Question
In $\triangle\text{ABC},\ \angle\text{C}=3\angle\text{B}=2(\angle\text{A}+\angle\text{B}).$ find the measure of each one of $\angle\text{A},\ \angle\text{B}$ and $\angle\text{C}.$
✓
Answer
Let $\angle\text{A}=\text{x}^\circ$ and $\angle\text{B}=\text{y}^\circ$
Then, $\angle\text{C}=3\angle\text{B}=(3\text{y})^\circ$
Now, we have:
$\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$
$\Rightarrow x + y + 3y = 180$
$\Rightarrow x + 4y = 180 ...(i)$
Also, $\angle\text{C}=2(\angle\text{A}+\angle\text{B})$
$\Rightarrow 3y = 2(x + y)$
$\Rightarrow 2x - y = 0 ...(ii)$
On multiplying $(ii)$ by $4$, we get:
$8x - 4y = 0 ...(iii)$
On adding $(i)$ and $(ii)$, we get:
$9x = 180$
$\Rightarrow x = 20$
On Substituting $x = 20$ in $(i)$, we get:
$20 + 4y = 180$
$\Rightarrow 4y = (180 - 20) = 160$
$\Rightarrow y = 40$
$\therefore$ $x = 20$ and $y = 40$
$\therefore\angle\text{A}=20^\circ,\ \angle\text{B}=40^\circ,$ $\angle\text{C}=(3\times40^\circ)=120^\circ$
Then, $\angle\text{C}=3\angle\text{B}=(3\text{y})^\circ$
Now, we have:
$\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$
$\Rightarrow x + y + 3y = 180$
$\Rightarrow x + 4y = 180 ...(i)$
Also, $\angle\text{C}=2(\angle\text{A}+\angle\text{B})$
$\Rightarrow 3y = 2(x + y)$
$\Rightarrow 2x - y = 0 ...(ii)$
On multiplying $(ii)$ by $4$, we get:
$8x - 4y = 0 ...(iii)$
On adding $(i)$ and $(ii)$, we get:
$9x = 180$
$\Rightarrow x = 20$
On Substituting $x = 20$ in $(i)$, we get:
$20 + 4y = 180$
$\Rightarrow 4y = (180 - 20) = 160$
$\Rightarrow y = 40$
$\therefore$ $x = 20$ and $y = 40$
$\therefore\angle\text{A}=20^\circ,\ \angle\text{B}=40^\circ,$ $\angle\text{C}=(3\times40^\circ)=120^\circ$
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