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Angles, triangles & polygons · Lesson 1 of 4

Angles and parallel lines

Angles on a straight line and at a point, vertically opposite angles, corresponding, alternate and co-interior angles, and the extra parallel line that solves a bend between two parallels.

15 min
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  3. 3
  4. 4

Every geometry answer is a chain of small steps, and each step needs a reason: a fact about angles that everyone agrees on. This lesson gives the facts about lines. Three need no parallel lines at all:

  • Angles on a straight line add up to 180∘180^\circ.
  • Angles at a point add up to 360∘360^\circ.
  • Vertically opposite angles (across the point where two lines cross) are equal.
ab
On a straight linea + b = 180°
abc
At a pointa + b + c = 360°
aabb
Vertically oppositeThe two a's are equal, and so are the two b's

Parallel lines

When a line (a transversal) crosses two parallel lines, it makes the same angles at both. Arrows on the lines mark them as parallel. Three pairs come up again and again:

  • Corresponding angles are in the same position at each crossing: both above the line and to the right of the transversal, for example. They are equal.
  • Alternate angles are both between the parallels, on opposite sides of the transversal. They are equal.
  • Co-interior angles are both between the parallels, on the same side of the transversal. They add up to 180∘180^\circ.
aa
CorrespondingEqual
aa
AlternateEqual
ba
Co-interiora + b = 180°

Try it

Angles and parallel linesDrag the gold point to tilt the line
62°62°
62°first angle62°second angleequalthey are
Corresponding angles are in the same position at each crossing (here both above the line, on the same side of the transversal). They are equal. Here the transversal makes 62° and 118° with the parallels: every angle at either crossing is one of these two.

Tilt the transversal and try each chip. At both crossings, every angle is one of just two sizes, and those two add up to 180∘180^\circ. Once you know one angle, you know all eight.

A bend between parallel lines

Many questions join two parallel lines with a path that bends at a corner. The trick is to draw another line through the corner, parallel to the first two. It splits the corner angle into two parts, and each part is an alternate angle to an angle at one of the parallels. So the corner angle is the sum of the two angles at the parallels.

aabb
A bend between parallelsDraw a parallel through the corner: the corner angle is a + b

Switch the board to “Bent line” and drag the corner: the angle there always equals the other two added together.

Worked example · WAEC 2014

WAEC 2014 · Paper 2 · Q1 (b)

In the diagram (not drawn to scale), EF‾\overline{EF} is parallel to GH‾\overline{GH}. AEBAEB and BHCBHC are straight lines. If ∠AEF=3x∘\angle AEF = 3x^\circ, ∠ABC=120∘\angle ABC = 120^\circ and ∠CHG=7x∘\angle CHG = 7x^\circ, find the value of ∠GHB\angle GHB.

3x°120°7x°AEBHCFG
Not drawn to scale (as in the paper); here the angles are drawn at their true sizes, with x = 15.

  1. Draw the extra line

    Draw a line through BB parallel to EFEF and GHGH. It splits ∠ABC=120∘\angle ABC = 120^\circ into an upper part and a lower part.

    Think first. The corner is at B. What line should you add?

  2. The upper part

    The upper part equals ∠AEF=3x\angle AEF = 3x (alternate angles, EF∥EF \parallel the new line).

    Think first. Which angle at E is it equal to, and why?

  3. The lower part

    The lower part equals ∠GHB\angle GHB (alternate angles). BHCBHC is a straight line, so ∠GHB=180∘−7x\angle GHB = 180^\circ - 7x (angles on a straight line).

    Think first. It is alternate to an angle at H. Which one, and what is that angle in terms of x?

  4. Solve for x

    3x+(180−7x)=1203x + (180 - 7x) = 120, so 180−4x=120180 - 4x = 120, 4x=604x = 60 and x=15x = 15.

    Think first. The two parts make 120°. Write the equation.

  5. The angle asked for

    ∠GHB=180∘−7×15∘=180∘−105∘=75∘\angle GHB = 180^\circ - 7 \times 15^\circ = 180^\circ - 105^\circ = 75^\circ. (Check: 3x=45∘3x = 45^\circ and 45∘+75∘=120∘45^\circ + 75^\circ = 120^\circ ✓.)

    Think first. Put x = 15 into ∠GHB.

Your turn

WAEC 2020 · Paper 1 · Q19

In the diagram, PQ∥SRPQ \parallel SR. Find the value of xx.

x68°246°PQSR
Worked solution (try it first)
  1. Angles at a point: the angle between the two lines at the bend is 360∘−246∘=114∘360^\circ - 246^\circ = 114^\circ.
  2. Draw a line through the bend parallel to PQPQ.
  3. By alternate angles, the upper line makes xx with it and the lower line makes 68∘68^\circ with it.
  4. So x+68∘=114∘x + 68^\circ = 114^\circ, which gives x=46x = 46, option B.

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