# ARC AND ANGLE RELATIONSHIPS IN CIRCLES

When chords, secants, and tangents intersect in a circle (Figure 1), on a circle (Figure 2), or outside of a circle (Figure 3), special relationships exist between the angle and arc measures formed. ## Interior Intersections

If two secants or chords intersect inside a circle, then the measure of the angle formed is equal to half the sum of the measures of the intercepted arcs. In the diagram above,

m∠1  =  1/2 ⋅ (m∠arc AD + m∠arc BC)

m∠2  =  1/2 ⋅ (m∠arc AB + m∠arc CD)

Example 1 :

Find m∠AED. Solution :

m∠AED  =  1/2 ⋅ (m∠arc AD + m∠arc BC)

Substitute.

m∠AED  =  1/2 ⋅ (45° + 109°)

m∠AED  =  1/2 ⋅ 154°

m∠AED  =  77°

Example 2 :

Find m∠YWX. Solution :

Find m∠VWY :

m∠VWY  =  1/2 ⋅ (m∠arc VY + m∠arc XZ)

m∠VWY  =  1/2 ⋅ (43° + 81°)

m∠VWY  =  1/2 ⋅ 124°

m∠VWY  =  62°

Find m∠YWX :

m∠VWY and m∠YWX are linear pair.

Then,

m∠VWY + m∠YWX  =  180°

Substitute m∠VWY  =  62°.

62° m∠YWX  =  180°

Subtract 62° from each side.

m∠YWX  =  118°

Example 3 :

Find m∠arc CDE. Solution :

1/2 ⋅ (m∠arc GF + m∠arc CDE)  =  m∠CHE

Substitute.

1/2 ⋅ (73° + m∠arc CDE)  =  128°

Multiply each side by 2.

73° + m∠arc CDE  =  256°

Subtract 73° from each side.

m∠arc CDE  =  183°

## On the Circle Intersections

If a secant and a tangent intersect at the point of tangency, then the measure of each angle formed is equal to half the measure of its intercepted arc. In the diagram above,

m∠1  =  1/2 ⋅ m∠arc AB

m∠2  =  1/2 ⋅ m∠arc ACB

Example 4 :

Find m∠DEG. Solution :

m∠DEG  =  1/2 ⋅ m∠arc EHG

Substitute.

m∠DEG  =  1/2 ⋅ 308°

m∠DEG  =  154°

Example 5 :

Find m∠arc KNM. Solution :

Find m∠arc KM :

1/2 ⋅ m∠arc KM  =  m∠JKM

Substitute.

1/2 ⋅ m∠arc KM  =  23°

Multiply each side by 2.

m∠arc KM  =  46°

In the circle above,

m∠arc KM + m∠arc KNM  =  360°

Substitute.

46° + m∠arc KNM  =  360°

Subtract 46° from each side.

m∠arc KNM  =  314°

## Exterior Intersections

If secants and/or tangents intersect on the exterior of a circle, then the measure of the angle formed is equal to half the difference of the intercepted arcs.

## Two Secants m∠A  =  1/2 ⋅ (m∠arc CE - m∠arc BD)

## Secant and Tangent m∠A  =  1/2 ⋅ (m∠arc BD - m∠arc BC)

## Two Tangents m∠A  =  1/2 ⋅ (m∠arc BDC - m∠arc BC)

Example 6 :

Find m∠KLM. Solution :

m∠KLM  =  1/2 ⋅ (m∠arc JN - m∠arc KM)

Substitute.

m∠KLM  =  1/2 ⋅ (140° - 66°)

m∠KLM  =  1/2 ⋅ 74°

m∠KLM  =  37°

Example 7 :

Find m∠BCD. Solution :

In the circle above,

m∠arc BD + marc DE + m∠arc EA  =  90°

Substitute.

m∠arc BD + 94° + 67°  =  180°

m∠arc BD + 161°  =  180°

Subtract 161° from each side.

m∠arc BD  =  19°

Finding m∠BCD :

m∠BCD  =  1/2 ⋅ (m∠arc AE - m∠arc BD)

Substitute.

m∠BCD  =  1/2 ⋅ (67° - 19°)

m∠BCD  =  1/2 ⋅ 48°

m∠BCD  =  24°

Example 8 :

Find m∠KLM. Solution :

m∠KLM  =  1/2 ⋅ (m∠arc KN - m∠arc KM)

Substitute.

m∠KLM  =  1/2 ⋅ (135° - 33°)

m∠KLM  =  1/2 ⋅ 102°

m∠KLM  =  51°

Example 9 :

Find m∠CDE. Solution :

In the circle above,

m∠arc CFE + marc CE  =  360°

Substitute.

m∠arc CFE + 106°  =  360°

Substitute 106° from each side.

m∠arc CFE  =  254°

Finding m∠CDE :

m∠CDE  =  1/2 ⋅ (m∠arc CFE - m∠arc CE)

Substitute.

m∠CDE  =  1/2 ⋅ (254° - 106°)

m∠KLM  =  1/2 ⋅ 148°

m∠KLM  =  74°

Example 10 :

Find m∠arc MK. 1/2 ⋅ (m∠arc NK - m∠arc MK)  =  m∠MLK

Substitute.

1/2 ⋅ (160° - m∠arc MK)  =  56°

Multiply each side by 2.

160° - m∠arc MK  =  112°

Subtract 160° from each side.

- m∠arc MK  =  - 48°

Multiply each side by -1.

m∠arc MK  =  48°

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