Equilateral triangle ABC is inscribed in a circle. Point P lies on this circle's (circumference), and PB = 8. If AB = 7 and PA < PC, find
the ordered pair PA and PC.
( I'm reposting someone's question because I believe it should be answered.)
Chord > PB = 8
Triangle side > AB = 7
Let the circle's diameter be BD ( Point D is 180° from point B)
Diameter > BD = ? BD = AB / cos(30°) BD = 8.083
Let the circle's center be an "O"
Chord > PD = ? PD = sqrt [(BD)² - (PB)²] PD = 1.155
Angle PBD > b cos(b) = PB / BD ∠ b = 8.213°
Angle ABP > a a = ∠ABD - ∠b ∠ a = 21.787°
Angle PBC > c c = ∠ABC - ∠a ∠ c = 38.213°
PA = ? (PA)² = (AB)²+(PB)² - 2(AB)(PB) cos(a) PA = 3
PC = ? (PC)² = (BC)²+(PB)² - 2(BC)(PB) cos(c) PC = 5