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

Let C be incircle of DeltaABC. If the tangents of lengths t_(1),t_(2)

1.	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
Answer» 0 1 2 3 Solution : `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`

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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

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