MHT CET · Maths · Sequences and Series
Angles of a triangle are in the ratio \(4: 1: 1\). Then the ratio of its greatest side to its perimeter is
- A \(3:(2+\sqrt{3})\)
- B \(\sqrt{3}:(2+\sqrt{3})\)
- C \(\sqrt{3}:(2-\sqrt{3})\)
- D \(1:(2+\sqrt{3})\)
Answer & Solution
Correct Answer
(B) \(\sqrt{3}:(2+\sqrt{3})\)
Step-by-step Solution
Detailed explanation
Let the angles of the triangle be \(4 x, x\) and \(x\).
\(\therefore 4 x+x+x=180^{\circ} \Rightarrow 6 x=180^{\circ} \Rightarrow x=30^{\circ}\)
By sine rule,
\(\frac{\sin 120^{\circ}}{\mathrm{a}}=\frac{\sin 30^{\circ}}{\mathrm{b}}=\frac{\sin 30^{\circ}}{\mathrm{c}} \)
\(\therefore \mathrm{a}:(\mathrm{a}+\mathrm{b}+\mathrm{c}) \)
\(=\left(\sin 120^{\circ}\right):\left(\sin 120^{\circ}+\sin 30^{\circ}+\sin 30^{\circ}\right) \)
\(\frac{\sqrt{3}}{2}: \frac{\sqrt{3}+2}{2}=\sqrt{3}: \sqrt{3}+2\)
\(\therefore 4 x+x+x=180^{\circ} \Rightarrow 6 x=180^{\circ} \Rightarrow x=30^{\circ}\)
By sine rule,
\(\frac{\sin 120^{\circ}}{\mathrm{a}}=\frac{\sin 30^{\circ}}{\mathrm{b}}=\frac{\sin 30^{\circ}}{\mathrm{c}} \)
\(\therefore \mathrm{a}:(\mathrm{a}+\mathrm{b}+\mathrm{c}) \)
\(=\left(\sin 120^{\circ}\right):\left(\sin 120^{\circ}+\sin 30^{\circ}+\sin 30^{\circ}\right) \)
\(\frac{\sqrt{3}}{2}: \frac{\sqrt{3}+2}{2}=\sqrt{3}: \sqrt{3}+2\)
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