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)

$2:\sqrt{6}:\sqrt{3}+1$

Answer 2

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, $\sin \, A = \sin \, 45^{\circ} = \frac{1}{\sqrt{2}}$
$\sin\,B = \sin \,60^{\circ} = \frac{\sqrt{3}}{2}$
$\sin \, C = \sin \, 75^{\circ} = \frac{\sqrt{3} +1}{ 2 \sqrt{2}}$
$\therefore \; a : b : c = \sin \, A : \sin \, B : \sin\, C$
$= \frac{1}{\sqrt{2}}: \frac{\sqrt{3}}{2}: \frac{\sqrt{3}+1}{2\sqrt{2}}$
$= 2: \sqrt{6}: \sqrt{3} +1$