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

 Jun 5, 2024
 #1
avatar+129401 
+1

 

 

Note that angle PAD = 150

AD = AP

So

angles ADP  and APD are equal

 

APD =  (180 -150) /2    = 15°

 

Therefore  angle BPD    = 60  -15    = 45°

 

 

cool cool cool

 Jun 5, 2024

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