Question
In $\text{PQR},$ if $3\angle\text{P}=4\angle\text{Q}=6\angle\text{R},$ calculate the angles of the triangle.
✓
Answer
Given, $3\angle\text{p}=4\angle\text{Q}=6\angle\text{R}$
Then, $\angle\text{P}=\frac{6}{3}\angle\text{R}=2\angle\text{R}$
$\angle\text{Q}=\frac{6}{4}\angle\text{R}=\frac{3}{2}\angle\text{R}$
In $\triangle\text{PQR},$
$\angle\text{P}+\angle\text{Q}+\angle\text{R}=180^{\circ}[$ angle sum property of a triangle$]$
$\Rightarrow2\angle\text{R}+\frac{3}{2}\angle\text{R}+\angle\text{R}=180^{\circ}$
$\Rightarrow3\angle\text{R}+\frac{3}{2}\angle\text{R}=180^{\circ}$
$\Rightarrow6\angle\text{R}+3\angle\text{R}=180^{\circ}\times2 [$On taking $LCM$ in $LHS]$
$\Rightarrow9\angle\text{R}=360^{\circ}$
$\Rightarrow\angle\text{R}=\frac{360^{\circ}}{9}=40^{\circ}$
$\therefore\angle\text{P}=2\angle\text{R}=2\times40^{\circ}=80^{\circ}$
and $\angle\text{Q}=\frac{3}{2}\angle\text{R}=\frac{3}{2}\times40^{\circ}=60^{\circ}$
Hence, all the angles of the triangles are $80^\circ , 60^\circ $ and $40^\circ $
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