If $r : R : r_{1} = 2 : 5 : 12$ then prove that the triangle is right angled at A.

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Detailed Solution

Given $r : R : r_{1} = 2 : 5 : 12 ∵ \frac{r}{2} = \frac{R}{5} = \frac{r_{1}}{12} = k$

$\Rightarrow r = 2 k, R = 5 k, r_{1} = 12 k$

$r_{1} - r = 12 k - 2 k = 10 k = 2 (5 k) = 2 R$

$\Rightarrow r_{1} - r = 2 R$

$\Rightarrow 4 R \sin \frac{A}{2} (\cos \frac{B}{2} \cos \frac{C}{2} - \sin \frac{B}{2} \sin \frac{C}{2}) = 2 R$

$∵ r_{1} = 4 R \sin \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}, r = 4 R \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}$

$\Rightarrow 4 R \sin \frac{A}{2} \cos (\frac{B + C}{2}) = 2 R \Rightarrow 2 \sin^{2} \frac{A}{2} = 1$

$\Rightarrow \sin^{2} \frac{A}{2} = \frac{1}{2} \Rightarrow \sin \frac{A}{2} = \frac{1}{\sqrt{2}}$

$\frac{A}{2} = 45^{0} \Rightarrow A = 90^{0}$

$∴$ The triangle is right angled at ‘A’.

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