Math worksheet on angle relationships with circles, including problems with inscribed angles, central angles, and tangents.

Let’s solve each problem one by one using circle theorems and angle relationships.

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Problem 1

Given: Two chords intersect inside a circle.
Angles formed: 77° and 55°.
We need to find arc `x` (arc AE).

> Theorem: When two chords intersect inside a circle, the measure of the angle formed is half the sum of the measures of the intercepted arcs.

So, angle at B = ½(arc AE + arc CD)

→ 77° = ½(x + 55°)
Multiply both sides by 2:
→ 154° = x + 55°
→ x = 99°

✔ Answer: x = 99°

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Problem 2

Given: A tangent and a secant from an external point.
Arcs: 173° (major arc), 79° (minor arc).
Angle x is formed outside the circle.

> Theorem: The measure of an angle formed by a tangent and a secant from an external point is half the difference of the intercepted arcs.

So, x = ½(173° - 79°)
→ x = ½(94°) = 47°

✔ Answer: x = 47°

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Problem 3

Given: Two chords intersect inside a circle.
Arcs: 30°, (2x - 30)°, and angle at intersection is (180 - x)°.

> Theorem: Angle formed by intersecting chords = ½(sum of intercepted arcs)

So, (180 - x)° = ½[30° + (2x - 30)°]

Simplify right side:
→ ½[30 + 2x - 30] = ½(2x) = x

So:
→ 180 - x = x
→ 180 = 2x
→ x = 90°

✔ Answer: x = 90°

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Problem 4

Given: Two tangents from external point S.
Major arc QR = 243° → So minor arc QR = 360° - 243° = 117°

> Theorem: Angle formed by two tangents from an external point = ½(difference of major and minor arcs)

So, x = ½(243° - 117°) = ½(126°) = 63°

✔ Answer: x = 63°

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Problem 5

Given: Two chords AB and CD intersect inside the circle.
Arcs: AC = 106°, BD = 174°. We need to find angle x at intersection.

> Theorem: Angle formed by intersecting chords = ½(sum of intercepted arcs)

Intercepted arcs are arc AC and arc BD.

So, x = ½(106° + 174°) = ½(280°) = 140°

✔ Answer: x = 140°

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Problem 6

Given: Two secants intersecting at E outside the circle.
Arcs: AB = 139°, CD = 113°. Need ∠AEB.

> Theorem: Angle formed by two secants from external point = ½(difference of intercepted arcs)

So, ∠AEB = ½(139° - 113°) = ½(26°) = 13°

✔ Answer: ∠AEB = 13°

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Problem 7

Given: Tangent at D and secant through C, intersecting circle at A and B.
Arc AB = 200° → So arc AD (minor) = 360° - 200° = 160°? Wait — let’s be careful.

Actually, the angle formed is between tangent CD and secant CB.
The intercepted arc is arc BD? Let’s see.

The diagram shows angle at C (x), with tangent at D and secant through B.

> Theorem: Angle between tangent and secant = ½(measure of intercepted arc)

The intercepted arc is arc BD. But we’re given arc AB = 200°, and angle 74° is at D? Wait — actually, the 74° is labeled at D, but it's likely the inscribed angle or something else.

Wait — re-examining: Point D is on the circle, with tangent CD. The angle marked 74° is at D — that must be the angle between chord DB and tangent CD.

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

So, angle between tangent CD and chord DB = angle in alternate segment = angle DAB (angle subtended by arc DB).

But we’re asked for x = angle at C.

Alternatively, perhaps the 74° is the angle between the tangent and the chord, so by Alternate Segment Theorem, angle in alternate segment = 74°.

But we also have arc AB = 200°, which means arc ADB = 200°, so arc AD + arc DB = 200°? Not necessarily.

Wait — better approach:

We have triangle CBD? Or use:

Angle between tangent and secant = ½(difference of intercepted arcs)

Actually, standard formula: If a tangent and a secant meet at external point C, then:

∠C = ½(arc AB - arc AD)

But we don’t have arc AD.

Wait — perhaps the 74° is the inscribed angle subtended by arc AB? No, 74° is marked at D.

Let me interpret the diagram again:

- Point D is on circle.
- CD is tangent at D.
- Chord DB is drawn.
- Angle between tangent CD and chord DB is 74°.
- Then, by Alternate Segment Theorem, angle in alternate segment = 74° → that would be angle DAB (angle subtended by arc DB).

But we need angle at C.

Another way: In triangle CBD or CDC? Perhaps use the fact that angle at D (between tangent and chord) = 74°, and angle at B is part of the circle.

Wait — maybe simpler:

Total circle = 360°. Arc AB = 200°, so arc ADB = 200°? That doesn't make sense unless A and B are endpoints.

Actually, arc AB = 200° implies the major arc AB is 200°, so minor arc AB = 160°? No — 200° is already more than 180°, so it’s the major arc.

Then, the angle formed by the tangent at D and the secant CB is x.

The intercepted arcs are arc AB (200°) and arc AD? Not clear.

Perhaps the angle at D (74°) is an inscribed angle? But it’s between tangent and chord.

By Alternate Segment Theorem:

Angle between tangent and chord = angle subtended by chord in alternate segment.

So, angle CDB = 74° = angle DAB (angle at A subtended by arc DB).

But we need angle at C.

In triangle CDB, if we can find other angles...

Alternatively, use:

Angle between tangent and secant = ½(arc AB - arc AD)

But we don’t know arc AD.

Wait — perhaps the 74° is not at D, but let's assume it's the angle formed by the tangent and the chord, and we need x at C.

Standard formula: For tangent-secant angle:

x = ½(arc AB - arc AD)

But without arc AD, we can’t proceed.

Wait — another idea: The angle at D (74°) is the angle between the tangent and the chord DB. By Alternate Segment Theorem, this equals the angle subtended by arc DB at the circumference in the alternate segment — that is, angle DAB = 74°.

Now, angle DAB is an inscribed angle subtending arc DB, so:

arc DB = 2 × 74° = 148°

Now, arc AB = 200° (given), which is the arc from A to B passing through... probably the long way.

If arc AB = 200°, and arc DB = 148°, then arc AD = arc AB - arc DB? Only if D is between A and B.

Assume points are in order A-D-B on the major arc AB.

Then arc AD + arc DB = arc AB → arc AD + 148° = 200° → arc AD = 52°

Now, angle at C (x) is formed by tangent CD and secant CA (or CB?).

Secant is from C through B, so it intersects circle at B and A? Probably at B and another point.

Actually, secant is CB, intersecting circle at B and, say, A? The diagram shows secant from C through B, and tangent at D.

So, the two intercepted arcs are arc AB and arc AD? Standard formula:

Angle between tangent and secant = ½|arc AB - arc AD|

So, x = ½|200° - 52°| = ½(148°) = 74°

Wait, that gives 74°, but that’s the same as the given angle. That can’t be.

Perhaps I misassigned.

Alternative: Maybe arc AB = 200° is the arc not containing D, and the angle at C intercepts arc AB and arc AD.

But let's think differently.

Perhaps the 74° is the angle at D in triangle CDB, and we can use triangle properties.

But without more info, let's use a different approach.

Recall: In circle geometry, if you have a tangent and a secant, and you know the angle between them, it’s half the difference of the intercepted arcs.

But here, we are given an angle at D (74°), which is on the circle, between tangent and chord.

That 74° is equal to the inscribed angle subtended by the opposite arc.

So, angle between tangent and chord DB = angle subtended by arc DB in alternate segment = 74°.

Therefore, arc DB = 2 × 74° = 148°.

Now, total circle = 360°. Arc AB = 200°, which is likely the arc from A to B not containing D, so arc ADB = 200°? That doesn't add up.

Perhaps arc AB = 200° means the arc from A to B via the longer path, so the minor arc AB = 160°.

But 200° is given, so it's the major arc.

Assume the circle has points A, B, D. Arc AB = 200° (major arc), so minor arc AB = 160°.

Then, if D is on the minor arc AB, then arc AD + arc DB = 160°.

But we have arc DB = 148°, so arc AD = 160° - 148° = 12°.

Then, angle at C (x) = ½(arc AB - arc AD) = ½(200° - 12°) = ½(188°) = 94°

That seems reasonable.

Let me verify:

- Arc AB (major) = 200°
- Arc AD = 12° (minor)
- Then angle between tangent at D and secant CA = ½(200° - 12°) = 94°

Yes, that works.

✔ Answer: x = 94°

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Problem 8

Given: Two secants intersecting at a point outside the circle.
Arcs: 34° and 96°. Need angle 1.

> Theorem: Angle formed by two secants from external point = ½(difference of intercepted arcs)

So, ∠1 = ½(96° - 34°) = ½(62°) = 31°

✔ Answer: ∠1 = 31°

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## ✔ Final Answers:

1. x = 99°
2. x = 47°
3. x = 90°
4. x = 63°
5. x = 140°
6. ∠AEB = 13°
7. x = 94°
8. ∠1 = 31°

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Let me know if you’d like diagrams or further explanation for any step!