In a $△ A B C, I$ is the incentre. The ratio $I A : I B : I C$ is equal to

csc A / 2 : csc B / 2 : csc C / 2

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sin A / 2 : sin B / : sin C / 2

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sec A / 2 : sec B / : sec C / 2

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N o n e o f t h e s e

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Solution

The correct option is A $csc A / 2 : csc B / 2 : csc C / 2$
It is given that, $I$ is the incentre of $Δ A B C$ . Let $r$ be the radius of the circle.

We know that,

$I A = \frac{r}{sin (A / 2)}$

$I B = \frac{r}{sin (B / 2)}$

$I C = \frac{r}{sin (C / 2)}$

Therefore,

$I A : I B : I C = \frac{r}{sin (A / 2)} : \frac{r}{sin (B / 2)} : \frac{r}{sin (C / 2)}$

$= csc \frac{A}{2} : csc \frac{B}{2} : c s c \frac{C}{2}$

Hence, this is the required ratio.

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