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December 19, 2014

December 19, 2014

Posted by **rajat garg** on Wednesday, August 15, 2012 at 2:35am.

- maths -
**Reiny**, Wednesday, August 15, 2012 at 8:27ammake a sketch.

let angle C = x , then angle B = 2x

also, let angle BAD = y = angle CAD , (Angle A was bisected by AD)

then in triangle ABC , 2x + 2y + x = 180

y = (180-3x)/2 = 90 - 3x/2

by exterior angle theorem,

angle BDA = x+y and angle CDA = 2x+y

let AB = m = DC , (given)

by the sine law:

m/sin(x+y) = AD/sin 2x and m/siny = AD/sinx

m = ADsin(x+y)/sin 2x and m = ADsiny/sinx

so ADsin(x+y)/sin2x = ADsiny/sinx

divide by AD

sin(x+y)/sin2x = siny/sinx

sin(x + 90-3x/2)/(2sinxcosx) = siny/sinx

sin(90 - x/2)/(2cosx) = siny , after dividing by sinx

cross-multiply

sin(90-x/2) = 2sinxcosx

sin(90-x/2) = sin 2x

so 90 - x/2 = 2x

times 2

180 - x = 4x

180=5x

x=36

then y = 90-3(36)/2 = 36

and angle A = 2y = 72°

- maths -
**Reiny**, Wednesday, August 15, 2012 at 8:31amThere is a wonderful field of math lying ahead from this problem.

notice that cos 36° = .809016.. which is 1/2 of the golden ratio of (1+√5)/2 or 1.61803....

Look up the "pentagon"

Draw the central triangles, your triangle is one of these.

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