Saturday, 17 October 2015

geometry - Given Triangle a,b,c and angles theta,alpha,beta.Find fraccos(alpha)sin(theta)



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,



Find: cos(α)sin(θ)enter image description here




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)=cacos(β)=accos(α)


cos(θ)=cos(α+β)=cos(α)cos(β)sin(α)sin(β)



Let ac=K




cos(θ)=Kcos(α)2sin(α)sin(β)


cos(θ)Kcos(α)2=1cos(α)21cos(β)2

(cos(θ)Kcos(α)2)2=(1cos(α)2)(1K2cos(α)2)



cos(θ)22Kcos(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)1K2=cos(α)2


1ksin(θ)22cos(θ)1KK=casin(θ)22cos(θ)caac=c2sin(θ)2(2ca)cos(θ)a2c2=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


Answer



Introduce the second altitude as shown.




enter image description here



Use the areas of the same triangle but with different altitudes to calculate the ratio. (Or doing the above in the reverse order.) The proofing process will then be much simpler.



For example, red =ccosθ and area =reda2.


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