Mathematics • Year 7 • Unit 3 • Lesson 1

Angles, Lines, and Geometry Foundations

Build fluency with classifying angles by size (acute, right, obtuse, straight, reflex) and finding unknown angles using complementary (sum 90°), supplementary (sum 180°), vertically opposite (equal) and angles at a point (sum 360°) relationships. Always state a reason.

Build · I Do / We Do / You Do

1. I do — fully worked example

Read every line. Each step shows the geometric reason on the right — every angle answer needs one.

Problem. Two straight lines cross at a point. One of the four angles formed measures 118°. Find the other three angles and state a reason for each.

Step 1 — The vertically opposite angle.

Angle directly across the X = 118°

Reason: vert. opp. ∠s are equal.

Step 2 — An adjacent angle along the straight line.

180 − 118 = 62°

Reason: ∠s on a str. line sum to 180°.

Step 3 — The fourth angle (vertically opposite the 62°).

Fourth angle = 62°

Reason: vert. opp. ∠s are equal.

Step 4 — Check using "angles at a point".

118 + 62 + 118 + 62 = 360° ✓

Answer: The four angles are 118°, 62°, 118°, 62°.

Stuck? Revisit lesson § "Watch Me Solve It · Vertically opposite" — across the X is equal; along a line is supplementary.

2. We do — fill in the missing steps

Same structure as Section 1, with the working faded. Fill the blanks. 4 marks

Problem. Three angles meet at a single point. They measure 130°, 95° and x°. Find x.

Step 1 — State the rule: all angles at a single point sum to ______°.

Step 2 — Set up the equation:

130 + 95 + x = _______

Step 3 — Add the two known angles:

130 + 95 = _______

Step 4 — Solve for x:

x = 360 − _______ = _______°

Step 5 — Write the reason: ( ___________________ )

Stuck? Reason shorthand: (∠s at a pt.). Always include it next to the calculation.

3. You do — independent practice

Show working AND a geometric reason for every answer. First four are foundation, middle two standard, last two extension.

Foundation — single step

3.1 Classify the angle 72°. 1 mark

3.2 Find the complement of 28°. 1 mark

3.3 Find the supplement of 113°. 1 mark

3.4 Two straight lines cross. One angle is 47°. State the size of the angle vertically opposite, and give the reason. 1 mark

Standard — combine two ideas

3.5 Two straight lines cross. One of the four angles is 64°. Find the other three angles. Give a reason for each. 2 marks

3.6 Four angles meet at a single point: 90°, 75°, 80° and x°. Find x. State the reason. 2 marks

Extension — push your thinking

3.7 In the angle name ∠PQR, which letter is the vertex? Then explain in one sentence why ∠PQR and ∠RQP describe the same angle. 2 marks

3.8 Two angles ∠AOB and ∠BOC sit side-by-side on a straight line AOC. ∠AOB = (2x + 10)°, ∠BOC = (3x − 5)°. Set up an equation, solve for x, and find both angles. 3 marks

Stuck on 3.8? Two angles forming a straight line are supplementary, so they must add to 180°.

How did this worksheet feel?

What I'll revisit before next class:

Answers — Do not peek before attempting

Section 2 — Three angles at a point (130°, 95°, x°)

Step 1: sum to 360°. Step 2: 130 + 95 + x = 360. Step 3: 130 + 95 = 225.
Step 4: x = 360 − 225 = 135°. Step 5: (∠s at a pt.).

3.1 — Classify 72°

72° < 90° → acute.

3.2 — Complement of 28°

Complementary angles sum to 90°. 90 − 28 = 62°.

3.3 — Supplement of 113°

Supplementary angles sum to 180°. 180 − 113 = 67°.

3.4 — Vertically opposite to 47°

47° (vert. opp. ∠s are equal).

3.5 — Four angles from a crossing (one = 64°)

Vertically opposite the 64° = 64° (vert. opp. ∠s).
Adjacent along the line = 180 − 64 = 116° (∠s on str. line).
Fourth angle = 116° (vert. opp. to the 116°).
Check: 64 + 116 + 64 + 116 = 360° ✓

3.6 — 90 + 75 + 80 + x = 360

Sum of knowns: 90 + 75 + 80 = 245. So x = 360 − 245 = 115° (∠s at a pt.).

3.7 — Vertex of ∠PQR

The middle letter is the vertex, so the vertex is Q. ∠PQR and ∠RQP describe the same angle because both have vertex Q with the same two rays (one to P and one to R) — the order you read the outside letters doesn't matter.

3.8 — (2x + 10)° + (3x − 5)° on a straight line

∠s on a str. line sum to 180°: (2x + 10) + (3x − 5) = 180.
Combine: 5x + 5 = 180. So 5x = 175 and x = 35.
∠AOB = 2(35) + 10 = 80°. ∠BOC = 3(35) − 5 = 100°.
Check: 80 + 100 = 180° ✓.