Question

in the adjoining figure, O is the centre of the circle. if angle PQR=100 , then find angle OPR

Answer 1

❤❤Hey mate...here is ur ans...❤❤

⏩GIVEN => <PQR = 100, O is the centre of the circle

⏩TO FIND => angle <OPR

✔We know that angle subtended in the centre of the circle is double the angle subtended by it on the remaining part of the circle

✔So <POR = 2×100°=200° ( reflex )................... ( 1 )

✔Now reflex POR + obtuse POR = 360° ( complete angle )

✔From ( 1 )
200° + <POR = 360°
POR = 160°( obtuse )........................( 2 )

✔Now we know that <OPR = <ORP ( angle subtended on the remaining part of the circle are equal ) ..............................( 3 )

✔Now in triangle POR, By angle sum property

<POR + <OPR + <ORP = 180°
160° + 2OPR = 180° ( from 2 & 3 )
2OPR = 180° - 160°
2OPR = 20°
OPR = 20°/2
OPR = ORP = 10°

✔So <OPR = 10°

❤❤Hope it helps u..thanks..❤❤

Answer 2

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.

Thank you______❤

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