MHT CET · Maths · Properties of Triangles
If the angles of a triangle are in the ratio \(4: 1: 1\), then the ratio of the longest side to the perimeter is
- A \(1: 6\)
- B \(\sqrt{3}:(2+\sqrt{3})\)
- C \(1:(2+\sqrt{3})\)
- D \(2: 3\)
Answer & Solution
Correct Answer
(B) \(\sqrt{3}:(2+\sqrt{3})\)
Step-by-step Solution
Detailed explanation
Let the angles of the triangle be \(4 \dot{x}, x\) and \(x\).
\(\begin{aligned}
\therefore \quad & 4 x+x+x=180^{\circ} \Rightarrow 6 x=180^{\circ} \Rightarrow x=30^{\circ} \\
& \frac{\sin 120^{\circ}}{\mathrm{a}}=\frac{\sin 30^{\circ}}{\mathrm{b}}=\frac{\sin 30^{\circ}}{\mathrm{c}} \\
\therefore \quad & \mathrm{a}:(\mathrm{a}+\mathrm{b}+\mathrm{c}) \\
& =\left(\sin 120^{\circ}\right):\left(\sin 120^{\circ}+\sin 30^{\circ}+\sin 30^{\circ}\right) \\
\quad & \frac{\sqrt{3}}{2}: \frac{\sqrt{3}+2}{2}=\sqrt{3}: \sqrt{3}+2
\end{aligned}\)
\(\begin{aligned}
\therefore \quad & 4 x+x+x=180^{\circ} \Rightarrow 6 x=180^{\circ} \Rightarrow x=30^{\circ} \\
& \frac{\sin 120^{\circ}}{\mathrm{a}}=\frac{\sin 30^{\circ}}{\mathrm{b}}=\frac{\sin 30^{\circ}}{\mathrm{c}} \\
\therefore \quad & \mathrm{a}:(\mathrm{a}+\mathrm{b}+\mathrm{c}) \\
& =\left(\sin 120^{\circ}\right):\left(\sin 120^{\circ}+\sin 30^{\circ}+\sin 30^{\circ}\right) \\
\quad & \frac{\sqrt{3}}{2}: \frac{\sqrt{3}+2}{2}=\sqrt{3}: \sqrt{3}+2
\end{aligned}\)
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