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)