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Trigonometric Ratios of Half Angles

In triangle A...

Question

In triangle $A B C$ , let $∠ C = \frac{π}{2}$ . If $r$ is the inradius and $R$ is circumradius of the triangle, then $2 (r + R)$ is equal to:

a + b

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b + c

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c + a

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a + b + c

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Solution

The correct option is A $a + b$
Given, $Δ A B C$ is right angled at vertex $C$ ,
$\Rightarrow$ circum-radius, $R = \frac{c}{2}$
$\Rightarrow 2 R = c \dots (i)$
Now,
$r = (s - c) \cdot tan \frac{C}{2}$
$= (s - c) tan \frac{π}{4} = (s - c)$
Solving the equation
$\Rightarrow 2 (r + R) = 2 r + 2 R$
$\Rightarrow 2 (r + R) = 2 (s - c) + 2 R = 2 s - 2 c + c = a + b$

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