11-Secant-Tangent and Tangent-Tangent Angles - Kuta Software - Free Printable

Let's solve each problem step by step using the Secant-Tangent and Tangent-Tangent Angle Theorems.

We'll use the following key theorems:

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🔷 Tangent-Secant Angle Theorem (Outside the Circle):

> The measure of an angle formed by a tangent and a secant drawn from a point outside the circle is half the difference of the measures of the intercepted arcs.

$$
\text{Angle} = \frac{1}{2}(\text{larger arc} - \text{smaller arc})
$$

---

🔷 Tangent-Tangent Angle Theorem (Outside the Circle):

> The measure of an angle formed by two tangents drawn from a point outside the circle is half the difference of the measures of the intercepted arcs.

$$
\text{Angle} = \frac{1}{2}(\text{major arc} - \text{minor arc})
$$

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🔷 Tangent-Chord Angle Theorem:

> The angle between a tangent and a chord is equal to the measure of the inscribed angle on the opposite side of the chord.

But in most of these problems, we’ll use the outside angle formulas.

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Let’s go through each problem:

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1)

Given: A tangent at $ E $, a secant from $ F $ through $ G $, angle $ \angle FEG = 76^\circ $

We are to find the measure of arc $ EG $.

This is a tangent-secant angle, so:
$$
\angle FEG = \frac{1}{2}(\text{arc } EG - \text{arc } FG)
$$

Wait — actually, look closely: the angle is formed between tangent and secant from point $ E $. But since the secant goes from $ F $ to $ G $, and $ E $ is on the circle, the tangent touches at $ E $, and the secant goes from $ E $ to $ G $, passing through $ F $? Wait — let's interpret.

Actually, the diagram shows:

- Tangent at $ E $
- Secant from $ F $ through $ E $ and $ G $ → so $ F $ is outside, $ E $ and $ G $ on circle.
- So $ \angle FEG $ is formed at point $ E $ by tangent and secant.

But that can't be — because the tangent and secant meet at $ E $, which is on the circle. That makes it a tangent-chord angle.

So this is a tangent-chord angle: the angle between tangent and chord $ EG $.

Then, by the Tangent-Chord Angle Theorem:
> The angle formed between a tangent and a chord is equal to the measure of the inscribed angle subtended by the same arc.

But more directly:
$$
\angle FEG = \frac{1}{2} \times \text{arc } EG
$$

Wait — no! Actually, the tangent-chord angle equals half the intercepted arc.

Yes:
$$
\angle FEG = \frac{1}{2} \times \text{arc } EG
$$

Given $ \angle FEG = 76^\circ $, then:
$$
76^\circ = \frac{1}{2} \times \text{arc } EG \Rightarrow \text{arc } EG = 152^\circ
$$

✔ Answer: 152°

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2)

Triangle with $ D $ outside, tangents to circle at $ T $ and $ S $, $ DT = DS $, $ \angle D = 50^\circ $

Find arc $ ST $

Two tangents from external point $ D $ → tangent-tangent angle.

The angle formed by two tangents is:
$$
\angle D = \frac{1}{2}(\text{major arc } ST - \text{minor arc } ST)
$$

Let $ x = \text{minor arc } ST $

Then major arc $ ST = 360^\circ - x $

So:
$$
50^\circ = \frac{1}{2}((360^\circ - x) - x) = \frac{1}{2}(360^\circ - 2x) = 180^\circ - x
$$

Solve:
$$
50 = 180 - x \Rightarrow x = 130^\circ
$$

So minor arc $ ST = 130^\circ $

✔ Answer: 130°

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3)

Point $ Y $ outside, tangent from $ Y $ touching at $ X $, secant from $ Y $ through $ Z $ and $ Q $, arc $ QZ = 146^\circ $

Find angle $ \angle XYZ $

This is tangent-secant angle from point $ Y $.

Intercepted arcs:
- Major arc: $ QXZ $ (but not labeled), but we know arc $ QZ = 146^\circ $
- The other arc is $ QXZ $, which is the rest of the circle?

Wait: the secant goes from $ Y $ through $ Z $ and $ Q $, so it intersects circle at $ Z $ and $ Q $. The tangent touches at $ X $. So the angle $ \angle XYZ $ intercepts arc $ ZX $ and arc $ QX $? Let's think.

Standard rule:
$$
\angle = \frac{1}{2}(\text{far arc} - \text{near arc})
$$

From point $ Y $, tangent to $ X $, secant to $ Q $ and $ Z $, with $ Z $ closer to $ Y $ than $ Q $? Usually, the order is $ Y $–$ Z $–$ Q $, so $ Z $ is near, $ Q $ is far.

But the arc intercepted by the angle is the arc between the two points where the secant hits, i.e., arc $ QZ $, and the arc from $ X $ to $ Z $?

Wait — better: the angle formed by tangent and secant from external point $ Y $, with tangent touching at $ X $, secant intersecting at $ Z $ and $ Q $, then the intercepted arcs are:

- The far arc: arc $ QX $ (not including $ Z $)
- The near arc: arc $ ZX $

But we're told arc $ QZ = 146^\circ $

That means arc $ QZ $ is the minor arc from $ Q $ to $ Z $, which is 146°.

But we need the arc from $ Q $ to $ X $ and $ Z $ to $ X $? Not clear.

Wait — perhaps the arc $ QZ $ is the one not containing $ X $, and $ X $ is on the other side.

Assume the circle has points $ Q $, $ Z $, and $ X $. Arc $ QZ = 146^\circ $, so the remaining arc $ ZXQ $ (going the other way) is $ 360^\circ - 146^\circ = 214^\circ $

Now, the tangent is at $ X $, and secant is from $ Y $ through $ Z $ and $ Q $. So the angle $ \angle XYZ $ is formed by tangent $ YX $ and secant $ YZQ $.

Then, the intercepted arcs are:
- The arc between the two intersection points: $ QZ $
- And the arc from $ X $ to $ Z $, but we need to identify which arc is intercepted.

Actually, standard formula:

$$
\angle = \frac{1}{2}(\text{intercepted arc} - \text{opposite arc})
$$

More precisely:
$$
\angle = \frac{1}{2}(\text{arc } QX - \text{arc } ZX)
$$

But we don’t have those.

Wait — better: when you have a tangent from $ Y $ at $ X $, and a secant from $ Y $ through $ Z $ and $ Q $, the angle $ \angle XYZ $ intercepts arc $ QX $ (the arc from $ Q $ to $ X $ not containing $ Z $), and the near arc is $ ZX $?

No — actually, the correct approach:

The angle formed by tangent and secant is half the difference of the intercepted arcs:
$$
\angle = \frac{1}{2}(\text{arc } QX - \text{arc } ZX)
$$

But we don’t know those.

Wait — maybe arc $ QZ = 146^\circ $ is the arc not containing $ X $, and $ X $ is on the other side.

Then, the far arc is the one not containing $ Z $ and $ Q $, i.e., arc $ QXZ $ going around the other way.

But actually, the two arcs between $ Q $ and $ Z $ are:
- Minor arc $ QZ = 146^\circ $
- Major arc $ QZ $ (through $ X $) = $ 360^\circ - 146^\circ = 214^\circ $

Now, for tangent-secant angle at $ Y $, the formula is:
$$
\angle = \frac{1}{2}(\text{major arc } QZ - \text{minor arc } QZ)
$$

Wait — no. The angle formed by tangent and secant intercepts one arc (the arc between the two points), and the formula is:

$$
\angle = \frac{1}{2}(\text{measure of intercepted arc} - \text{measure of opposite arc})
$$

Actually, the correct rule is:

> The measure of an angle formed by a tangent and a secant drawn from a point outside the circle is half the difference of the measures of the intercepted arcs.

In this case, the tangent touches at $ X $, the secant passes through $ Z $ and $ Q $, so the two intercepted arcs are:
- Arc $ QX $ (from $ Q $ to $ X $, not passing through $ Z $)
- Arc $ ZX $ (from $ Z $ to $ X $, not passing through $ Q $)

But we only know arc $ QZ = 146^\circ $, which is from $ Q $ to $ Z $, not through $ X $.

So arc $ QZ = 146^\circ $, which is the minor arc between $ Q $ and $ Z $, so the major arc $ QZ $ (passing through $ X $) is $ 360^\circ - 146^\circ = 214^\circ $

Now, the tangent is at $ X $, so the angle $ \angle XYZ $ intercepts:
- The arc from $ Q $ to $ Z $ not containing $ X $: that's 146°
- The arc from $ Q $ to $ Z $ containing $ X $: 214°

But actually, the angle formed by tangent and secant intercepts only one arc: the arc between the two points where the secant intersects the circle — but the tangent adds another point.

Wait — the correct interpretation is:

The tangent touches at $ X $, the secant intersects at $ Z $ and $ Q $. So the angle $ \angle XYZ $ is formed between tangent $ YX $ and secant $ YZQ $. The intercepted arcs are:
- The arc from $ X $ to $ Q $ along the circle (not passing through $ Z $)
- The arc from $ X $ to $ Z $ along the circle (not passing through $ Q $)

But without knowing where $ X $ is, we assume it's on the major arc or minor arc.

But given arc $ QZ = 146^\circ $, and $ X $ is the point of tangency, likely $ X $ is on the other side, so the arc from $ Q $ to $ Z $ passing through $ X $ is the major arc = $ 360^\circ - 146^\circ = 214^\circ $

Then, the angle $ \angle XYZ $ is formed by tangent at $ X $ and secant to $ Z $ and $ Q $, so it intercepts:
- The far arc: arc $ QXZ $ (the long way) = 214°
- The near arc: arc $ QZ $ = 146°

Wait — no: the formula is:

$$
\angle = \frac{1}{2}(\text{far arc} - \text{near arc})
$$

Where:
- Far arc = arc from $ Q $ to $ Z $ not containing $ X $ = 146°
- Near arc = arc from $ Q $ to $ Z $ containing $ X $ = 214°

No — actually, it's the opposite.

The intercepted arc is the arc between the two points on the circle, and the angle is half the difference between the major and minor arcs between the two points.

But here, the two points are $ Q $ and $ Z $, so the two arcs are:
- Minor arc $ QZ = 146^\circ $
- Major arc $ QZ = 214^\circ $

The angle formed by tangent and secant from $ Y $ is half the difference of these two arcs.

But wait — the tangent is at $ X $, which is on the major arc, so the angle intercepts the minor arc $ QZ $.

Actually, the formula is:

$$
\angle = \frac{1}{2}(\text{arc } QXZ - \text{arc } QZ)
$$

But we don’t know arc $ QXZ $.

Wait — let’s use the standard formula:

For a tangent from $ Y $ touching at $ X $, and a secant from $ Y $ intersecting the circle at $ Z $ and $ Q $ (with $ Z $ closer to $ Y $), then:

$$
\angle = \frac{1}{2}(\text{arc } QX - \text{arc } ZX)
$$

But we don’t know those.

Alternatively, the angle is half the difference of the intercepted arc and the opposite arc.

But I recall: the angle formed by a tangent and a secant is:

$$
\angle = \frac{1}{2}(\text{arc } QXZ - \text{arc } QZ)
$$

But arc $ QXZ $ is the arc from $ Q $ to $ Z $ passing through $ X $, which is the major arc = 214°, and arc $ QZ $ = 146°, so:

$$
\angle = \frac{1}{2}(214^\circ - 146^\circ) = \frac{1}{2}(68^\circ) = 34^\circ
$$

Yes! This is the standard formula.

So:
$$
\angle XYZ = \frac{1}{2}(\text{major arc } QZ - \text{minor arc } QZ) = \frac{1}{2}(214^\circ - 146^\circ) = \frac{1}{2}(68^\circ) = 34^\circ
$$

✔ Answer: 34°

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4)

Tangents from $ Q $ to $ R $ and $ P $, $ \angle RQP = 90^\circ $? Wait — no.

Wait: $ QR $ and $ QP $ are tangents, and arc $ RP = 120^\circ $

Find angle $ \angle RQP $

Two tangents from external point $ Q $, touching at $ R $ and $ P $, and arc $ RP = 120^\circ $

So, this is tangent-tangent angle.

The angle is:
$$
\angle RQP = \frac{1}{2}(\text{major arc } RP - \text{minor arc } RP)
$$

Minor arc $ RP = 120^\circ $, so major arc $ RP = 360^\circ - 120^\circ = 240^\circ $

Then:
$$
\angle = \frac{1}{2}(240^\circ - 120^\circ) = \frac{1}{2}(120^\circ) = 60^\circ
$$

✔ Answer: 60°

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5)

Tangent from $ M $ to $ N $, secant from $ M $ through $ J $ and $ K $, arc $ JK = 130^\circ $

Find angle $ \angle NMJ $

Tangent at $ N $, secant from $ M $ through $ J $ and $ K $

So, tangent-secant angle at $ M $.

Intercepted arcs:
- Minor arc $ JK = 130^\circ $
- Major arc $ JK = 360^\circ - 130^\circ = 230^\circ $

The angle is:
$$
\angle = \frac{1}{2}(\text{major arc } JK - \text{minor arc } JK) = \frac{1}{2}(230^\circ - 130^\circ) = \frac{1}{2}(100^\circ) = 50^\circ
$$

✔ Answer: 50°

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6)

Tangent from $ R $ to $ Q $, secant from $ R $ through $ G $ and $ H $, arc $ GH = 64^\circ $

Find angle $ \angle GRH $

Tangent at $ Q $, secant from $ R $ through $ G $ and $ H $

So, tangent-secant angle at $ R $

Arc $ GH = 64^\circ $, so minor arc = 64°, major arc = 360° - 64° = 296°

Then:
$$
\angle GRH = \frac{1}{2}(296^\circ - 64^\circ) = \frac{1}{2}(232^\circ) = 116^\circ
$$

Wait — but is that right? The angle is $ \angle GRH $, which is at $ R $, between tangent $ RQ $ and secant $ RG $, so yes.

But wait — the arc $ GH $ is the arc between $ G $ and $ H $, so the angle intercepts arc $ GH $, and the opposite arc.

But if arc $ GH = 64^\circ $, and the tangent is at $ Q $, which is on the other side, then the major arc $ GH $ (passing through $ Q $) is $ 360^\circ - 64^\circ = 296^\circ $

So:
$$
\angle = \frac{1}{2}(296^\circ - 64^\circ) = 116^\circ
$$

✔ Answer: 116°

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7)

Tangents from $ K $ to $ C $ and $ J $, arc $ CJ = 110^\circ $

Find angle $ \angle CKJ $

Two tangents from $ K $, touching at $ C $ and $ J $, arc $ CJ = 110^\circ $

So, tangent-tangent angle.

Minor arc $ CJ = 110^\circ $, so major arc $ CJ = 360^\circ - 110^\circ = 250^\circ $

Then:
$$
\angle CKJ = \frac{1}{2}(250^\circ - 110^\circ) = \frac{1}{2}(140^\circ) = 70^\circ
$$

✔ Answer: 70°

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8)

Tangent from $ L $ to $ K $, secant from $ L $ through $ M $ and $ N $, arc $ MN = 126^\circ $, arc $ MK = 61^\circ $

Find angle $ \angle MLK $

Wait — angle $ \angle MLK $ — at $ L $, between tangent $ LK $ and secant $ LM $, which goes through $ M $ and $ N $

So, tangent-secant angle at $ L $

The secant intersects at $ M $ and $ N $, with $ M $ closer to $ L $, $ N $ farther.

The intercepted arcs are:
- Arc $ MN = 126^\circ $
- Arc $ MK = 61^\circ $ — but $ K $ is the point of tangency.

Wait — the tangent is at $ K $, so the arc from $ M $ to $ K $ is 61°, and arc $ MN = 126^\circ $

But the two points on the secant are $ M $ and $ N $, so the arc between them is $ MN = 126^\circ $

But the tangent is at $ K $, so the angle intercepts:
- The arc from $ M $ to $ N $ not containing $ K $: that’s 126°
- The arc from $ M $ to $ N $ containing $ K $: $ 360^\circ - 126^\circ = 234^\circ $

But we also know arc $ MK = 61^\circ $

Wait — probably arc $ MK $ is part of the circle.

Let’s suppose the points are arranged as: $ M $, $ K $, $ N $ on the circle, with arc $ MK = 61^\circ $, arc $ KN = ? $, arc $ MN = 126^\circ $

But arc $ MN = 126^\circ $, and arc $ MK = 61^\circ $, so if $ K $ is between $ M $ and $ N $, then arc $ MN = arc $ MK + arc $ KN $, so:

$$
126^\circ = 61^\circ + arc $ KN $ \Rightarrow arc $ KN = 65^\circ $
$$

But we need the major arc $ MN $, which is $ 360^\circ - 126^\circ = 234^\circ $

Now, the tangent is at $ K $, and the secant is from $ L $ through $ M $ and $ N $

So the angle $ \angle MLK $ is formed by tangent $ LK $ and secant $ LMN $

Then, the intercepted arcs are:
- The arc from $ M $ to $ N $ not containing $ K $: 126°
- The arc from $ M $ to $ N $ containing $ K $: 234°

But the formula is:
$$
\angle = \frac{1}{2}(\text{major arc } MN - \text{minor arc } MN) = \frac{1}{2}(234^\circ - 126^\circ) = \frac{1}{2}(108^\circ) = 54^\circ
$$

Wait — but we have arc $ MK = 61^\circ $, and arc $ KN = 65^\circ $, so the arc from $ M $ to $ N $ through $ K $ is $ 61^\circ + 65^\circ = 126^\circ $? No — that would make arc $ MN $ = 126°, but that’s the same as the minor arc.

Wait — contradiction.

If arc $ MK = 61^\circ $, arc $ KN = 65^\circ $, then arc $ MN $ (via $ K $) = $ 61^\circ + 65^\circ = 126^\circ $, so that’s the minor arc.

Then the major arc $ MN $ = $ 360^\circ - 126^\circ = 234^\circ $

But now, the tangent is at $ K $, and the secant goes from $ L $ through $ M $ and $ N $

So the angle $ \angle MLK $ intercepts:
- The arc from $ M $ to $ N $ not containing $ K $: that’s the major arc = 234°
- The arc from $ M $ to $ N $ containing $ K $: 126°

But the formula is:
$$
\angle = \frac{1}{2}(\text{far arc} - \text{near arc})
$$

Far arc = arc from $ M $ to $ N $ not containing $ K $ = 234°

Near arc = arc from $ M $ to $ N $ containing $ K $ = 126°

Wait — no: the far arc is the one not between the points, but in tangent-secant, it’s the arc opposite to the angle.

Actually, the standard formula is:

$$
\angle = \frac{1}{2}(\text{arc } MN_{\text{major}} - \text{arc } MN_{\text{minor}})
$$

But both arcs are between $ M $ and $ N $, so:

$$
\angle = \frac{1}{2}(234^\circ - 126^\circ) = \frac{1}{2}(108^\circ) = 54^\circ
$$

But we also have arc $ MK = 61^\circ $, which may be used to verify.

Wait — perhaps the angle intercepts arc $ MK $ and arc $ NK $?

No — the angle is at $ L $, formed by tangent $ LK $ and secant $ LM $, so the intercepted arc is from $ M $ to $ N $, but the tangent is at $ K $, so the arc between $ M $ and $ N $ that does not contain $ K $ is the major arc, and the one that does is the minor arc.

But since arc $ MN $ = 126°, and $ K $ is on the circle, and arc $ MK = 61^\circ $, arc $ KN = 65^\circ $, then $ K $ is on the minor arc $ MN $, so the minor arc contains $ K $.

Therefore, the far arc (not containing $ K $) is $ 360^\circ - 126^\circ = 234^\circ $

Then:
$$
\angle = \frac{1}{2}(234^\circ - 126^\circ) = 54^\circ
$$

✔ Answer: 54°

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✔ Final Answers:

1) 152°
2) 130°
3) 34°
4) 60°
5) 50°
6) 116°
7) 70°
8) 54°

Let me know if you want diagrams or further explanation!