $\angle\text{Q}+\angle\text{R}=21^\circ,$ find $\angle\text{P},\angle\text{Q}$ and $\angle\text{R}.$
✓
Answer
Since, $\angle\text{P},\angle\text{Q}$ and $\angle\text{R}$ are the angles of a triangle. So, $\angle\text{P}+\angle\text{Q}+\angle\text{R}=180^\circ\ ...(\text{i)}$ Now, $\angle\text{P}-\angle\text{Q}=42^\circ$ [Given]$\Rightarrow\angle\text{P}=42^\circ+\angle\text{Q}\ ....(\text{ii)}$
and $\angle\text{Q}-\angle\text{R}=21^\circ$ [Given]$\angle\text{R}=\angle\text{Q}-21^\circ\ ....(\text{iii)}$
Substituting the value of $\angle\text{P}$ and $\angle\text{R}$ from (ii) and (iii) in (i), we get,$42^\circ+\angle\text{Q}+\angle\text{Q}+\angle\text{Q}-21^\circ=180^\circ$
$\Rightarrow3\angle\text{Q}+21^\circ=180^\circ$
$\Rightarrow3\angle\text{Q}=180^\circ-21^\circ=159^\circ$
$\Rightarrow\angle\text{Q}=\frac{159^\circ}{3}=53^\circ$
$\therefore\angle\text{P}=42^\circ+\angle\text{Q}$
$=42^\circ+53^\circ=95^\circ$
$\angle\text{R}=\angle\text{Q}-21^\circ$
$=53^\circ-21^\circ=32^\circ$
$\therefore\angle\text{P}=95^\circ,\angle\text{Q}=53^\circ$ and $\angle\text{R}=32^\circ.$
and $\angle\text{Q}-\angle\text{R}=21^\circ$ [Given]$\angle\text{R}=\angle\text{Q}-21^\circ\ ....(\text{iii)}$
Substituting the value of $\angle\text{P}$ and $\angle\text{R}$ from (ii) and (iii) in (i), we get,$42^\circ+\angle\text{Q}+\angle\text{Q}+\angle\text{Q}-21^\circ=180^\circ$
$\Rightarrow3\angle\text{Q}+21^\circ=180^\circ$
$\Rightarrow3\angle\text{Q}=180^\circ-21^\circ=159^\circ$
$\Rightarrow\angle\text{Q}=\frac{159^\circ}{3}=53^\circ$
$\therefore\angle\text{P}=42^\circ+\angle\text{Q}$
$=42^\circ+53^\circ=95^\circ$
$\angle\text{R}=\angle\text{Q}-21^\circ$
$=53^\circ-21^\circ=32^\circ$
$\therefore\angle\text{P}=95^\circ,\angle\text{Q}=53^\circ$ and $\angle\text{R}=32^\circ.$
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