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Angles of a triangle are in the ratio $4: 1: 1$. Then the ratio of its greatest side to its perimeter is
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Verified Answer
The correct answer is:
$\sqrt{3}:(2+\sqrt{3})$
Let the angles of the triangle be $4 x, x$ and $x$.
$$
\therefore 4 x+x+x=180^{\circ} \Rightarrow 6 x=180^{\circ} \Rightarrow x=30^{\circ}
$$
By sine rule,
$$
\begin{aligned}
& \frac{\sin 120^{\circ}}{\mathrm{a}}=\frac{\sin 30^{\circ}}{\mathrm{b}}=\frac{\sin 30^{\circ}}{\mathrm{c}} \\
\therefore \quad & \mathrm{a}:(\mathrm{a}+\mathrm{b}+\mathrm{c}) \\
& =\left(\sin 120^{\circ}\right):\left(\sin 120^{\circ}+\sin 30^{\circ}+\sin 30^{\circ}\right) \\
& =\frac{\sqrt{3}}{2}: \frac{\sqrt{3}+2}{2}=\sqrt{3}: \sqrt{3}+2
\end{aligned}
$$
$$
\therefore 4 x+x+x=180^{\circ} \Rightarrow 6 x=180^{\circ} \Rightarrow x=30^{\circ}
$$
By sine rule,
$$
\begin{aligned}
& \frac{\sin 120^{\circ}}{\mathrm{a}}=\frac{\sin 30^{\circ}}{\mathrm{b}}=\frac{\sin 30^{\circ}}{\mathrm{c}} \\
\therefore \quad & \mathrm{a}:(\mathrm{a}+\mathrm{b}+\mathrm{c}) \\
& =\left(\sin 120^{\circ}\right):\left(\sin 120^{\circ}+\sin 30^{\circ}+\sin 30^{\circ}\right) \\
& =\frac{\sqrt{3}}{2}: \frac{\sqrt{3}+2}{2}=\sqrt{3}: \sqrt{3}+2
\end{aligned}
$$
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