ABC is a triangle with BC/(AB-BC) = (AB+BC)/AC.
Find k, given that ∠ACB = k*∠BAC
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)
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Posted by Jer
on 2015-05-27 08:55:59 |
Solution