One rule, then the next
An angle chase finds an unknown angle through a chain of steps. Each step uses one rule to find one angle, and the angle it finds becomes a known angle for the next step. The chase ends when the angle asked for is found.
These are the rules to chain. Angles on a straight line add up to 180°, and angles around a point add up to 360°. Vertically opposite angles are equal. Where a line crosses two parallel lines, corresponding angles are equal, alternate angles are equal, and co-interior angles add up to 180°. The angles of a triangle add up to 180°, and an exterior angle of a triangle equals the sum of the two opposite interior angles.
Step 1: across the parallel lines
A line crosses two parallel lines, and one angle at the upper crossing is 64°. The angle on the other side of the crossing line, between the parallels at the lower crossing, makes a Z shape with it. Alternate angles are equal, so it is also 64°.
The given 64° at the upper crossing, and its alternate angle at the lower crossing, also 64°.
Step 2: along the straight line
Now use the 64° just found. Beside it, on the lower parallel line, is the next angle. Angles on a straight line add up to 180°, so that angle is 180° − 64° = 116°.
The chase went through the 64° at the lower crossing to reach the 116°. That is the shape of every chase: the angle found in one step is the given of the next.
The 64° found in Step 1 and the angle beside it on the parallel line add up to 180°, so that angle is 116°.
the eight angles are only two sizes, θ and 180° − θ, four of each: alternate, corresponding and co-interior are three names for that
Slide the upper crossing down onto the lower one
Drag the upper crossing sideways to tilt the crossing line: each of the eight angles is either or . Slide the upper crossing down onto the lower one and its four angles land on the lower four. So in a chase across parallel lines, once one angle is known, all eight are.
Step 3: through a triangle
In this triangle two angles are 38° and 74°, and one side is carried on past the third corner. The angle x between that carried-on side and the triangle is wanted.
First find the third angle inside the triangle. The angles of a triangle add up to 180°, so it is 180° − 38° − 74° = 68°. Then x and that 68° lie on a straight line, so x = 180° − 68° = 112°.
The exterior angle rule gives the same answer in one step: x = 38° + 74° = 112°. The two routes agree, which is a good check on a chase.
The triangle with angles of 38° and 74°, and the exterior angle x at its third corner: x = 180° − 68° = 112°.
Planning a chase
Write every angle you find on the diagram as soon as you find it, with the name of the rule that gave it. A chase that is written down can be checked, and each new angle on the page may be the one the next rule needs.
If there is no rule that reaches x directly, find any angle you can, near x or not. Often the chase has to pass through angles the question never mentions.
Stopping too early is the usual mistake: the 64° in Step 1 is a step, not the answer. Using the wrong total is the other one: angles on a straight line add up to 180°, not 360°.
When the angles are written with a letter
The rules still apply when angles are given as expressions, and then each rule gives an equation. Two angles on a straight line are 3x + 10° and 2x + 20°. They add up to 180°, so 3x + 10 + 2x + 20 = 180. Collect the terms: 5x + 30 = 180, so 5x = 150 and x = 30. The angles are 3 × 30 + 10 = 100° and 2 × 30 + 20 = 80°, and 100° + 80° = 180° checks.
Worked example: Three Parallel Lines with Two Transversals
Question AB, CD and EF are three parallel lines. G is on AB, H on CD and K on EF; GH and HK are straight. ∠ AGH = 112° and ∠ HKE = 47°. Find ∠ GHK.
1.AB ∥ CD ∥ EF. GH crosses the top two lines; HK crosses the bottom two. At H the path bends.
Three parallels. The path G-H-K bends on the middle line. 2.On the transversal GH: ∠ AGH and ∠ GHC are co-interior, so ∠ GHC = 180° − 112° = 68°.
C shape on GH: ∠ GHC = 180° − 112° = 68°. 3.CD is a straight line: ∠ GHD = 180° − 68° = 112°.
On the straight line CD: ∠ GHD = 180° − 68° = 112°. 4.On the transversal HK: ∠ DHK and ∠ HKE are alternate angles, so ∠ DHK = 47°.
Z shape on HK: ∠ DHK = ∠ HKE = 47°. 5.∠ GHK runs from HG through HD to HK: 112° + 47° = 159°.
∠ GHK = 112° + 47° = 159°, from HG through HD to HK. 6.Check: the angle on the other side of the bend is 360° − 159° = 201°, a reflex angle, so 159° is the one inside the bend.
The other way round H is 360° − 159° = 201°, a reflex angle.
Answer: ∠ GHK = 159°
Common mistakes
- Using ∠ GHC = 68° on the wrong side of H and giving 68° + 47° = 115°: HC and HK are on opposite sides of the transversal GH.
- Reading ∠ HKE as corresponding to ∠ GHD: they belong to different transversals, so no F, Z or C shape joins them.
The angles of an isosceles triangle
A triangle with two equal sides is called isosceles. The corner where the two equal sides meet is its apex, and the side opposite the apex is its base. The two angles at the ends of the base are equal.
The reason is that the apex is the same distance from both ends of the base, so it lies on the perpendicular bisector of the base. Fold the triangle along that line. The fold carries one end of the base onto the other, and the apex stays where it is, so the triangle lands exactly on itself and each base angle lands on the other. In an isosceles triangle whose apex angle is 50°, the two base angles share 180° − 50° = 130°, so each is 130° ÷ 2 = 65°.
Worked example: Hidden Isosceles Triangle from a Shared Side
Question B, C and D lie on a straight line. AB = AC and AC = CD. ∠ BAC = 40°. Find ∠ ADC and ∠ BAD.
1.AB = AC, so triangle ABC is isosceles with apex A: ∠ ABC = ∠ ACB.
Ticks mark the equal sides: AB = AC, and AC = CD. 2.∠ ABC = ∠ ACB = (180° − 40°) ÷ 2 = 70°.
Triangle ABC, apex A: base angles (180° − 40°) ÷ 2 = 70°. 3.BCD is a straight line, so ∠ ACD = 180° − 70° = 110°.
On the straight line BCD, ∠ ACD = 180° − 70° = 110°. 4.AC = CD, so triangle ACD is isosceles with apex C: ∠ CAD = ∠ CDA.
Triangle ACD has its equal sides meeting at C, so C is its apex. 5.∠ ADC = (180° − 110°) ÷ 2 = 35°.
∠ ADC = (180° − 110°) ÷ 2 = 35°. 6.∠ BAD = ∠ BAC + ∠ CAD = 40° + 35° = 75°.
∠ BAD = 40° + 35° = 75°.
Answer: ∠ ADC = 35°; ∠ BAD = 75°
Common mistakes
- Taking ∠ ACD = 70° because it is next to the 70°: it is the other angle on the straight line, 110°.
- Putting the apex of the second triangle at A and making ∠ ACD a base angle: the equal sides are AC and CD, which meet at C.