Question
In $\triangle\text{ABC},$ if $3\angle\text{A}=4\angle\text{B}=6\angle\text{C},$ calculate $\angle\text{A},\angle\text{B}$ and $\angle\text{C}.$
✓
Answer
Let $3\angle\text{A}=4\angle\text{B}=6\angle\text{C}=\text{x}$ (say) Then, $3\angle\text{A}=\text{x}$$\Rightarrow\angle\text{A}=\frac{\text{x}}{3}$
$4\angle\text{B}=\text{x}$
$\Rightarrow\angle\text{B}=\frac{\text{x}}{4}$
and $6\angle\text{C}=\text{x}$$\Rightarrow\angle\text{C}=\frac{\text{x}}{6}$
As $\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$$\Rightarrow\frac{\text{x}}{3}+\frac{\text{x}}{4}+\frac{\text{x}}{6}=180$
$\Rightarrow\frac{4\text{x }+\ 3\text{x }+2\text{x}}{12}=180$
$\Rightarrow9\text{x}=180\times12$
$\Rightarrow\text{x}=\frac{180\times12}{9}=240$
$\therefore\angle\text{A}\frac{\text{x}}{3}=\frac{240}{3}=80^\circ$
$\angle\text{B}=\frac{\text{x}}{4}=\frac{240}{4}=60^\circ$
$\angle\text{C}=\frac{\text{x}}{6}=\frac{240}{6}=60^\circ$
$4\angle\text{B}=\text{x}$
$\Rightarrow\angle\text{B}=\frac{\text{x}}{4}$
and $6\angle\text{C}=\text{x}$$\Rightarrow\angle\text{C}=\frac{\text{x}}{6}$
As $\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$$\Rightarrow\frac{\text{x}}{3}+\frac{\text{x}}{4}+\frac{\text{x}}{6}=180$
$\Rightarrow\frac{4\text{x }+\ 3\text{x }+2\text{x}}{12}=180$
$\Rightarrow9\text{x}=180\times12$
$\Rightarrow\text{x}=\frac{180\times12}{9}=240$
$\therefore\angle\text{A}\frac{\text{x}}{3}=\frac{240}{3}=80^\circ$
$\angle\text{B}=\frac{\text{x}}{4}=\frac{240}{4}=60^\circ$
$\angle\text{C}=\frac{\text{x}}{6}=\frac{240}{6}=60^\circ$
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