MHT CET · Maths · Properties of Triangles
The angles of a triangle are in the ratio 5:1:6, then ratio of the smallest side to the greatest side is
- A \(\sqrt{3}+1: 2 \sqrt{2}\)
- B \(2 \sqrt{2}: \sqrt{3}+1\)
- C \(2 \sqrt{2}: \sqrt{3}-1\)
- D \(\sqrt{3}-1: 2 \sqrt{2}\)
Answer & Solution
Correct Answer
(D) \(\sqrt{3}-1: 2 \sqrt{2}\)
Step-by-step Solution
Detailed explanation
Let the angles of the triangle be \(5 x, x, 6 x\)
\(\begin{array}{ll}
\therefore & 5 x+x+6 x=180^{\circ} \\
\therefore & 12 x=180^{\circ} \\
\therefore & x=15^{\circ}
\end{array}\)
\(\therefore \quad\) Three angles are \(75^{\circ}, 15^{\circ}, 90^{\circ}\).
\(\begin{aligned}
& \frac{\sin 75^{\circ}}{a}=\frac{\sin 15^{\circ}}{b}=\frac{\sin 90^{\circ}}{c}=k \\
\therefore \quad & \frac{b}{c}=\frac{\frac{\sin 15^{\circ}}{k}}{\frac{\sin 90^{\circ}}{k}}=\frac{\frac{\sqrt{3}-1}{2 \sqrt{2}}}{1}
\end{aligned}\)
\(\therefore \quad\) Required ratio \(=\mathrm{b}: \mathrm{c}=\sqrt{3}-1: 2 \sqrt{2}\)
\(\begin{array}{ll}
\therefore & 5 x+x+6 x=180^{\circ} \\
\therefore & 12 x=180^{\circ} \\
\therefore & x=15^{\circ}
\end{array}\)
\(\therefore \quad\) Three angles are \(75^{\circ}, 15^{\circ}, 90^{\circ}\).
\(\begin{aligned}
& \frac{\sin 75^{\circ}}{a}=\frac{\sin 15^{\circ}}{b}=\frac{\sin 90^{\circ}}{c}=k \\
\therefore \quad & \frac{b}{c}=\frac{\frac{\sin 15^{\circ}}{k}}{\frac{\sin 90^{\circ}}{k}}=\frac{\frac{\sqrt{3}-1}{2 \sqrt{2}}}{1}
\end{aligned}\)
\(\therefore \quad\) Required ratio \(=\mathrm{b}: \mathrm{c}=\sqrt{3}-1: 2 \sqrt{2}\)
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