In the adjoining figure is an equilateral triangle.
QR is produced to S , such that QR=RS .Calculate ∠PRS and ∠PSR
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in ∆ PQR, QR = RS
so all angle is 60° in∆ PQR
<PRQ = 60°
<PRQ + <PRS = 180°
60° <PRS = 180°
< PRS = 120°
so in ∆ PSR, PR = RS
<PRS+ <RPS+<PSR = 180
so <RPS = < PSR= 60
<PSR = 30°
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