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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

 Apr 4, 2016
 #1
avatar+2498 
+3


\(\angle AOC=\angle POB=66 \)

.
 Apr 4, 2016
 #2
avatar+129852 
0

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

 

 

cool cool cool

 Apr 4, 2016

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