This kinda supports Red Fox' claim that every rational fraction is reachable with simple unique configurations like the above one. So one down {infinite minus one} to go!
The two inclined tangents to the circle and the vertical tangent together make an isosceles triangle, the base of which is assumed to be 2a. Hence diameter of the circle = √3a. If Φ be the base angle of this isosceles triangle then tan(Φ/2)=(√3a)/2a=√3/2 and tan(Φ)= 2(√3/2)/(1- 3/4)= 4√3 or tan(Ω/2)= 1/(4√3). This gives us (sec(Ω/2))^2 =1+1/(16*3)=49/48 or cos(Ω/2)=√(48/49). So cos(Ω)= 2(48/49)-1= 47/49, Option A
This kinda supports Red Fox' claim that every rational fraction is reachable
The two inclined tangents to the circle and the vertical tangent together make an isosceles triangle, the base of which is assumed to be 2a. Hence diameter of the circle = √3a. If Φ be the base angle of this isosceles triangle then tan(Φ/2)=(√3a)/2a=√3/2 and tan(Φ)= 2(√3/2)/(1- 3/4)= 4√3 or tan(Ω/2)= 1/(4√3). This gives us (sec(Ω/2))^2 =1+1/(16*3)=49/48 or cos(Ω/2)=√(48/49). So cos(Ω)= 2(48/49)-1= 47/49, Option A
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