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If the angles of a triangle are in the ratio $3: 4: 5$, then the sides are in the ratio
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Verified Answer
The correct answer is:
$2: \sqrt{6}: \sqrt{3}+1$
Let the angles of triangle are $3 \theta, 4 \theta, 5 \theta$. As we know that,
$\angle A+\angle B+\angle C=180^{\circ}$
$\begin{array}{ll}\Rightarrow & 3 \theta+4 \theta+5 \theta=180^{\circ} \\ \Rightarrow & 12 \theta=180^{\circ} \Rightarrow \theta=15^{\circ}\end{array}$
So, angle are $45^{\circ}, 60^{\circ}, 75^{\circ}$.
Now, $\sin A=\sin 45^{\circ}=\frac{1}{\sqrt{2}}$
$\sin B=\sin 60^{\circ}=\frac{\sqrt{3}}{2}$
$\sin C=\sin 75^{\circ}=\frac{\sqrt{3}+1}{2 \sqrt{2}}$
So, $\quad a: b: c=\sin A: \sin B: \sin C$
$\begin{aligned}
&=\frac{1}{\sqrt{2}}: \frac{\sqrt{3}}{2}: \frac{\sqrt{3}+1}{2 \sqrt{2}} \\
&=2: \sqrt{6}: \sqrt{3}+1
\end{aligned}$
$\angle A+\angle B+\angle C=180^{\circ}$
$\begin{array}{ll}\Rightarrow & 3 \theta+4 \theta+5 \theta=180^{\circ} \\ \Rightarrow & 12 \theta=180^{\circ} \Rightarrow \theta=15^{\circ}\end{array}$
So, angle are $45^{\circ}, 60^{\circ}, 75^{\circ}$.
Now, $\sin A=\sin 45^{\circ}=\frac{1}{\sqrt{2}}$
$\sin B=\sin 60^{\circ}=\frac{\sqrt{3}}{2}$
$\sin C=\sin 75^{\circ}=\frac{\sqrt{3}+1}{2 \sqrt{2}}$
So, $\quad a: b: c=\sin A: \sin B: \sin C$
$\begin{aligned}
&=\frac{1}{\sqrt{2}}: \frac{\sqrt{3}}{2}: \frac{\sqrt{3}+1}{2 \sqrt{2}} \\
&=2: \sqrt{6}: \sqrt{3}+1
\end{aligned}$
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