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In a triangle $\mathrm{ABC}$, if $\mathrm{a}=2 \mathrm{~b}$ and $\mathrm{A}=3 \mathrm{~B}$ then which one of the following is correct?
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Verified Answer
The correct answer is:
The triangle is right-angled
We know from the Sine law that
$\frac{\mathrm{a}}{\sin \mathrm{A}}=\frac{\mathrm{b}}{\sin \mathrm{B}}$
$\Rightarrow \frac{2 b}{\sin 3 B}=\frac{b}{\sin B}$
$\Rightarrow 2 \sin B=\sin 3 B$
$\Rightarrow 2 \sin B=3 \sin B-4 \sin ^{3} B$
$\Rightarrow \sin B-4 \sin ^{3} B=0$
$\Rightarrow \sin B\left(1-4 \sin ^{2} B\right)=0$
$\Rightarrow \sin B=0$ or $1-4 \sin ^{2} B=0 \Rightarrow B=0$ or $B=30^{\circ}$
$\Rightarrow \mathrm{B}=30^{\circ}$ and $\mathrm{A}=3 \times 30^{\circ}=90^{\circ}$
$\Rightarrow \mathrm{B}=0$ is not possible so, $\mathrm{B}=30^{\circ}$ and $\mathrm{A}=3 \times 30^{\circ}=90^{\circ}$
$\Rightarrow$ The triangle is right angled triangle.
$\frac{\mathrm{a}}{\sin \mathrm{A}}=\frac{\mathrm{b}}{\sin \mathrm{B}}$
$\Rightarrow \frac{2 b}{\sin 3 B}=\frac{b}{\sin B}$
$\Rightarrow 2 \sin B=\sin 3 B$
$\Rightarrow 2 \sin B=3 \sin B-4 \sin ^{3} B$
$\Rightarrow \sin B-4 \sin ^{3} B=0$
$\Rightarrow \sin B\left(1-4 \sin ^{2} B\right)=0$
$\Rightarrow \sin B=0$ or $1-4 \sin ^{2} B=0 \Rightarrow B=0$ or $B=30^{\circ}$
$\Rightarrow \mathrm{B}=30^{\circ}$ and $\mathrm{A}=3 \times 30^{\circ}=90^{\circ}$
$\Rightarrow \mathrm{B}=0$ is not possible so, $\mathrm{B}=30^{\circ}$ and $\mathrm{A}=3 \times 30^{\circ}=90^{\circ}$
$\Rightarrow$ The triangle is right angled triangle.
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