Let C be incircle of `DeltaABC`. If the tangents of lengths `t_(1),t_(2) and t_(3)` are drawn inside the given triangle parallel to side a,b, and c, respectively, then `(t_(1))/(a) + (t_(2))/(b) + (t_(3))/(c)` is equal to

Options

A. 0
B. 1
C. 2
D. 3

Answer

B

Explanation

`DeltaAP_(1) P_(2) ~ DeltaABC`
`rArr (t_(1))/(a) = (h-2r)/(h) = 1 -(2r)/(h)`
`rArr (t_(1))/(a) = 1 -(2Delta)/(h)`(where `Delta ~= ar (DeltaABC)`)
`rArr (t_(1))/(a) = 1 -(2(1)/(2) ah)/(sh)`
`rArr (t_(1))/(a) = 1 - (a)/(s)`
Similarly, `(t_(2))/(b) = 1 - (b)/(s) and, (t_(3))/(c) = 1 -(c)/(s)`
`:. (t_(1))/(a) + (t_(2))/(b) + (t_(3))/(c) = 3 - ((a+b+c))/(s) = 3 -2 =1`