Take a look at the points in the circle in the figure:
The chord bar(AB)¯¯¯¯¯¯AB can be calculated the distance formula:
AB = sqrt((5-1)^2 + (2-4)^2) = sqrt(16+4)=sqrt(20)=2sqrt(5)AB=√(5−1)2+(2−4)2=√16+4=√20=2√5
Now we know the outside angle between bar(OB)¯¯¯¯¯¯OB and bar(OA)¯¯¯¯¯¯OA
angle alpha=BhatOA = (5pi)/3α=BˆOA=5π3 thus the inside angle theta=BhatOA = (pi)/3θ=BˆOA=π3
From the picture we can see that bar(BO) = bar(AO) = r = Radius¯¯¯¯¯¯BO=¯¯¯¯¯¯AO=r=Radius
Also bar(AB) = bar(AH) + bar(HB) = sqrt(5)¯¯¯¯¯¯AB=¯¯¯¯¯¯AH+¯¯¯¯¯¯HB=√5
Now triangle AOHAOH form a right angle triangle with angles:
/_HAO = beta = pi/3∠HAO=β=π3
/_AOH = theta/2 = pi/6∠AOH=θ2=π6
This the 30-60-90 right angle triangle with sides relationship of
(2:1 :sqrt(3))(2:1:√3)
Using (AOH)_(Delta) we write bar(AO) = r = bar(AH)/sin(beta) = sqrt(5)/sinbeta
Now =>sinbeta = sqrt(3)/2
Thus r = 2 sqrt(5)/sqrt(3)= 2sqrt(5/3)
Now Arc Length, L = r theta = 2sqrt(5/3)*pi/3 = (2pi)/3*sqrt(5/3)
![enter image source here]()
