Given acute Triangle A,B,C and angle θ opposite b and angles α and β where α+β=θ created by the perpendicular from B to the opposite side,
Finding this is not too bad the work is shown below.
My question is what happens if it is not a perpendicular line. Suppose the ratio of the divided sides was k. Is there a simple form solution?
Solution: Let h be the length of the height.
Then: cos(α)=hc and cos(β)=ha. Therefore
cos(α)cos(β)=ca⇒cos(β)=accos(α)
cos(θ)=cos(α+β)=cos(α)cos(β)−sin(α)sin(β)
Let ac=K
cos(θ)=Kcos(α)2−sin(α)sin(β)
cos(θ)−Kcos(α)2=√1−cos(α)2√1−cos(β)2
(cos(θ)−Kcos(α)2)2=(1−cos(α)2)(1−K2cos(α)2)
cos(θ)2−2Kcos(α)2cos(θ)+K2cos(α)=1−(1+K2)cos(α)2+K2cos(α)
−sin(θ)2=2Kcos(α)2cos(θ)−(1+K2)cos(α)2
−sin(θ)22Kcos(θ)−1−K2=cos(α)2
−1ksin(θ)22cos(θ)−1K−K=−casin(θ)22cos(θ)−ca−ac=−c2sin(θ)2(2ca)cos(θ)−a2−c2=cos(α)2
c2sin(θ)2b2=cos(α)2
cos(α)=csin(θ)b
By similar methods we get:
cos(β)=asin(θ)b
cos(α)sin(θ)=cbcos(β)sin(θ)=ab
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