Come on now, check your book on the "product to sum" identities
![\bf cos({{ \alpha}})sin({{ \beta}})=\cfrac{1}{2}[sin({{ \alpha}}+{{ \beta}})\quad -\quad sin({{ \alpha}}-{{ \beta}})]](https://tex.z-dn.net/?f=%5Cbf%20cos%28%7B%7B%20%5Calpha%7D%7D%29sin%28%7B%7B%20%5Cbeta%7D%7D%29%3D%5Ccfrac%7B1%7D%7B2%7D%5Bsin%28%7B%7B%20%5Calpha%7D%7D%2B%7B%7B%20%5Cbeta%7D%7D%29%5Cquad%20-%5Cquad%20sin%28%7B%7B%20%5Calpha%7D%7D-%7B%7B%20%5Cbeta%7D%7D%29%5D)
Given CS=2x+1, SB=6x, CR=7.5, and RA=18. What must the value of x be in order to prove SR || BA? Justify your answer
1 answer:
The picture in the attached figure
we know that
In similar triangles. The ratio of the lengths of the sides CS and CB must be equal to the ratio of the lengths of sides CR and CA. CS / CB = CR / CA
which can also be written as,
CS / (CS + SB) = CR / (CR + RA)
CS*(CR+RA)=CR*(CS+RA)
CS=2x+1
SB=6x
CR=7.5
RA=18
(2x+1)*[7.5+18]=7.5*[2x+1+18]
(2x+1)*[25.5]=7.5*[2x+19]
(51x+25.5)=15x+142.5
51x-15x=142.5-25.5
36x=117
x=117/36
x=3.25
the answer is
x=3.25
we know that
In similar triangles. The ratio of the lengths of the sides CS and CB must be equal to the ratio of the lengths of sides CR and CA. CS / CB = CR / CA
which can also be written as,
CS / (CS + SB) = CR / (CR + RA)
CS*(CR+RA)=CR*(CS+RA)
CS=2x+1
SB=6x
CR=7.5
RA=18
(2x+1)*[7.5+18]=7.5*[2x+1+18]
(2x+1)*[25.5]=7.5*[2x+19]
(51x+25.5)=15x+142.5
51x-15x=142.5-25.5
36x=117
x=117/36
x=3.25
the answer is
x=3.25
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