# 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.

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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)

Angle ROQ= angle POR+ anglePOE+angleEOQ

Angle ROQ=110+90+90

Angle ROQ=270

HENCE found