Geometry

high importance~4 Q in Tier 136 formulas⚡ 19 shortcuts6 subtopics

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Subtopic 1 of 6·Triangles and their centres →

Lines and angles

At a point on a straight line the total angle is $180^\circ$ ; around a point it is $360^\circ$ . When two straight lines cross, the vertically opposite angles are equal.

For parallel lines cut by a transversal (the standard CGL figure):

Corresponding angles are equal (both "top-right", etc.).
Alternate angles are equal (Z-shape).
Co-interior (allied) angles add to $180^\circ$ — this gives a small equation in $x$ .

An angle bisector splits an angle into two equal parts; two bisectors of a triangle meet at the incentre.

📖 Notes 🧩 5 question types ⚡ Formulas 🎯 12 practice Q

Detailed notes

The angle family you must see instantly

An angle measures the turn between two rays. The two facts every question uses:

Complementary angles add to $90^\circ$ ; supplementary angles add to $180^\circ$ .
If the supplement of an angle exceeds the angle by $d$ , the angle is $\frac{180^\circ - d}{2}$ . If the angle is $n$ times its complement, write $x = n(90^\circ - x)$ and solve — the unknown always cancels to one linear equation.

A useful fixed result: supplement $-$ complement $= (180^\circ - x) - (90^\circ - x) = 90^\circ$ , always, whatever $x$ is. Questions test this fact directly, so keep it ready.

Angles on a line and around a point

Angles on one straight line add to $180^\circ$ ; angles filling one full turn around a point add to $360^\circ$ . When two straight lines cross, four angles appear — the vertically opposite pairs are equal, and any two neighbouring angles are supplementary. So if one of the four angles is $a$ , the set is $a,\ 180-a,\ a,\ 180-a$ . Given three angles around a point, the fourth is just the remainder after subtracting from $360^\circ$ — the whole subtopic is subtraction once the picture is right.

Parallel lines cut by a transversal

A transversal is a line crossing two (parallel) lines. It creates 8 angles, but only 2 distinct values. Three name-tags describe the same two facts:

Pair	Picture	Rule
Corresponding	F-shape, same corner	equal
Alternate	Z-shape, opposite sides	equal
Co-interior (allied)	C-shape, same side inside	add to $180^\circ$

So the exam question is really: pick the correct relation, get one linear equation in $x$ , solve, substitute back. Two checks catch almost every slip: a co-interior pair must sum to exactly $180^\circ$ , and a corresponding/alternate pair must be exactly equal.

Reading the question, not just solving it

The classic lost mark: the question asks for the angle and the options include $x$ itself. After solving for $x$ , re-read which angle was asked — the larger of the pair, the smaller, or an angle in a different position (a corresponding partner of a co-interior mate, say). Another common twist: a ratio like $2:3$ for co-interior angles — let them be $2k$ and $3k$ so that $5k = 180^\circ$ .

Bisectors

An angle bisector splits an angle into two equal halves. Two results get asked:

If two supplementary angles are bisected, the halves differ by half the original difference.
The bisectors of a pair of co-interior angles meet at $90^\circ$ — the halves sum to half of $180^\circ$ , so the third angle of the triangle they form is exactly $90^\circ$ .

Quick revision

Complement $= 90^\circ - x$ ; supplement $= 180^\circ - x$ ; their difference is always $90^\circ$ .
On a line $180^\circ$ ; around a point $360^\circ$ ; vertically opposite angles are equal.
Parallel + transversal: corresponding = alternate; co-interior sums to $180^\circ$ .
Expressions in $x$ : choose the relation first, solve the linear equation, then answer the angle actually asked.
Supplement exceeds angle by $d$ $\Rightarrow$ angle $= \frac{180^\circ - d}{2}$ .
Bisectors of co-interior angles meet at $90^\circ$ ; bisecting halves any difference.

Types of questions asked

Every way this subtopic shows up in exams — how to recognise it, the formula or logic to use, and a solved example.

Type 1: Co-interior angles given as expressions in xvery common3 practice Q

How to spot it:

Parallel lines with a transversal; two angles on the same side inside are $(ax+b)^\circ$ and $(cx+d)^\circ$ — find $x$ or one of the angles.

(ax+b) + (cx+d) = 180

Co-interior (allied) angles between parallels add to $180^\circ$ — set the sum.
Solve the linear equation for $x$ .
Substitute back into the angle asked; check the pair really sums to $180^\circ$ .

Why: only the allied pair is supplementary — corresponding and alternate pairs are equal, not added.

Example: Two co-interior angles between parallel lines are $(2x+30)^\circ$ and $(x+15)^\circ$ . Find the larger angle.

$3x + 45 = 180 \Rightarrow x = 45$ . Angles: $120^\circ$ and $60^\circ$ — larger $= 120^\circ$ . Check: $120+60=180$ .

Type 2: Corresponding or alternate angles set equalvery common2 practice Q

How to spot it:

Angles in F-shape (corresponding) or Z-shape (alternate) positions are both given as expressions in $x$ — or one is given and its partner asked.

ax+b = cx+d \ \text{(equal pairs)}

Identify the pair shape: F (corresponding) or Z (alternate) — both are EQUAL.
Equate the two expressions; solve for $x$ .
Answer the angle asked; the other angle values follow from $180^\circ -$ the found angle.

Why: a transversal across parallels makes only two distinct angle values — an acute one $a$ and its supplement $180^\circ-a$ .

Example: Two corresponding angles between parallel lines are $(3x-10)^\circ$ and $(2x+15)^\circ$ . Find the angle.

$3x-10 = 2x+15 \Rightarrow x = 25$ ; angle $= 3(25)-10 = 65^\circ$ . (Its supplement is $115^\circ$ . )

Type 3: Complement and supplement of an anglevery common2 practice Q

How to spot it:

'Find the angle whose supplement is k times its complement', 'the complement of x°', or a difference between supplement and complement.

180 - x = n(90 - x)

Write complement $= 90^\circ - x$ , supplement $= 180^\circ - x$ .
Translate the sentence into one equation and solve for $x$ .
Remember supplement $-$ complement is always $90^\circ$ — use it as an instant check.

Why: both are defined from the same $x$ , so every 'k times' statement is a linear equation.

Example: The supplement of an angle is three times its complement. Find the angle.

$180 - x = 3(90 - x) \Rightarrow 180 - x = 270 - 3x \Rightarrow 2x = 90 \Rightarrow x = 45^\circ$ . Check: supplement $135 = 3 \times 45$ .

Type 4: Angles on a line, around a point, vertically oppositecommon2 practice Q

How to spot it:

Several angles at a point or on a line are given (some as expressions); the missing one is asked — or two crossing lines give one angle and ask the rest.

\text{on a line}=180^\circ,\quad \text{round a point}=360^\circ

Decide the total: straight line $180^\circ$ , full turn $360^\circ$ .
Add the known/expression angles, equate to the total, solve.
For crossing lines: vertically opposite pairs are equal and neighbours are supplementary.

Why: turn totals are fixed, so the missing angle is just the remainder.

Example: Two straight lines intersect. One of the four angles formed is $75^\circ$ . The obtuse angle among the remaining three is:

Neighbour angles are supplementary: $180 - 75 = 105^\circ$ ; the vertically opposite mate repeats $75^\circ$ . Obtuse answer $= 105^\circ$ .

Type 5: Angle bisectors in line configurationsoccasional2 practice Q

How to spot it:

An angle is bisected and the halves (or the angle between two bisectors) are asked, often inside a parallel-lines or triangle figure.

\text{bisected half} = \frac{\theta}{2}

A bisector splits the angle into two equal parts — write each as $\frac{\theta}{2}$ .
Carry the halves into the triangle/pair the figure forms.
Known bonus: bisectors of a co-interior pair meet at $90^\circ$ .

Why: halving both of two angles that sum to $180^\circ$ leaves exactly $90^\circ$ for the triangle the bisectors form.

Example: Two supplementary angles differ by $40^\circ$ . After both are bisected, what is the difference between the two bisected halves?

Halving scales every difference by $\frac12$ : $\frac{40}{2} = 20^\circ$ . (Angles $110^\circ, 70^\circ$ → halves $55^\circ, 35^\circ$ → gap $20^\circ$ .)

Formulas

Angles on a line / around a point

\text{on a line}=180^\circ,\quad \text{round a point}=360^\circ

Co-interior angles

(a+b)^\circ=180^\circ\ \text{for allied angles between parallels}

Complement / supplement

x+(90^\circ-x)=180^\circ\cdot\tfrac12,\quad \text{supplement}-\text{complement}=180^\circ-2\cdot\text{angle}

Parallel line pairs

\text{corresponding}=\text{alternate}=\text{vertically opposite (each pair equal)}

Shortcut tricks

⚡ Co-interior gives the equation

If two allied angles are given as expressions in $x$ , set their sum to $180^\circ$ , solve for $x$ , then substitute back into the angle asked.

Example: Two co-interior angles between parallel lines are $(3x+20)^\circ$ and $(2x+10)^\circ$ . Find the larger angle.

$3x+20+2x+10=180\Rightarrow x=30$ . Angles: $110^\circ$ and $70^\circ$ ; larger $=110^\circ$ .

⚡ Supplement − complement one-liner

For an angle $x$ : supplement $=180-x$ , complement $=90-x$ , so the difference is always $90^\circ$ , and the angle itself is $\frac{180-d}{2}$ if the difference $d$ is given.

Example: The supplement of an angle is four times its complement. Find the angle.

$180-x=4(90-x)\Rightarrow 180-x=360-4x\Rightarrow 3x=180\Rightarrow x=60^\circ$ .

Where students lose marks

Forgetting that co-interior angles are supplementary only between parallel lines.
Reading the wrong pair: alternate vs corresponding in the transversal figure.
Solving for $x$ and answering $x$ instead of the angle asked.

Practice sets — 12 questions

Sets of 10, mixed across the question types above. Each answer comes with a step-by-step explanation.

Topic test · 12 questions

Suggested time 7 min · wrong answers go to your mistake notebook automatically.