Math worksheet on angle properties of circles featuring eight geometric diagrams with labeled angles and sides to solve for unknown angles.

Explanation:
Let’s solve each problem one by one using circle geometry rules:

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1. ∠ACB = ?
We have a circle with center O. Points A, B, C, D lie on the circle (cyclic quadrilateral). ∠AOB = 172° (central angle).
∠ACB is an inscribed angle that subtends the same arc AB as central angle ∠AOB.
Inscribed angle = ½ × central angle for the same arc.
But note: ∠ACB subtends arc AB *the long way around*? Wait — look at diagram: point C is on the major arc AB (since ∠AOB = 172° is the minor central angle), so ∠ACB subtends the minor arc AB, which corresponds to central angle 172°.
So inscribed angle = ½ × 172° = 86°.

✔ Check: In a circle, opposite angles of a cyclic quadrilateral sum to 180°, but we don’t need that here. Direct inscribed angle rule applies.

Answer: 86°

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2. ∠ACD = ?
Circle with points A, B, C, D. Given: ∠BAD = 28°, ∠ABD = 54°, and AD = CD (marked with tick marks → chords equal ⇒ arcs equal ⇒ inscribed angles subtending those arcs are equal). So ∠ABD = ∠CBD? Wait — tick marks on AD and CD mean AD = CD ⇒ triangle ADC is isosceles with AD = CD ⇒ base angles at A and C are equal: ∠DAC = ∠DCA.

But let’s use triangle ABD first:
In △ABD, angles: ∠BAD = 28°, ∠ABD = 54°, so ∠ADB = 180 − 28 − 54 = 98°.

Now, ∠ADB and ∠ACB are inscribed angles subtending same arc AB → they should be equal? No — both subtend arc AB, so yes: ∠ACB = ∠ADB = 98°? But we need ∠ACD.

Alternative approach: Since AD = CD, arcs AD and CD are equal ⇒ inscribed angles subtending them are equal. So ∠ABD (subtends arc AD) = ∠CBD (subtends arc CD). So ∠CBD = 54°.

Then in triangle ABC, we know ∠ABC = ∠ABD + ∠DBC = 54° + 54° = 108°.
Also, ∠BAC = ∠BAD = 28° (same angle).
So in △ABC: ∠ACB = 180 − 108 − 28 = 44°.

But we want ∠ACD. Note: points A–C–D: ∠ACD is part of triangle ACD.

Since AD = CD, triangle ACD is isosceles with AD = CD ⇒ ∠DAC = ∠DCA = ∠ACD.

What is ∠DAC? That’s angle between DA and AC. We know ∠BAC = 28°, and if we find ∠BAD = 28°, then point D is such that A–D–B is triangle. Actually, ∠DAC = ∠BAC − ∠BAD? No — D is not on AB.

Better: Use cyclic quadrilateral ABCD. Opposite angles sum to 180°:
∠BAD + ∠BCD = 180° → 28° + ∠BCD = 180° → ∠BCD = 152°.
But ∠BCD = ∠BCA + ∠ACD. We found earlier ∠BCA = ∠ACB = 44° (from triangle ABC). So:
152° = 44° + ∠ACD → ∠ACD = 108°.

Wait — does that make sense? Let's verify with another method:

From earlier, ∠ADB = 98°. Since ABCD is cyclic, ∠ACB = ∠ADB = 98° (same arc AB). So ∠ACB = 98°, not 44°! I made a mistake above.

Yes! Key rule: Angles subtended by same chord (AB) at the circumference are equal. So ∠ADB and ∠ACB both subtend chord AB → they are equal. So ∠ACB = 98°.

Now, in cyclic quadrilateral ABCD: ∠BAD + ∠BCD = 180° → 28° + ∠BCD = 180° → ∠BCD = 152°.

But ∠BCD = ∠BCA + ∠ACD = ∠ACB + ∠ACD = 98° + ∠ACD
So: 98° + ∠ACD = 152° → ∠ACD = 54°.

Also, AD = CD ⇒ triangle ACD is isosceles ⇒ ∠DAC = ∠DCA = ∠ACD = 54°. Then ∠ADC = 180 − 54 − 54 = 72°. Does that match? ∠ADB = 98°, and ∠BDC = ? Not needed.

So final: ∠ACD = 54°

✔ Confirmed.

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3. ∠PAQ = ?
Points P, Q, R, S on circle, center A? Wait — label shows A inside, not center. Actually, point A is inside, and angles given: ∠PSR = 40°, ∠QRP = 32°. We’re to find ∠PAQ.

Observe: ∠PSR = 40° is inscribed angle subtending arc PR. So arc PR = 2 × 40° = 80°.
∠QRP = 32° is inscribed angle subtending arc QP. So arc QP = 2 × 32° = 64°.

Now ∠PAQ is a central angle? No — A is not marked as center. But in diagram, A is intersection of chords PQ and SR? Actually, looks like A is intersection point of diagonals PR and QS? Wait — labels: triangle PAQ, with A inside circle. Likely A is intersection of chords PR and QS.

Standard theorem: Angle formed by two intersecting chords = ½ (sum of intercepted arcs).
∠PAQ is formed by chords AP and AQ, intercepting arcs PQ and SR? Let's identify:

When two chords intersect inside a circle (at A): angle between them = ½ (arc opposite + arc opposite). Specifically, ∠PAQ intercepts arcs PQ and SR.

Yes: chords PS and QR intersect at A? Hmm.

Given angles at circumference: ∠PSR = 40° → subtends arc PR.
∠QRP = 32° → subtends arc QP.

We want ∠PAQ. If A is intersection of chords PR and QS, then ∠PAQ is angle between PA and QA, i.e., between chords PR and QS. Then it intercepts arcs PQ and SR.

So ∠PAQ = ½ (arc PQ + arc SR).

We have arc PQ = 2 × ∠PRQ? Wait, ∠PRQ is same as ∠QRP = 32°, which subtends arc QP → arc QP = 64°.

What about arc SR? ∠SPR or ∠SQR? We know ∠PSR = 40°, which subtends arc PR → arc PR = 80°.

We need arc SR. Note that full circle = 360°. Arcs: PQ + QR + RS + SP = 360°.

We know arc PQ = 64°.
Arc PR = arc PQ + arc QR = 80° → so arc QR = 80° − 64° = 16°.

Now, ∠PSR = 40° subtends arc PR = 80°, correct.

What about arc SR? That’s arc from S to R passing through Q? Or minor arc SR? Likely minor arc SR = arc SQ + QR? Not clear.

Alternative: Maybe A is the center! Look again: In diagram 3, point A is labeled near center, and there’s a small dot — possibly center. And angles given at circumference: ∠PSR = 40°, ∠QRP = 32°. If A is center, then ∠PAQ is central angle subtending arc PQ.

Then arc PQ = 2 × inscribed angle subtending it. Which inscribed angle subtends arc PQ? ∠PRQ = 32° (given as ∠QRP) — yes, ∠QRP is at R, looking at Q and P, so it subtends arc QP. So arc QP = 2×32° = 64°, so central angle ∠PAQ = 64°.

That’s simplest and matches typical worksheet design.

Also, ∠PSR = 40° subtends arc PR → arc PR = 80°, consistent.

So assume A is center → ∠PAQ = 64°

✔ We'll go with 64°.

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4. ∠OFG = ?
Circle, center O. Points E, F, G on circle. Given ∠FOG = 65° (central angle). Triangle OFG is isosceles (OF = OG = radius). So base angles ∠OFG = ∠OGF.

Sum of angles in triangle: ∠FOG + 2×∠OFG = 180°
→ 65° + 2x = 180° → 2x = 115° → x = 57.5°

Answer: 57.5°

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5. ∠ODB = ?
Circle, center O. Points B, D on circle. Triangle ODB: OD = OB = radius ⇒ isosceles. Given external angle at F: ∠BFD = 77°, and F is outside, with FB and FD tangent? Wait — diagram shows triangle BFD with F outside, BD chord, and O connected to B and D. Likely FB and FD are tangents? But not marked.

Actually, common configuration: If two tangents from point F touch circle at B and D, then ∠BFD = 77°, and angle between tangents = 180° − central angle ∠BOD. So ∠BOD = 180° − 77° = 103°.

Then in triangle OBD, isosceles, ∠ODB = ∠OBD = (180 − 103)/2 = 77/2 = 38.5°

Yes, standard result: angle between two tangents = 180° − central angle.

So ∠ODB = 38.5°

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6. ∠PRQ = ?
Circle, center O. Point R on circle, P on circle, Q outside on tangent line at R. Given ∠PRO = 80°, where O is center, so OR ⟂ RQ (radius perpendicular to tangent). So ∠ORQ = 90°.

In triangle PRO: ∠PRO = 80°, ∠POR = ? Not given. But we want ∠PRQ = angle between PR and RQ.

Since ∠ORQ = 90°, and ∠PRO = 80°, then ∠PRQ = ∠ORQ − ∠ORP = 90° − 80° = 10°.

Because at point R, angles around: ∠PRQ + ∠PRO = ∠ORQ = 90°, if O–R–Q is right angle and P is on other side.

Yes: PR and OR form 80°, OR and RQ form 90°, so PR and RQ form 10°.

Answer: 10°

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7. ∠OAB = ?
Circle, center O. OA = 6 cm (radius), AB = 8 cm, and OB is not given, but B is outside? Actually, diagram shows OA radius, AB tangent? Because AB touches circle at A, and OA ⟂ AB (radius to tangent). So ∠OAB = 90°.

But wait: if AB is tangent at A, then yes, radius OA ⟂ tangent AB → ∠OAB = 90°.

However, length AB = 8 cm, OA = 6 cm — maybe triangle OAB is right-angled at A, so ∠OAB = 90°.

Yes, standard.

Answer: 90°

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8. ∠BOC = ? and ∠BDC = ?
Circle, center O. Given ∠BAC = 95° (inscribed angle). ∠BAC subtends arc BC. So arc BC = 2 × 95° = 190°.

Central angle ∠BOC subtends same arc BC → ∠BOC = 190°.

But central angles are usually taken as the smaller one (<180°)? However, arc can be major arc. Since inscribed angle is 95° > 90°, it must subtend major arc (>180°), so central angle is 190° — acceptable.

Now ∠BDC: point D on circle, so ∠BDC is inscribed angle subtending arc BC as well? Let's see: points B–D–C, so ∠BDC subtends arc BC (the opposite arc, not containing D). If D is on the opposite side of BC from A, then ∠BDC subtends the *minor* arc BC.

Total circle = 360°, major arc BC = 190°, so minor arc BC = 360 − 190 = 170°.

Inscribed angle subtending minor arc BC = ½ × 170° = 85°.

Is D on the minor arc side? Diagram shows quadrilateral ABCD with A and D on opposite sides of BC, and ∠BAC = 95°, so A is on major arc side, D on minor arc side → ∠BDC = 85°.

Thus:
∠BOC = 190°
∠BDC = 85°

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Now compile all answers:

1. 86°
2. 54°
3. 64°
4. 57.5°
5. 38.5°
6. 10°
7. 90°
8. ∠BOC = 190°, ∠BDC = 85°

Double-check #1: Central angle 172°, inscribed angle on same side = 86° — correct.
#2: Verified via cyclic quadrilateral → 54° — correct.
#3: Assuming A is center → 64° — most plausible.
Others standard.

Final Answer:
1. 86°
2. 54°
3. 64°
4. 57.5°
5. 38.5°
6. 10°
7. 90°
8. ∠BOC = 190°, ∠BDC = 85°