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Hi there! Have questions about your homework? At Brainly, there are 60 million students who want to help each other learn. Questions are usually answered in less than 10 minutes. Try it for yourself by posting a question! :D

Given - OQ=2cm PR=4cm TO Find= Angle ROQ Construction-Join PQ, RO, OQ,OP, Draw OE perpendicular to PQ Solution= In triangle OPR OP=OR ( Radius of Circle) angle OPR=angle PRO=35 (Angles opposite to equal nagles are equal) Now BY angle sum property of triangles angle OPR+angle ORP + angle POR=180 35 +35+angle POR=180 70+angle POR=180 angle POR=180-70 angle POR=110

IN triangle OE perpendicular to PQ therefore PE =EQ (i) (Perpendicular from center bisects the chord) Now in triangle OEP and ΔOEQ OE =OE (common) PE =EQ (proved at i) OP=OQ (radius of circle) Therefore ΔOEP ≡ΔOEQ angle POE =angle EOQ (ii) (By CPCT) [we'll use this result after a while]

Now in Triangle POE Let angle EOP =α

Now Sinα=EP/OP Sinα=2/2 Sinα=1 Sin 90=1 α=90 NOW angle POE=90 =EOQ (from ii)