Circle Angles and Arcs Worksheet Answer Key – Geometry

Problem 1 Central Angle and Arc

A central angle has the same measure as its intercepted arc.

m∠AOB = 118°

Problem 2 Inscribed Angle from an Arc

The inscribed angle is half of its arc: ½ · 134° = 67°.

m∠ACB = 67°

Problem 3 Arc from an Inscribed Angle

The arc is twice the inscribed angle: 2 · 38° = 76°.

76°

Problem 4 Why Circles Are Similar

(a) Translate circle A so that its center lands on the center of B, then dilate about that center with scale factor 156 = 52.

(b) Circumferences scale by the factor 52 (30π ÷ 12π). Areas scale by the square: 254 (225π ÷ 36π).

(a) translation, then dilation with factor 5/2 (b) 5 : 2 and 25 : 4

Problem 5 Angle in a Semicircle

An angle inscribed in a semicircle is a right angle, so m∠C = 90°.

m∠B = 180° − 90° − 35° = 55°.

∠C = 90°, ∠B = 55°

Problem 6 Tangent Length

OT ⊥ PT, so triangle OTP is a right triangle with hypotenuse OP.

OP = √(8² + 15²) = √289 = 17.

OP = 17 cm

Problem 7 Angle Between Two Tangents

Quadrilateral OAPB has right angles at A and B, so ∠P + ∠AOB = 180°.

m∠P = 180° − 118° = 62°.

m∠P = 62°

Problem 8 Inscribed Quadrilateral Angles

Opposite angles are supplementary. ∠C = 180° − 95° = 85°. ∠D = 180° − 72° = 108°.

Check: 95 + 72 + 85 + 108 = 360.

∠C = 85°, ∠D = 108°

Problem 9 Solving for x with an Arc

Inscribed angle = ½ arc: 4x + 3 = ½(12x − 18) = 6x − 9.

12 = 2x, so x = 6. The angle is 4(6) + 3 = 27° and the arc is 54°.

x = 6; angle = 27°

Problem 10 Angles on the Same Arc

Inscribed angles on the same arc are equal: 5x − 12 = 3x + 14, so 2x = 26 and x = 13.

The angle is 5(13) − 12 = 53°.

x = 13; angle = 53°

Problem 11 Congruent Chords

Congruent chords cut off congruent arcs: 7x + 9 = 9x − 17, so 26 = 2x and x = 13.

Arc AB = 7(13) + 9 = 100°. Check: arc CD = 9(13) − 17 = 100°.

x = 13; each arc is 100°

Problem 12 Distance from the Center to a Chord

The perpendicular from the center bisects the chord, so a right triangle has legs 7 and d and hypotenuse 11.

d = √(11² − 7²) = √72 = 6√2 ≈ 8.49.

6√2 ≈ 8.5 in

Problem 13 Four Arcs Around a Circle

Arc DA = 360° − (88° + 102° + 71°) = 99°.

∠ABC intercepts arc ADC = 71° + 99° = 170°, so ∠ABC = 85°.

∠ADC intercepts arc ABC = 88° + 102° = 190°, so ∠ADC = 95°.

85° + 95° = 180°, as expected for opposite angles of an inscribed quadrilateral.

DA = 99°, ∠ABC = 85°, ∠ADC = 95°, sum 180°

Problem 14 Tangents with a Major Arc

Minor arc = 360° − 238° = 122°.

m∠P = ½(238° − 122°) = ½(116°) = 58°. Check: 180° − 122° = 58°.

Minor arc = 122°, ∠P = 58°

Problem 15 Guard at the Pond

The tangent segment is a leg of a right triangle with hypotenuse 22 and other leg 9: √(22² − 9²) = √403 ≈ 20.07.

Half the angle at P satisfies sin(½∠P) = 922, so ½∠P ≈ 24.15° and ∠P ≈ 48.3°.

Tangent ≈ 20.1 ft; angle ≈ 48.3°

Problem 16 Triangle in a Window

(a) AB is a diameter, so the inscribed angle at C intercepts a semicircle of 180° and measures half: 90°.

(b) BC = √(29² − 20²) = √441 = 21.

(c) tan A = 2120, so ∠A = 46.4° ≈ 46.4°.

(b) BC = 21 in (c) ∠A ≈ 46.4°

Problem 17 Lights on a Circular Track

Opposite angles are supplementary. (6x + 4) + (4x + 16) = 180 gives 10x = 160, so x = 16. ∠A = 100°, ∠C = 80°.

(7y − 5) + (3y + 25) = 180 gives 10y = 160, so y = 16. ∠B = 107°, ∠D = 73°.

Check: 100 + 80 + 107 + 73 = 360.

A = 100°, B = 107°, C = 80°, D = 73°

Problem 18 Roundabout Tangent Roads

The line from P to the center bisects the 54° angle, giving 27°, and the radius to the tangent point is perpendicular to the road.

OP = 60sin 27° ≈ 132.16 ft. Road length = 60tan 27° ≈ 117.76 ft.

OP ≈ 132.2 ft; each road ≈ 117.8 ft

Problem 19 Clock Face Triangle

Adjacent hour marks are 30° apart. Arc 12 to 4 = 120°, arc 4 to 9 = 150°, arc 9 to 12 = 90°.

The angle at 12 intercepts the arc from 4 to 9: ½ · 150° = 75°. The angle at 4 intercepts the arc from 9 to 12: ½ · 90° = 45°. The angle at 9 intercepts the arc from 12 to 4: ½ · 120° = 60°.

75 + 45 + 60 = 180.

75° at 12, 45° at 4, 60° at 9

Problem 20 Triangle from Three Arcs

∠C intercepts arc AB: 50°. ∠A intercepts arc BC: 75°. ∠B intercepts arc CA: 55°. Sum: 180°.

The longest side is opposite the largest angle, ∠A, so it is BC.

A = 75°, B = 55°, C = 50°; longest side BC