document.write( "Question 1043143: In the diagram is triangle BAP, AP=AB (isoceles triangle) and from pt A, extend a line to meet PB at pt R. Angle BAR=60degr
\n" ); document.write( "From pt A, extend the line from A to R to futher to ptQ.This will form another triangle BAQ, triangle ABQ is equilateral and angle PAB = 82 degr.
\n" ); document.write( "Then from pt P and pt Q, further extend the 2 line to meet at the peak pt C.This will form external triangle ABC.\r
\n" ); document.write( "\n" ); document.write( "Calculate,
\n" ); document.write( "a) angle, ACB, pls show steps ( there is Excel drawing but not able to upload here)
\n" ); document.write( "b) angle PQR, pls show step
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Algebra.Com's Answer #658527 by solver91311(24713)\"\" \"About 
You can put this solution on YOUR website!
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\n" ); document.write( "\n" ); document.write( "Sorry, but the description you give of your diagram is nonsense. If triangle BAP is isosceles with AP = AB, then the line segment AR where R lies on the segment P must perforce be the perpendicular bisector of segment PB and the bisector of angle A. Hence, if angle BAR = 60 degrees, then PAR must be 60 degrees also and therefore angle PAB must be 120 degrees, not 82 degrees as you stated.\r
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\n" ); document.write( "\n" ); document.write( "John
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\n" ); document.write( "My calculator said it, I believe it, that settles it
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