+188 In the figure, PA and PB are tangent to circle O and PD bisects ∠BPA . The figure is not drawn to scale.
for m angle AOC = 66, find m angle POB
A. 78
B. 33
C. 90
D. 66
+129849 Let's prove Solveit's answer :
m<OBP = m<OAP tangents drawn to a circle meet radii at right angles
OB = OA radii in the same circle are congruent
PB = PA tangents drawn to a circle from a point are congruent
ΔBOP = ΔAOP SAS
m<POB = m<POA by definition of congruent triangles
m<POA = m<AOC they are the same angle
m<POB = m<AOC = 66 substitution
