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avatar+1075 

Let $ABCD$ be a square.  Let $P$ be a point outside square $ABCD$ such that triangle $ABP$ is equilateral.  Find $\angle BPD$ in degrees.

 Dec 20, 2023
 #1
avatar+129197 
+1

         P

 

 

A                  B

 

 

C                 D

 

Angle ABP  = 60

Angle ABD = 90

Angle PBD = 60 + 90  = 150

 

In triangle PBD , BP = BD

So angle BDP  = angle BPD

 

So angle BPD =   (180 - 150) / 2 =  30 / 2  =  15°

 

 

cool cool cool

 Dec 20, 2023

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