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In a circle with centre O, OQ is the radius=2 cm. PR is a chord of the circle and PQ=4 cm, and angle PRO=35°. Find angle ROQ. Please answer fast.



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
Sin 90=1
NOW angle POE=90 =EOQ                     (from ii)

Angle ROQ= angle POR+ anglePOE+angleEOQ
Angle ROQ=110+90+90
Angle ROQ=270
HENCE found

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