Point P lies on minor arc AB of the circle circumscribing square ABCD. If AB = 5 and PA = 4, find the length of PD.
+23245 I have an answer for this, but it takes a lot of steps -- I welcome a quicker solution.
1) Find the length of the diameter AC:
Since AC is a diameter, I can use triangle ABC to find this length.
Both AB and BC = 5 and triangle ABC is an isosceles right triangle, AC = 5·sqrt(2).
2) Find the length PC:
Triangle ABC is a right triangle (angle ABC is inscribed in a semicircle), so use the Pythagorean Theorem.
PC2 + PA2 = AC2 ---> PC2 + 42 = [5·sqrt(2)]2 ---> PC2 + 16 = 50 ---> PC2 = 34 ---> PC = sqrt(34).
3) Find the size of angle(PAC):
Use the Law of Cosines for triangle PAC).
PC2 = PA2 + AC2 - 2·PA·AC·cos(PAC) ---> [sqrt(34)]2 = (4)2 + [5·sqrt(2)]2 - 2·(4)·5(sqrt(2)·cos(PAC)
---> 34 = 16 + 50 - 40·sqrt(2)·cos(PAC) ---> -32 = -40·sqrt(2)·cos(PAC) ---> PAC = 55.5501o
4) angle(PAD) = 45o + 55.5501o = 100.5501o
5) Use the Law of Cosines on Triangle(APD):
AP = 5; AD = 5; angle(PAD) = 100.501o ---> DP2 = AP2 + AD2 - 2·AD·AP·cos(PAD)
---> DP2 = (4)2 + (5)2 - 2·(4)·(5)·cos(100.5501) ---> DP2 = 48.3238 ---> DP = 6.95
Two comments:
1) I hope that I haven't made a mistake!
2) I hope that someone can find an easier way!
+1486 AB = 5
PA = 4
BD = ? BD = sqrt [(AB)² + (AD)²] BD = 7.072 r = 3.536
Let the circle's origin be an "O", and a midpoint of PA an "M" ∠POM = q sin(q) = (PM)/r q = 34.44°
∠POA = 2*q ∠POA= 68.88°
∠AOB = 90°
∠BOP = ∠AOB - ∠POA ∠BOP = 21.12°
∠(BOP)/2 = w w = 10.56°
PB = ? sin(w) =[(PB)/2] / r (PB)/2 = 0.648 PB = 1.296
Finally... PD = sqrt [(BD)² - (PB)²] PD = 6.952 
