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Let ABC be a triangle with angleBAC = 2p...

Let ABC be a triangle with `angleBAC = 2pi//3 and AB = x` such that (AB) (AC) = 1. If x varies, then find the longest possible length of the angle bisector AD

Text Solution

Verified by Experts

The correct Answer is:
`(1)/(2)` unit

`AD = y = (2bc)/(b + c) cos.(A)/(2) = (bx)/(b + x)`
But `bx = 1 " or " b = (1)/(x)`
`:. y = (x)/(1 + x^(2)) = (1)/(x + (1)/(x))`
Thus, `y_("max") = (1)/(2)`, since the minimum value of the denominator is `2 " if " x gt 0`
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