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in Co-ordinate geometry by (63.5k points)

In a right angled triangle ABC, ∠C = 90°  and sides AC, AB are roots of the equation 2 + y2 = 3y. If the internal angle bisector of angle A intersects BC at D such that BD : CD = x2 + 1 : 2x , then find the sum of all possible values of tan(A/2).

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y2 + 2 = 3y

y = 1, 2 

AC = 1, AB = 2

BD/CD = (x2 + 1)/2x

(BD + CD)/CD = (x2 + 1 + 2x)/2x

√3/CD = (x + 1)2/2x

CD = 2√3x/(x + 1)2 = b/(c + b)(√3)

Similarly BD = √3(x2 + 1)/(x + 1)2

2√3x/(x + 1)2 = 1/√3

6x = (x + 1)2 ⇒x2 + 1 – 4x = 0 

⇒ x1 + x2 = 4

tan(A1/2) + tan(A2/2) = 2x1/1 + 2x2/1 = 8

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