Angles In A Circle Worksheet Awesome 14 Best Of Different Types ... - Free Printable

Let’s solve each problem one by one using circle geometry theorems. The key theorems we’ll use are:

- Inscribed Angle Theorem: An inscribed angle is half the measure of its intercepted arc.
- Angle formed by two chords intersecting inside a circle: Half the sum of the intercepted arcs.
- Angle formed by two secants, two tangents, or a secant and a tangent from a point outside the circle: Half the difference of the intercepted arcs.
- Tangent-Radius Theorem: A radius drawn to a point of tangency is perpendicular to the tangent line (90°).
- Total degrees in a circle = 360°.

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Problem ①

We have an angle x formed outside the circle by two secants. The intercepted arcs are 100° and 30°.

> Formula for external angle:
> \( x = \frac{1}{2} (\text{larger arc} - \text{smaller arc}) \)

So:
\[
x = \frac{1}{2}(100^\circ - 30^\circ) = \frac{1}{2}(70^\circ) = \boxed{35^\circ}
\]

✔ Answer: x = 35°

---

Problem ②

Two chords intersect inside the circle. The intercepted arcs are 70° and 120°.

> Formula for internal angle:
> \( x = \frac{1}{2} (\text{arc}_1 + \text{arc}_2) \)

So:
\[
x = \frac{1}{2}(70^\circ + 120^\circ) = \frac{1}{2}(190^\circ) = \boxed{95^\circ}
\]

✔ Answer: x = 95°

---

Problem ③

This is a triangle with vertex C outside the circle, and sides CA and CB are tangents touching the circle at A and B. The arc AB (not containing C) is 240°, so the minor arc AB is:

\[
360^\circ - 240^\circ = 120^\circ
\]

> Formula for angle between two tangents from external point:
> \( x = \frac{1}{2} (\text{major arc} - \text{minor arc}) \)

Wait — actually, since both tangents touch the circle, the angle at C is:

\[
x = \frac{1}{2} (\text{measure of major arc AB} - \text{measure of minor arc AB})
\]

But note: the major arc is 240°, minor arc is 120°, so:

\[
x = \frac{1}{2}(240^\circ - 120^\circ) = \frac{1}{2}(120^\circ) = \boxed{60^\circ}
\]

✔ Answer: x = 60°

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Problem ④

External angle x formed by a secant and a tangent. The intercepted arcs are 130° and 60°.

> Formula: \( x = \frac{1}{2}(\text{larger arc} - \text{smaller arc}) \)

Here, the 130° arc is intercepted by the secant, and the 60° arc is likely the one “between” the tangent and secant? Actually, let’s interpret carefully.

Actually, in this diagram, the angle is formed by a tangent and a secant. The intercepted arcs are the far arc (130°) and the near arc (60°).

So:

\[
x = \frac{1}{2}(130^\circ - 60^\circ) = \frac{1}{2}(70^\circ) = \boxed{35^\circ}
\]

✔ Answer: x = 35°

---

Problem ⑤

This shows a central angle (vertex at center) and an inscribed angle sharing one side. The inscribed angle intercepts an arc of 110°.

The angle marked x is formed by a radius and a chord — it's an inscribed angle? Wait — no, the vertex is on the circle, and one side is a radius? That doesn’t make sense unless it’s part of a triangle.

Actually, looking closely: there’s a vertical radius and another chord forming angle x with the 110° arc.

Wait — the 110° arc is intercepted by the inscribed angle. But angle x appears to be adjacent to that.

Actually, if the 110° arc is intercepted by an inscribed angle, then that inscribed angle would be half of 110° = 55°.

But angle x seems to be part of a right triangle? No — perhaps it’s the angle between the radius and the chord.

Alternative interpretation: the angle x is formed by a radius and a chord, and the intercepted arc is 110°. Then, since the radius to the endpoint makes an isosceles triangle, but that might overcomplicate.

Wait — simpler: the 110° arc is opposite angle x, which is an inscribed angle. So:

\[
x = \frac{1}{2} \times 110^\circ = \boxed{55^\circ}
\]

Yes — because any inscribed angle intercepting a 110° arc is 55°.

✔ Answer: x = 55°

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Problem ⑥

Two chords intersect inside the circle. The intercepted arcs are 50° and x. The angle formed is 40°.

> Formula: \( \text{angle} = \frac{1}{2}(\text{arc}_1 + \text{arc}_2) \)

So:

\[
40^\circ = \frac{1}{2}(50^\circ + x)
\]

Multiply both sides by 2:

\[
80^\circ = 50^\circ + x
\Rightarrow x = 80^\circ - 50^\circ = \boxed{30^\circ}
\]

✔ Answer: x = 30°

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Problem ⑦

Angle x is an inscribed angle intercepting an arc. The other two arcs are 140° and 100°, so the intercepted arc for angle x is:

\[
360^\circ - 140^\circ - 100^\circ = 120^\circ
\]

Then, inscribed angle = half the intercepted arc:

\[
x = \frac{1}{2} \times 120^\circ = \boxed{60^\circ}
\]

✔ Answer: x = 60°

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Problem ⑧

This shows a circle with a horizontal diameter and two vertical lines tangent to the circle at left and right ends. The angle marked is 2x + 10°, and it’s formed between the tangent and the diameter.

> Tangent-Radius Theorem: Radius ⊥ Tangent → 90°

So, the angle between the tangent and the diameter (which is a radius extended) must be 90°.

Therefore:

\[
2x + 10^\circ = 90^\circ
\Rightarrow 2x = 80^\circ
\Rightarrow x = \boxed{40^\circ}
\]

✔ Answer: x = 40°

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## ✔ Final Answers:

① 35°
② 95°
③ 60°
④ 35°
⑤ 55°
⑥ 30°
⑦ 60°
⑧ 40°

Let me know if you want diagrams or step-by-step visuals for any!