Question
A chord PQ subtends an angle of 70° at the centre O of a circle. What is the angle subtended by the same arc PQ at any point R on the major arc (i.e., the remaining part of the circle)?
Solution — Step by Step
We have chord PQ with central angle ∠POQ = 70°. Point R lies on the major arc. We need ∠PRQ — the inscribed angle subtending the same minor arc PQ.
The Inscribed Angle Theorem (also called the Central Angle Theorem) tells us:
So the central angle is always twice the inscribed angle subtending the same arc. This is one of the highest-weightage theorems in CBSE Class 9 and 10 circles.
We know ∠POQ = 70°. Plugging in:
∠PRQ = 35°
This holds for any point R on the major arc — whether R is near P, near Q, or anywhere in between. All inscribed angles on the same arc are equal.
Why This Works
Let’s understand the proof — this comes up in board exams as a “prove that” question, so knowing the logic matters.
Join OR and extend it to a point, say T, on the other side of the circle. Now we have two triangles: △OPR and △OQR. Since OP = OR = OQ (all radii), both triangles are isosceles.
In △OPR, ∠OPR = ∠ORP (base angles of isosceles triangle). Exterior angle ∠POT = ∠OPR + ∠ORP = 2∠ORP. Similarly, ∠QOT = 2∠ORQ. Adding both sides gives us ∠POQ = 2∠PRQ. That’s the entire proof — clean and elegant.
The proof has two cases: when the centre O lies inside angle PRQ (as above), and when O lies outside angle PRQ. CBSE examiners sometimes ask you to prove both cases. For the “outside” case, subtract instead of add: ∠POQ = ∠QOT − ∠POT = 2∠QOR − 2∠POR = 2∠PRQ.
Alternative Method
There’s a corollary worth memorising for faster MCQ solving:
Angle in a semicircle = 90°. This is the central angle theorem in disguise. If PQ is a diameter, central angle = 180°. Inscribed angle = 180° ÷ 2 = 90°. Every angle in a semicircle is a right angle.
For the original problem, you can also work backwards: if the inscribed angle were given as 35°, you’d find the central angle by multiplying by 2. This reverse approach is common in problems where the arc is given indirectly through a reflex angle or through the major arc’s central angle.
Common Mistake
Confusing which arc the point lies on. If R is on the minor arc (same side as the chord’s “short arc”), the angle formula changes. The angle at R would be:
That’s because R on the minor arc subtends the major arc PQ, whose central angle is the reflex angle = 360° − 70° = 290°. Many students apply 70°/2 = 35° regardless of where R sits — that’s the most common error on this theorem in board exams.
Quick check: Angles subtended on the major and minor arcs by the same chord always add up to 180°. Here, 35° + 145° = 180° ✓ — use this to verify your answer instantly.