SOLUTION: 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
From pt A, extend the line from A to R to futher to ptQ
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-> SOLUTION: 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
From pt A, extend the line from A to R to futher to ptQ
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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
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.
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.
Calculate,
a) angle, ACB, pls show steps （ there is Excel drawing but not able to upload here)
b) angle PQR, pls show step Answer by solver91311(24713) (Show Source):
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.
John
My calculator said it, I believe it, that settles it
