+911 Equilateral triangle ABC is inscribed in a circle. Let P be a point on arc BC. Given that $PB = 18$ and $PC = 18,$ find $PA.$
+130042 
ABPC is a quadrilateral inscribed in the circle
Angle BAC = 60°
So angle BPC is supplemental to angle BAC = 120°
Using the Law of Cosines
BC^2 = PB^2 + PC^2 - 2(PB * PC) cos (120°)
BC^2 = 18^2 + 18^2 - 2(18*18( (-1/2)
BC^2 = 18^2 + 18^2 + 18^2
BC^2 = 3*18^2
BC = 18sqrt (3) = side of the equilateral triangle
Triangle AEC is a 30-60-90 right triangle
AC = 18sqrt (3) = the hypotenuse
AE is opposite the 60° angle = 18sqrt 3 * (sqrt 3 /2) = 27
Triangle PEC is also a 30-60-90 right triangle
EC = 9sqrt 3 is opposite the 60° angle
PE is opposite the 30° angle = 9sqrt 3/sqrt 3 = 9
PA = PE + AE = 9 + 27 = 36
