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: X² + Y*Z = Z²

I will call∠ BAC just A and ∠ ABC just B
The the given info becomes
3A+2B=180
or B = 90-1.5A    [1]
and since A+B+C=180 we can subtract to get
2A+B-C=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(90-1.5A)/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 2011-02-15 11:11:14