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

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

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)

Problem:

ReplyDeleteLet 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

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

ReplyDeleteThank you Gouravji

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

ReplyDeleteDraw 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)