Angle Relationships Worksheet Answer Key – Grade 7

Problem 1 Find the complement

Complement = 90° − 38° = 52°.

52°

Problem 2 Find the supplement

Supplement = 180° − 127° = 53°.

53°

Problem 3 Vertical and adjacent

The vertical angle is equal: 64°. An adjacent angle is supplementary: 180° − 64° = 116°.

vertical 64°, adjacent 116°

Problem 4 Complementary equation

x + 41 = 90, so x = 90 − 41 = 49.

x = 49

Problem 5 Straight line equation

118 + x = 180, so x = 62.

x = 62

Problem 6 Four angles at a crossing

∠2 is adjacent to ∠1: 180° − 72° = 108°. ∠3 is vertical to ∠1: 72°. ∠4 is vertical to ∠2: 108°.

∠2 = 108°, ∠3 = 72°, ∠4 = 108°

Problem 7 Classify the pairs

(a) 68 + 22 = 90, complementary. (b) 99 + 81 = 180, supplementary. (c) 45 + 135 = 180, supplementary. (d) 30 + 50 = 80, neither.

(a) complementary, (b) supplementary, (c) supplementary, (d) neither

Problem 8 Triple angle

Let the smaller angle be x. Then x + 3x = 180, so 4x = 180 and x = 45. The larger is 135.

45° and 135°

Problem 9 Complementary expressions

(2x + 4) + (x + 17) = 90, so 3x + 21 = 90, 3x = 69, x = 23.

Angles: 2(23) + 4 = 50 and 23 + 17 = 40. Check: 50 + 40 = 90.

x = 23; angles 50° and 40°

Problem 10 Vertical angle expressions

Vertical angles are equal: 4x − 12 = 2x + 30, so 2x = 42 and x = 21.

Angle: 4(21) − 12 = 72; check 2(21) + 30 = 72.

x = 21; each angle is 72°

Problem 11 Linear pair expressions

(6x − 5) + (4x + 15) = 180, so 10x + 10 = 180, x = 17.

Angles: 6(17) − 5 = 97 and 4(17) + 15 = 83. Check: 97 + 83 = 180.

x = 17; angles 97° and 83°

Problem 12 Less than its complement

Let the angle be x. Its complement is 90 − x. So x = (90 − x) − 24, giving 2x = 66 and x = 33.

Check: complement 57, and 57 − 33 = 24.

33°

Problem 13 More than its supplement

Let the angle be x. Then x = (180 − x) + 30, so 2x = 210 and x = 105.

Check: supplement 75, and 75 + 30 = 105.

105°

Problem 14 Three angles on a line

x + 2x + (3x + 30) = 180, so 6x + 30 = 180, 6x = 150, x = 25.

Angles: 25°, 50° and 3(25) + 30 = 105°. Check: 25 + 50 + 105 = 180.

x = 25; angles 25°, 50°, 105°

Problem 15 Leaning ladder

The wall and ground meet at 90°, so the ladder-wall angle is the complement of 74°: 90 − 74 = 16°.

The angle on the other side of the ladder along the ground is supplementary to 74°: 180 − 74 = 106°.

16° with the wall and 106° on the other side

Problem 16 Crossing roads

Vertical angles are equal: 3x + 15 = 5x − 25, so 40 = 2x and x = 20.

Each of the two vertical angles is 3(20) + 15 = 75°. Each neighbor is 180 − 75 = 105°.

x = 20; the angles are 75°, 105°, 75°, 105°

Problem 17 Split right angle

The angles are complementary: (5x + 4) + (4x + 14) = 90, so 9x + 18 = 90 and x = 8.

∠AOB = 5(8) + 4 = 44° and ∠BOC = 4(8) + 14 = 46°. Check: 44 + 46 = 90.

x = 8; angles 44° and 46°

Problem 18 Supplement versus complement

Let the angle be x. Supplement: 180 − x. Complement: 90 − x. Equation: 180 − x = 4(90 − x).

180 − x = 360 − 4x, so 3x = 180 and x = 60.

Check: supplement 120, complement 30, and 4 × 30 = 120.

60°

Problem 19 Bridge cables

Adjacent angles at a crossing are supplementary: (7x − 4) + (2x + 13) = 180, so 9x + 9 = 180 and x = 19.

7(19) − 4 = 129 and 2(19) + 13 = 51. The vertical angles repeat these values.

x = 19; angles 129°, 51°, 129°, 51°

Problem 20 Sign board angle

The post is perpendicular to the ground, so the two angles are complementary: (4x + 3) + (x + 12) = 90, so 5x + 15 = 90 and x = 15.

Angles: 4(15) + 3 = 63° and 15 + 12 = 27°. Check: 63 + 27 = 90.

x = 15; angles 63° and 27°