Question

In Fig. 9.24, ∠PQR = 100°, where P, Q and R are points on a circle with centre O. Find ∠OPR.

Answer 1

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 centre is double the angle subtended by it at any point on the remaining part of the circle. \(POR = 2 PSR = 2 (80°) = 160° \)

In \(∆POR,\) 
OP = OR (Radii of the same circle) 
\(∠OPR = ∠ORP \) (Angles opposite to equal sides of a triangle)
\(∠OPR + ∠ORP + ∠POR = 180°\) (Angle sum property of a triangle) 
\(∠OPR + 160° = 180° 2 \)
\(∠OPR = 180° − 160° = 20º 2 \)
\(∠OPR = 10°\)

Answer 2

Let's think of PR as a line inside a circle.

Now, pick any spot S on the longer part of the circle. When you join PQRS, you make a shape called a cyclic quadrilateral.
We know that when you add up opposite angles of a cyclic quadrilateral, it equals 180 degrees. So, we can say:
\(∠ PQR + ∠ PSR = 180°\)

So, \(∠ PSR = 180° - 100° = 80°\)

Another thing we know is that when you have an arc and measure its angle at the middle of the circle, it's twice the angle you get when you measure it at any point on the circle. So, we get:

\(∠ POR = 2 \times ∠ PSR = 2  \times  80° = 160°\)

In triangle POR, OP is the same as OR (they are both radii of the circle). Because in a triangle, when you have equal sides, the angles opposite those sides are equal. So, we have:

\(∠ OPR = ∠ ORP\)

And using the rule that angles in a triangle add up to 180 degrees, we get:

\(2  \times  ∠ OPR + 160° = 180°\)

So, \(2  \times  ∠ OPR = 180° - 160° = 20°\)

Therefore, \(∠ OPR = 10°\)