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In a triangle $\mathrm{ABC}, \mathrm{a}=2 \mathrm{~b}$ and $\angle \mathrm{A}=3 \angle \mathrm{B}$. Which one of the following is correct?
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Verified Answer
The correct answer is:
The triangle is right-angled
From properties of triangle we know that
$\frac{\mathrm{a}}{\sin \mathrm{A}}=\frac{\mathrm{b}}{\sin \mathrm{B}}=\frac{\mathrm{c}}{\sin \mathrm{C}}$
Given that $\mathrm{a}=2 \mathrm{~b}$ and $\mathrm{A}=3 \mathrm{~B}$.
We get $\frac{2 b}{\sin 3 B}=\frac{b}{\sin B}$
$\Rightarrow \frac{\sin 3 B}{\sin B}=2 \Rightarrow \frac{3 \sin B-4 \sin ^{3} B}{\sin B}=2$
$\Rightarrow 3-4 \sin ^{2} B=2$
$\Rightarrow \sin ^{2} B=\left(\frac{1}{2}\right)^{2} \Rightarrow \sin B=\frac{1}{2}=\sin \frac{\pi}{6}$
$\Rightarrow B=\frac{\pi}{6}$ since $A=3 B$
So, $\angle A=\frac{\pi}{2}$
Thus, triangle is right angled triangle.
$\frac{\mathrm{a}}{\sin \mathrm{A}}=\frac{\mathrm{b}}{\sin \mathrm{B}}=\frac{\mathrm{c}}{\sin \mathrm{C}}$
Given that $\mathrm{a}=2 \mathrm{~b}$ and $\mathrm{A}=3 \mathrm{~B}$.
We get $\frac{2 b}{\sin 3 B}=\frac{b}{\sin B}$
$\Rightarrow \frac{\sin 3 B}{\sin B}=2 \Rightarrow \frac{3 \sin B-4 \sin ^{3} B}{\sin B}=2$
$\Rightarrow 3-4 \sin ^{2} B=2$
$\Rightarrow \sin ^{2} B=\left(\frac{1}{2}\right)^{2} \Rightarrow \sin B=\frac{1}{2}=\sin \frac{\pi}{6}$
$\Rightarrow B=\frac{\pi}{6}$ since $A=3 B$
So, $\angle A=\frac{\pi}{2}$
Thus, triangle is right angled triangle.
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