Angles in a Circle Worksheets - Math Monks - Free Printable

Let’s solve each problem one by one using angle properties of circles — including:

- Angle at the center is twice the angle at the circumference subtended by the same arc.
- Angles in the same segment are equal.
- Angle in a semicircle is 90°.
- Tangent is perpendicular to radius at point of contact → 90°.
- Sum of angles in a triangle = 180°.
- Exterior angle of a triangle = sum of opposite interior angles.

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Problem 1

Given: ∠AOB = 172° (central angle)

We need: ∠ACB (angle at circumference subtended by arc AB)

✔ Rule: Angle at center = 2 × angle at circumference

So,

> ∠ACB = ½ × ∠AOB = ½ × 172° = 86°

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Problem 2

Given:
- ∠ABD = 28°
- ∠CBD = 54° → so ∠ABC = 28° + 54° = 82°
- Arc AD has equal marks → implies chords AD and CD are equal? Or arcs? But more importantly, we see that ∠ACD is an angle subtended by arc AD.

Wait — actually, look at triangle ABD and CBD. We’re to find ∠ACD.

Note: ∠ABD and ∠ACD are angles subtended by the same arc AD.

✔ Rule: Angles in the same segment are equal.

So, ∠ACD = ∠ABD = 28°

*(Since both are subtended by arc AD)*

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Problem 3

Given:
- ∠PSQ = 40°
- ∠QSR = 32°
- Need ∠PAQ

∠PSQ and ∠PAQ are both subtended by arc PQ.

✔ Rule: Angles in the same segment are equal.

So, ∠PAQ = ∠PSQ = 40°

*(They both subtend arc PQ)*

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Problem 4

Given:
- ∠EOF = 65° (central angle)
- Need ∠OFG

Note: Triangle OFG is isosceles because OF and OG are radii → OF = OG

So, ∠OFG = ∠OGF

Sum of angles in triangle OFG = 180°

So,

> ∠OFG + ∠OGF + ∠FOG = 180°
> 2 × ∠OFG + 65° = 180°
> 2 × ∠OFG = 115°
> ∠OFG = 57.5°

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Problem 5

Given:
- ∠BFD = 77°
- Need ∠ODB

Note: OB and OD are radii → triangle OBD is isosceles → ∠OBD = ∠ODB

Also, BF and DF are tangents from point F → so FB = FD, and FO bisects ∠BFD.

But more directly: In quadrilateral OBFD, angles at B and D are 90° each (tangent ⊥ radius).

So, ∠OBF = 90°, ∠ODF = 90°

In quadrilateral OBFD:

> Sum of angles = 360°
> ∠OBF + ∠BFD + ∠ODF + ∠BOD = 360°
> 90° + 77° + 90° + ∠BOD = 360°
> ∠BOD = 360° - 257° = 103°

Now, in triangle OBD (isosceles with OB = OD):

> ∠OBD + ∠ODB + ∠BOD = 180°
> 2 × ∠ODB + 103° = 180°
> 2 × ∠ODB = 77°
> ∠ODB = 38.5°

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Problem 6

Given:
- ∠POR = 68° (central angle)
- Need ∠PRQ

Note: RQ is tangent at R → so OR ⊥ RQ → ∠ORQ = 90°

We need ∠PRQ — which is part of triangle PRQ or angle between chord PR and tangent RQ.

✔ Alternate Segment Theorem: The angle between a tangent and a chord is equal to the angle subtended by the chord in the alternate segment.

So, ∠PRQ = angle subtended by arc PR in alternate segment → which is ∠PQR? Wait, no.

Actually, ∠PRQ = ∠RQP? Not quite.

Let me rephrase:

The angle between tangent RQ and chord RP is equal to the angle subtended by RP in the alternate segment — which is ∠RSP if S were on the circle... but here, we can use central angle.

Alternative approach:

In triangle ORP: OP = OR (radii) → isosceles → ∠OPR = ∠ORP

∠POR = 68° → so ∠OPR = ∠ORP = (180° - 68°)/2 = 56°

Now, since OR ⊥ RQ → ∠ORQ = 90°

Then, ∠PRQ = ∠ORQ - ∠ORP = 90° - 56° = 34°

✔ So, ∠PRQ = 34°

*(This also matches Alternate Segment Theorem: angle between tangent and chord = angle in alternate segment. The angle subtended by arc PR at circumference would be half of 68° = 34° — yes!)*

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Problem 7

Given:
- OA = 6 cm (radius)
- AB = 8 cm
- Need ∠OAB

Note: OA is radius, AB is tangent → so ∠OAB = 90°? Wait — is AB tangent?

Looking at diagram: Point A is on circle, OA is radius, and AB is drawn outward — likely tangent.

✔ If AB is tangent at A, then OA ⊥ AB → ∠OAB = 90°

But wait — the question gives lengths OA = 6, AB = 8 — perhaps it's not a tangent? Let me check.

If OA and AB are sides of triangle OAB, and OB is hypotenuse, then:

OA = 6, AB = 8 → if ∠OAB = 90°, then OB = √(6² + 8²) = √(36+64)=√100=10 — which is plausible.

But the diagram shows AB as a line from A to B outside the circle — very likely tangent.

✔ So, by definition of tangent: radius ⊥ tangent at point of contact → ∠OAB = 90°

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Problem 8

Given:
- ∠BAC = 95°
- ∠BOC = ? (central angle subtended by arc BC)
- ∠BDC = ?

First, ∠BAC is angle at circumference subtended by arc BC.

✔ Rule: Central angle = 2 × angle at circumference

So,

> ∠BOC = 2 × ∠BAC = 2 × 95° = 190°

Wait — that’s impossible in a circle! Maximum central angle is 360°, but 190° is possible if it’s reflex? But usually we take minor arc.

Actually, ∠BAC = 95° — this is an obtuse angle, so it must be subtended by the major arc BC, meaning the central angle for minor arc BC would be 2 × (180° - 95°)? No.

Wait — let’s think carefully.

In circle, angle at circumference = half the central angle subtended by the same arc.

If ∠BAC = 95°, then it must be subtended by arc BC — but 95° > 90°, so arc BC must be > 180° — i.e., major arc.

So, central angle for arc BC (major) = 2 × 95° = 190°

Then, the reflex angle at center is 190°, and the minor angle would be 360° - 190° = 170° — but the question asks for ∠BOC — which is typically the smaller angle unless specified.

But in diagram, point O is inside quadrilateral ABCD, and ∠BOC appears to be the reflex angle? Not sure.

Wait — perhaps ∠BAC is NOT subtended by arc BC? Let’s check.

Points: A, B, C, D on circle. ∠BAC is at A, between points B-A-C → so it subtends arc BC.

Yes.

So, if ∠BAC = 95°, then arc BC = 190° → so central angle ∠BOC = 190° (reflex).

But often in such problems, they expect the minor central angle — but 190° is valid.

Alternatively, maybe the diagram shows O as center, and ∠BOC is the angle inside the quadrilateral — which would be the reflex angle.

Let’s proceed.

Then, ∠BDC — this is angle at D subtended by arc BC.

Same arc BC → so ∠BDC should equal ∠BAC? Only if they are in the same segment.

But ∠BAC and ∠BDC are on opposite sides of chord BC — so they are in opposite segments.

✔ Rule: Angles subtended by same arc in opposite segments are supplementary.

So,

> ∠BDC + ∠BAC = 180°
> ∠BDC = 180° - 95° = 85°

And for ∠BOC — since ∠BAC = 95° subtends arc BC, central angle = 2 × 95° = 190°

So:

> ∠BOC = 190°
> ∠BDC = 85°

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## ✔ Final Answers:

1. ∠ACB = 86°
2. ∠ACD = 28°
3. ∠PAQ = 40°
4. ∠OFG = 57.5°
5. ∠ODB = 38.5°
6. ∠PRQ = 34°
7. ∠OAB = 90°
8. ∠BOC = 190°, ∠BDC = 85°

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Let me know if you’d like diagrams or step-by-step visuals for any problem!