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Angle Ratio Ascertainment (Posted on 2015-05-27) Difficulty: 2 of 5
ABC is a triangle with BC/(AB-BC) = (AB+BC)/AC.
Find k, given that ∠ACB = k*∠BAC

No Solution Yet Submitted by K Sengupta    
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Solution Solution Comment 1 of 1
A quick geometer's sketchpad drawing seemed to indicate k=2.

To prove this name the sides AB=c, BC=a, AC=b

the equation becomes a/(c-a)=(c+a)/b
solve for c to get c^2=a^2+ab

then by law of cosines

cos(ABC)=(a^2+b^2-c^2)/(2ac)=(b^2-ab)/(2ab)=(b-a)/(2a)

cos(BAC)=(b^2+c^2-a^2)/(2bc)=(b+a)/(2sqrt(a^2+ab))

cos(2*BAC)=cos^2(BAC)-1 =[algebra]= (b-a)/(2a)

cos(2*BAC)=cos(ABC)

2*BAC=ABC

(Incidentally the triangle is isosceles with angles 36-36-72)

  Posted by Jer on 2015-05-27 08:55:59
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