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EAMCET Engineering 2002 Solved Paper

Section: Mathematics

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Question : 200 of 200

Marks: +1, -0

If

\triangle ABC

is right angled at

A

r_2+r_3

$r_1-r$
$r_1+r$
$r-r_1$
$R$

Solution:

- Given that

\angle A=90^{\circ}

Now,

r_2+r_3=4 R \cos \;\frac{A}{2} \sin \;\;\frac{B}{2} \cos \;\frac{C}{2}+4 R \cos \;\frac{A}{2}

\;\cos \;\frac{B}{2} \sin \;\;\frac{C}{2}

\;=4 R \cos \;\frac{A}{2} \left[ \sin \;\;\frac{B}{2} \cos \;\frac{C}{2}+\cos \;\frac{B}{2} \sin \;\;\frac{C}{2}\right]

\;=4 R \cos 45^{\circ} \cdot \sin \; \left( \;\frac{B+C}{2} \right)

\;=4 R \cdot \;\frac{1}{\sqrt{2}} \cdot \sin \; \left( \;\frac{\pi}{2}-\;\frac{A}{2} \right)

\;=4 R \cdot \;\frac{1}{\sqrt{2}} \cdot \cos \left( \;\frac{A}{2} \right)

\;=4 R \cdot \;\frac{1}{\sqrt{2}} \cdot \cos 45^{\circ}

\;=4 R \cdot \;\frac{1}{\sqrt{2}} \cdot \;\frac{1}{\sqrt{2}}=2 R

\; \therefore r_2+r_3=2 R

\;\;\text{ Also, }\; \;\; r_1+r_2+r_3-r=4 R=2(2 R)

\;r_1+r_2+r_3-r=2 (r_2+r_3)

\;\Rightarrow \;\; r_2+r_3=r_1-r

\;

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