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In triangle $ABC$, $AB = AC$. Point $P$ lies on $\overline{AB}$ such that $CP = BC$. If $\angle APC = 115^\circ$, what is $\angle ACP$ (in degrees)?

 Jul 1, 2021
 #1
avatar+120966 
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                A

       P 115

B                             C

 

Angle BPC is supplemental  to  APC =    65°

In triangle BPC, since CP   = BC  then  angle  PBC =   65°

Then angle  PCB  =  180 - 2 (65)  =   50°

 

But AB =  AC....so  angle  ABC = 65° =  angle ACB

 

And

 

ACB  = PCB   +  ACP

65 = 50  +  ACP

65  -  50  =  ACP  =  15°

 

cool cool cool

 Jul 1, 2021

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