help asap thx

 

Find angle BPC     

 

angle PBC = 150^o which is the total of PBA (60^o cuz the triangle is equilateral) plus ABC (90^o cuz it says so)  

 

since PBC is 150^o that leaves 30^o for the total of both BPC and BCP cuz the total of angles in a triangle is 180^o  

 

triangle PBC is isoceles since sides PB and BC are equal  

 

therefore angles BPC and BCP are equal, so each one is half of their 30^o total  

 

so angle BPC is 15^o cuz 15 is half of 30

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