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:
$\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$
⇒ x + y + 3y = 180
⇒ x + 4y = 180 ...(i)
Also, $\angle\text{C}=2(\angle\text{A}+\angle\text{B})$
⇒ 3y = 2(x + y)
⇒ 2x - y = 0 ...(ii)
On multiplying (ii) by 4, we get:
8x - 4y = 0 ...(iii)
On adding (i) and (ii), we get:
9x = 180
⇒ x = 20
On Substituting x = 20 in (i), we get:
20 + 4y = 180
⇒ 4y = (180 - 20) = 160
⇒ 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:
$\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$
⇒ x + y + 3y = 180
⇒ x + 4y = 180 ...(i)
Also, $\angle\text{C}=2(\angle\text{A}+\angle\text{B})$
⇒ 3y = 2(x + y)
⇒ 2x - y = 0 ...(ii)
On multiplying (ii) by 4, we get:
8x - 4y = 0 ...(iii)
On adding (i) and (ii), we get:
9x = 180
⇒ x = 20
On Substituting x = 20 in (i), we get:
20 + 4y = 180
⇒ 4y = (180 - 20) = 160
⇒ 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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