In the given figure centre point O ,angle PQR=100° find angle OPR
Answers
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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Answer:
We will get the angle OPR=10°.
Step-by-step explanation:
- As per the question we have to evaluate the given data.
Given data:- angle PQR=100°.
To find:- angle OPR=?
Solution:-
- Consider PR as a chord of the circle.
- Take any point S on the major arc of the circle.
- PQRS is a cyclic quadrilateral.
∠PQR + ∠PSR = 180° (Opposite angles of a cyclic quadrilateral)
∠PSR = 180° − 100° = 80°
- We know that the angle subtended by an arc at the center is double the angle subtended by it at any point on the remaining part of the circle.
∠POR = 2∠PSR = 2 (80°) = 160°
OP = OR (Radii of the same circle)
∠OPR = ∠ORP (Angles opposite to equal sides of a triangle)
[tex]=>\angle O P R+\angle O R P+\angle P O R=180^{\circ}\\ =>2 \angle O P R+160^{\circ}=180^{\circ} \\ =>2 \angle O P R=180^{\circ}-160^{\circ}=20^{\circ} \\ =>\angle O P R=10^{\circ} \end{array}[/tex]