Sunday, August 14, 2011

KVPY Practice Questions - IV

4 comments:

  1. Problem:
    Let ABC be a triangle in which AB = AC and let I
    be its in-centre. Suppose BC = AB + AI. Find ∠BAC.
    Solution:
    Let AB = AC = b, BC = 2a
    AI = 2a - b
    AD^2 = AB^2 - BD^2 = b^2 - a^2
    BI bisects angle ABD in triangle ABD.
    So AI/ID = AB/BD = b/a and
    2a - b = AI = [b/(a + b)].AD
    Hence
    b^2 - a^2 = AD^2 = [(2a - b)^2.(a + b)^2]/b^2
    b^2.(b - a) = (2a - b)^2.(a + b)
    b^3 - ab^2
    = 4a^3 + 4a^2b + ab^2 + b^3 - 4a^2b - 4ab^2
    = 4a^3 - 3ab^2 + b^3
    Thus 2ab^2 = 4a^3 , b^2 = 2a^2
    sin∠BAD = a/b = 1/sqrt 2, ∠BAD = 45 deg
    ∠BAC = 2∠BAD = 90 deg

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  2. DEAR SIR I also like math like you .i think that you mase a mistake in above mentioned Q.1 that you write angle DAC in place of angle DAB in equation (ii) in the solution

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  3. Ref.:Soln.of (KVPY Practice) Q.1- equation(ii)

    Draw CE ∥ DA (meeting AB at E)
    ADCE is a parallelogram
    So CE = DA = CB,
    ∆CEB is isosceles
    So ∠CBA = ∠CBE = ∠CEB
    Also ∠CEB = ∠DAB (corresp angles, CE ∥ DA)
    Hence ∠DAB = ∠CBA as stated in (ii)

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