ABC is a triangle where the respective lengths of the sides BC, AC and AB are denoted by X, Y and Z. It is known that 3*∠ BAC + 2*∠ ABC = 180^{o}
Prove that: X^{2} + Y*Z = Z^{2}
I will call
∠ BAC just A and
∠ ABC just B
The the given info becomes
3A+2B=180
or B = 901.5A [1]
and since A+B+C=180 we can subtract to get
2A+BC=0
C = 2A+B
C = 90+.5A [2]
By the law of sines
sinB/y=sinC/z
substituting [1] and [2] gives
y/z = sin(901.5A)/sin(90+.5A)
y/z = cos(1.5A)/cos(.5A) [3]
By the law of cosines
x^2 = z^2 + y^2 2zycosA
x^2 = z^2(1  y^2/z^2  2(y/z)cosA)
x^2 = z^2(1 (y/z)(2cosA  y/z)) [4]
Looking more closely at the
2cosA  y/z
we can substitute [3] to get
2cosA  cos(1.5A)/(cos(.5A)
=(2cos(A)cos(.5A)  cos(1.5A))/cos(.5A)
=(cos(1.5A)  cos(.5A)  cos(1.5A))/cos(.5A)
=cos(.5A)/cos(.5A)
=1
so substituting this back into [4] we get
x^2 = z^2(1  (y/z))
x^2 = z^2  yz
x^2 + yz = z^2
QED

Posted by Jer
on 20110215 11:11:14 