Much Simpler solution: 1) extend quarter circle from point A to point C on extension of line BO forming a semi circle of diameter 60. 2) Define theta as 1/2 of angle PBO. Theta= atan(15/30) 3) PBC forms a right triangle (from theorem for a triangle inscribed in a semi-circle with the diameter as the hypotenuse) 4) PB=60*cos(2*theta)=36 (angle PBC=2*theta)
I got an alternative solution with trigonometry. Let C be on segment BP and OC is perpendicular to BP, then BP=2*BC …..(1) Let angle OBP=b tan(b/2)=(AO/2)/BO=15/30=1/2 -> cos(b/2)=2/sqrt(5) BC=BO*cos b=30*(2*cos(b/2)^2-1)=18 replacing to (1) BP=2*18=36
from Morocco thank you for this wonderful clear proof
You are welcome sir ❤️❤️❤️👍👍👍
Much Simpler solution:
1) extend quarter circle from point A to point C on extension of line BO forming a semi circle of diameter 60.
2) Define theta as 1/2 of angle PBO. Theta= atan(15/30)
3) PBC forms a right triangle (from theorem for a triangle inscribed in a semi-circle with the diameter as the hypotenuse)
4) PB=60*cos(2*theta)=36 (angle PBC=2*theta)
I got an alternative solution with trigonometry.
Let C be on segment BP and OC is perpendicular to BP, then BP=2*BC …..(1)
Let angle OBP=b
tan(b/2)=(AO/2)/BO=15/30=1/2 -> cos(b/2)=2/sqrt(5)
BC=BO*cos b=30*(2*cos(b/2)^2-1)=18 replacing to (1)
BP=2*18=36
Nice one 👍👍