Each of the points P and Q lie on a circle with its center at O and having a radius of √50.
Point R is inside the circle such that:
∠PQR = 90^{o}, PQ = 6, QR = 2
Find OR
First note that
∠POQ = 2*∠OQR
By the law of cosines
cos(POQ) = (50+5036)/(2*50)=16/25
cos(POQ)=cos(2*OQR)=2cos(OQR)^21
cos(OQR)=sqrt(41/50)
By the law of cosines again
OR^2 = 50+42*sqrt(50)*2*cos(OQR)
OR^2 = 544sqrt(41)
OR = sqrt(544sqrt(41))
or about 5.328

Posted by Jer
on 20160408 11:37:32 