MCQ
In a $\triangle\text{ABC},\ \angle\text{C}=3\angle\text{B}=2(\angle\text{A}+\angle\text{B}),$ then $\angle\text{B}=?$
- A$20^\circ$
- ✓$40^\circ$
- C$60^\circ $
- D$80^\circ$
✓
Answer
Correct option: B.
$40^\circ$
Give that in a $\triangle\text{ABC},$
$\angle\text{C}=3\angle\text{B}=2(\angle\text{A}+\angle\text{B})$
$\Rightarrow\angle\text{C}=3\angle\text{B}$ and $\angle\text{C}=2(\angle\text{A}+\angle\text{B})$
Consider, $\angle\text{C}=2(\angle\text{A}+\angle\text{B})$
$\Rightarrow3\angle\text{B}=2(\angle\text{A}+\angle\text{B})$
$\Rightarrow\angle\text{B}=2\angle\text{A}$
By the Angle Sum Property
$\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$
$\Rightarrow\angle\text{A}+2\angle\text{A}+2(\angle\text{A}+2\angle\text{A})=180^\circ$
$\Rightarrow\angle\text{A}+2\angle\text{A}+2\angle\text{A}+4\angle\text{A}=180^\circ$
$\Rightarrow9\angle\text{A}=180^\circ$
$\Rightarrow\angle\text{A}=20^\circ$
So, $\angle\text{B}=2\angle\text{A}$
$\Rightarrow\angle\text{B}=40^\circ$
$\angle\text{C}=3\angle\text{B}=2(\angle\text{A}+\angle\text{B})$
$\Rightarrow\angle\text{C}=3\angle\text{B}$ and $\angle\text{C}=2(\angle\text{A}+\angle\text{B})$
Consider, $\angle\text{C}=2(\angle\text{A}+\angle\text{B})$
$\Rightarrow3\angle\text{B}=2(\angle\text{A}+\angle\text{B})$
$\Rightarrow\angle\text{B}=2\angle\text{A}$
By the Angle Sum Property
$\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$
$\Rightarrow\angle\text{A}+2\angle\text{A}+2(\angle\text{A}+2\angle\text{A})=180^\circ$
$\Rightarrow\angle\text{A}+2\angle\text{A}+2\angle\text{A}+4\angle\text{A}=180^\circ$
$\Rightarrow9\angle\text{A}=180^\circ$
$\Rightarrow\angle\text{A}=20^\circ$
So, $\angle\text{B}=2\angle\text{A}$
$\Rightarrow\angle\text{B}=40^\circ$
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