In Fig.10.37, ∠PQR = 100° where P, Q and R are points on a circle with centre O. Find ∠OPR
Solution:
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.
A quadrilateral ABCD is called cyclic if all the four vertices of it lie on a circle and the sum of either pair of opposite angles of a cyclic quadrilateral is 180°.
Mark any point on the major arc side (opposite side to point Q) as S.
Since all points P, Q, R, S lie on the circle, PQRS becomes a cyclic quadrilateral.
We know that the sum of either pair of opposite angles of a cyclic quadrilateral is 180°.
Therefore,
∠PQR + ∠PSR = 180°
100° + ∠PSR = 180°
∠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.
Therefore,
∠POR = 2∠PSR
= 2 × 80°
= 160°
Consider the ∆OPR. It is an isosceles triangle as OP = OR = Radius of the circle.
Thus, ∠OPR = ∠ORP
Sum of all angles in a triangle is 180°.
Therefore,
∠OPR + ∠POR + ∠ORP = 180°
∠OPR + 160° + ∠OPR = 180°
2∠OPR = 180° - 160°
∠OPR = 10°
☛ Check: Class 9 Maths NCERT Solutions Chapter 10
Video Solution:
In Fig. 10.37, ∠PQR = 100° where P, Q and R are points on a circle with center O. Find ∠OPR
Maths NCERT Solutions Class 9 Chapter 10 Exercise 10.5 Question 3
Summary:
If, in the given figure, ∠PQR = 100° where P, Q, and R are three points on a circle with center O, then ∠OPR = 10°.
☛ Related Questions:
- In Fig. 10.38, ∠ABC = 69° and ∠ACB= 31°, find ∠BDC.
- In Fig. 10.39, A, B, C and D are four points on a circle. AC and BD intersect at a point E such that ∠BEC = 130° and ∠ECD = 20°. Find ∠BAC.
- ABCD is a cyclic quadrilateral whose diagonals intersect at a point E. If ∠DBC = 70°, ∠BAC is 30° find ∠BCD. Further if AB = BC, find ∠ECD.
- If diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of the quadrilateral, prove that it is a rectangle.