NCERT Solutions for Class 9th Maths Chapter 5 Exercise Set 5.6 — Angle subtended by an arc at a point on the circle outside the arc

Book page 110–111 Updated on2026-09-08

Q1.

In a circle with centre O, the central angle AOB is 60°. If the radius of the circle is 12 cm, what is the length of the chord AB?

Answer

AB = 12 cm — the chord equals the radius.

In ΔOAB: OA = OB = 12 cm (radii) ⇒ the triangle is isosceles
∠OAB = ∠OBA (base angles of an isosceles triangle)
∠OAB + ∠OBA + 60° = 180°
2 ∠OAB = 120° ⇒ ∠OAB = ∠OBA = 60°

All three angles are 60°, so ΔOAB is equilateral
∴ AB = OA = OB = 12 cm

Check with the chord formula: the distance of AB from O is d = 12 cos 30° = 6√3, so AB = 2√(144 − 108) = 2√36 = 12 cm ✓

Worth remembering: a central angle of 60° always gives a chord equal to the radius. That is why a regular hexagon inscribed in a circle has side equal to the radius — six 60° slices fill up 360°.

Q2.

Let A and B be two points on a circle with centre O. (i) Are there points X, Y on the circle, on the same side of AB, such that ∠AXB is different from ∠AYB? (ii) Is it true that if ∠AXB = ∠AYB, then X and Y lie on the same side of the circle? (iii) If ∠AXB = ∠AYB, and X and Y do not lie on the circle, does the circle through A, B and X also pass through Y?

Answer

(i) No. If X and Y are both on the circle and on the same side of the chord AB, they lie on the same arc, so ∠AXB = ∠AYB always.

Both X and Y lie outside the same arc AB, so by Theorem 9
∠AXB = ½ (angle subtended by that arc at O) = ∠AYB
There is no freedom left — the angle cannot vary.

(ii) No, this is not always true. Equal angles do not force X and Y onto the same side of AB.

If X is on the major arc and Y on the minor arc, then AXBY is a cyclic 4-gon, so
∠AXB + ∠AYB = 180° (Theorem 11)
These two can be equal only when each is 90°, i.e. when AB is a diameter.

Counterexample: let AB be a diameter. Take X above AB and Y below it.
∠AXB = ∠AYB = 90° (angle in a semicircle), yet X and Y are on opposite sides.

So the correct statement is: if ∠AXB = ∠AYB and AB is not a diameter, then X and Y must be on the same side; if AB is a diameter, they need not be.

(iii) Yes, provided X and Y lie on the same side of the line AB. This is exactly Theorem 10.

A, B, X are non-collinear ⇒ a circle passes through A, B, X (Theorem 1)
∠AXB = ∠AYB and X, Y are on the same side of AB
⇒ by Theorem 10, A, B, X, Y are concyclic
⇒ the circle through A, B, X also passes through Y

Why the "same side" condition cannot be dropped: reflect Y in the line AB to get Y′. Then ∠AY′B = ∠AYB, but Y′ generally lies on a different circle through A and B. Theorem 10's proof works by ruling out "Y inside" and "Y outside" using the exterior-angle theorem, and that argument needs Y on the same side as X.

Q3.

Find x in Fig. 5.26.

Answer

x = 80°.

In Fig. 5.26 the four points A, D, C, B lie on a circle, so ADCB is a cyclic quadrilateral. The angle 100° is marked at D and the angle x is marked at B, and D and B are opposite vertices.

By Theorem 11, opposite angles of a cyclic quadrilateral add up to 180°:
∠ADC + ∠ABC = 180°
100° + x = 180°
x = 180° − 100° = 80°

ADCB is cyclic, so the angles at the opposite vertices D and B add to 180°.

Why opposite angles add to 180°: ∠ADC stands on the arc ABC and ∠ABC stands on the arc ADC. Together those two arcs make up the whole circle, so the two central angles add to 360°. Each inscribed angle is half its central angle, so the two inscribed angles add to ½ × 360° = 180°.

Check it yourself: the other pair must also add to 180°. So ∠DAB + ∠DCB = 180° here as well, whatever those two individual values happen to be.

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