KVPY SX SB Exam 20141 Question Paper

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Question : 7 of 120

Marks: +1, -0

Let ABC be a triangle such that AB = BC . Let F be the midpoint of AB and T be a point on BC such that FXis perpendicular to AB . If BX = 3XC then the ratio

{BC}/{AC}

equals

$√3$
$√2$
$√{3/2}$
1

Solution:

∵

\cos B = 2/3

Also cos B =

{16y^2+ 16y^2- AC ^2 } /{2.4y.4y}

2/3= {32y^2 − AC^2}/{32y^2}

64y^2 = 96y^2 -3AC^2

3AC^2 = 32y^2

AC ={4√2}/√3y

∴ {BC}/{AC}={4y}/({4√2y}/√3)=√3/√2

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