MCQ
In a $\triangle\text{ABC},$ if $3\angle\text{A}=4\angle\text{B}=6\angle\text{C}$ then $A : B : C = ?$
- A$4 : 3 : 2$
- B$6 : 4 : 3$
- C$2 : 3 : 4$
- ✓$3 : 4 : 6$
✓
Answer
Correct option: D.
$3 : 4 : 6$
In the given figure, $\angle\text{ACB}+\angle\text{ACD}=180^\circ$ (Linear pair of angles)
$\therefore 5y^\circ + 7y^\circ = 180^\circ $
$\Rightarrow 12y^\circ = 180^\circ $
$\Rightarrow y = 15$
In $\triangle\text{ABC}$
$\angle\text{A}+\angle\text{B}+\angle\text{ACB}=180^\circ$ (Angle sum property)
$\therefore 3y^\circ + x^\circ + 5y^\circ = 180^\circ $
$\Rightarrow x^\circ + 8y^\circ = 180^\circ $
$\Rightarrow x^\circ + 8 \times 15^\circ = 180^\circ $
$\Rightarrow x^\circ + 120^\circ = 180^\circ $
$\Rightarrow x^\circ = 180^\circ − 120^\circ = 60^\circ $
Thus, the value of $x$ is $60^\circ .$
Hence the correct answer is $3 : 4 : 6$
$\therefore 5y^\circ + 7y^\circ = 180^\circ $
$\Rightarrow 12y^\circ = 180^\circ $
$\Rightarrow y = 15$
In $\triangle\text{ABC}$
$\angle\text{A}+\angle\text{B}+\angle\text{ACB}=180^\circ$ (Angle sum property)
$\therefore 3y^\circ + x^\circ + 5y^\circ = 180^\circ $
$\Rightarrow x^\circ + 8y^\circ = 180^\circ $
$\Rightarrow x^\circ + 8 \times 15^\circ = 180^\circ $
$\Rightarrow x^\circ + 120^\circ = 180^\circ $
$\Rightarrow x^\circ = 180^\circ − 120^\circ = 60^\circ $
Thus, the value of $x$ is $60^\circ .$
Hence the correct answer is $3 : 4 : 6$
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