angle pqr is equal 100 degrees where p q and R are points on a circle with Centre O find angle opr
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PQR = 100°∠POR = 2∠PQR = 200° (central angle is two times the angle in the same arc)Minor ∠POR = 360 −200 = 160°OP = OR (radii of circle)∠OPR =∠ORP (angles opposite to equal sides are equl)△OPR, by angle sum property of triangle∠OPR + ∠ORP +∠POR = 180∠OPR = (180 − 160)2 = 10°
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Hello mate =_=
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Solution:
P, Q and R are the points on a circle with centre O where ∠PQR=100°
Construction: S is a point on the major arc PR. Join P and S, R and S to form a cyclic quadrilateral.
∠PQR+∠PSR=180°
(Sum of opposite angles of a cyclic quadrilateral is equal to 180°)
⇒100°+∠PSR=180°
⇒∠PSR=180°−100°=80°
Also, ∠POR=2∠PSR
(Angle subtended by an arc at the centre is double the angle subtended by it at the circumference of the circle.)
⇒∠POR=2×80°=160°
In ∆POR, we have
∠POR+∠ORP+∠OPR=180°
But, we have ∠ORP=∠OPR (Angles opposite to the equal sides in a triangle are equal.)
⇒160°+∠OPR+∠OPR=180°
⇒2∠OPR=180°−160°=20°
⇒∠OPR=20/2=10°
hope, this will help you.
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