Hello sir i want to know the best solution of the question from exercise 10.5of math of Circles chapter of class 9th give me the best and easy for solving this question how i solve it of question no.3 In Fig. 10.37, PQR = 100°, where P, Q and R are points on a circle with centre O. Find OPR.
Since angle which is subtended by an arc at the centre of the circle is double the angle subtended by that arc at any point on the remaining part of the circle.
So, the reflex POR = 2×PQR
We know the values of angle PQR as 100°
So, POR = 2×100° = 200°
∴ POR = 360°-200° = 160°
Now, in ΔOPR,
OP and OR are the radii of the circle
So, OP = OR
Also, OPR = ORP
Now, we know sum of the angles in a triangle is equal to 180 degrees
So,
POR+OPR+ORP = 180°
OPR+OPR = 180°-160°
As OPR = ORP
2OPR = 20°
Thus, OPR = 10°