+0  
 
0
66
1
avatar

Circle $\Gamma$ with center $O$ is tangential to $PA$ at $A$. Line $PO$ extended intersects $\Gamma$ at $B$ (i.e. $PO < PB$). $C$ is a point on $AB$ such that $PC$ bisects $ \angle APB$. What is the measure (in degrees) of $ \angle PCA$?

 Jan 20, 2021
 #1
avatar+116126 
+1

Let AP= AO

Angle PAO = 90

So angle APO =Angle AOP = 45

Then since PC is an angle bisector, then Angle APC  = 22.5

And Angle AOP  = Angle  AOD  = 45

The angle ABD  is an inscribed angle intercepting an arc of 45 ....so its measure = 22.5 = angle OBA

And in triangle AOB, OA  = OB....so Angle OBA  = Angle  OAB  =  22.5

And angle CAP =  Angle PAO + Angle OAB =   90  + 22.5 =  112.5

So in triangle APC, angle PCA  = 180  - 112.5 - 22.5  =  180 - 135  =   45 (degrees)

 

 

cool cool cool

 Jan 20, 2021

71 Online Users

avatar
avatar
avatar
avatar