Question

If the angles of a triangle are in the ratio $3:4:5$, then the sides are in the ratio

• A. $3:4:5$
• B. $2:\sqrt{3}:\sqrt{3}+1$
• C. $\sqrt{2}:\sqrt{6}+\sqrt{3}+1$
• D. $2:\sqrt{6}:\sqrt{3}+1$

Answer 1

Answer: (D)

Let the angles of a triangle are $3\theta ,4\theta ,5\theta$.
We know
$\angle A+\angle B+\angle C=180^{\circ }$
$\Rightarrow 3\theta +4\theta +5\theta =180^{\circ }$
$\Rightarrow 12\theta =180^{\circ }$
$\Rightarrow \theta =15^{\circ }$
$\therefore$ Angles are $45^{\circ },60^{\circ },75^{\circ }$
Now, $\sin A=\sin 45^{\circ }=\frac{1}{\sqrt{2}}$
$\sin B=\sin 60^{\circ }=\frac{\sqrt{3}}{2}$

$\sin C=\sin 75^{\circ }=\frac{\sqrt{3}+1}{2\sqrt{2}}$
$\therefore a:b:c=\sin A:\sin B:\sin C$
$=\frac{1}{\sqrt{2}}:\frac{\sqrt{3}}{2}:\frac{\sqrt{3}+1}{2\sqrt{2}}$
$=2:\sqrt{6}:\sqrt{3}+1$