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Why a triangle's angles make a straight line, and what a transversal repeats.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
In this lesson you use two facts that do a great deal of work: the angles of a triangle add to $180^\circ$, and a line crossing two parallel lines makes the same angles at both crossings. You will see why each is true, and use them, with equations when you need to, to find missing angles.
You know that a straight angle is $180^\circ$ and a full turn is $360^\circ$. You know that when two lines cross, the angles opposite each other (vertical angles) are equal. You can solve an equation like $3x + 20 = 110$. In the last lesson you saw that sliding a figure (a translation) keeps every angle. This lesson uses all of that to explain two facts that do a great deal of work in geometry.
| Term | What it means |
|---|---|
| Straight angle | An angle of $180^\circ$: the angles along one side of a straight line add to this. |
| Vertical angles | The two opposite angles made where two lines cross. They are always equal. |
| Transversal | A line that crosses two or more other lines. |
| Corresponding angles | Angles in the same position at two crossings of a transversal, such as both above the line and to the right. Equal when the lines are parallel. |
| Alternate interior angles | Angles between the two lines, on opposite sides of the transversal. Equal when the lines are parallel. |
| Co-interior (same-side interior) angles | Angles between the two lines, on the same side of the transversal. They add to $180^\circ$ when the lines are parallel. |
| Exterior angle | The angle between one side of a triangle and the extension of the next side, outside the triangle. |
| Remote interior angles | The two angles inside a triangle that are not next to a given exterior angle. |
Fact 1: parallel lines copy angles. When a transversal crosses two parallel lines, it makes four angles at each crossing, and the two sets of four are identical. So:
The reason is the translation from the last lesson. Slide the top crossing along the transversal until it lands on the bottom one. The top line lands exactly on the bottom line, because the lines are parallel, so every angle lands on an equal angle.
Fact 2: the angles of a triangle add to $180^\circ$. This is true of every triangle: tall, flat, right-angled or not. It follows from Fact 1: draw a line through one corner parallel to the opposite side, and the three angles line up along it.
A result that comes straight from Fact 2 is the exterior angle theorem: an exterior angle of a triangle equals the sum of the two remote interior angles.
Another way: picture
Cut a triangle out of paper and tear off its three corners. Put the three points together, side by side, with their edges touching. However you drew the triangle, the three corners always make a straight edge: half a turn, $180^\circ$. Try it with a long, thin triangle and with a nearly equal-sided one; the pieces look very different but always fill the same straight line.
Another way: story
Picture walking all the way around a triangle. At each corner you turn through the exterior angle. When you are back where you started, facing the same way, you have turned once around: $360^\circ$. Each exterior angle is $180^\circ$ minus an interior angle, so three of them are $540^\circ$ minus the interior sum. Setting that equal to $360^\circ$ gives an interior sum of $180^\circ$ again.
Where a transversal crosses one line it makes four angles. Next to each other, two angles make a straight line, so they add to $180^\circ$. Opposite each other, two angles are vertical angles, so they are equal. That means one angle tells you all four: if one is $70^\circ$, the others are $110^\circ$, $70^\circ$ and $110^\circ$.
When the transversal crosses a second line parallel to the first, the second crossing is a copy of the first. So one angle tells you all eight. The pairs have names:
| Pair | Where they are | Rule |
|---|---|---|
| Corresponding | same corner at each crossing | equal |
| Alternate interior | between the lines, opposite sides of the transversal | equal |
| Alternate exterior | outside the lines, opposite sides of the transversal | equal |
| Co-interior | between the lines, same side of the transversal | add to $180^\circ$ |
A quick way to tell them apart: corresponding angles make an F shape, alternate interior angles make a Z shape, and co-interior angles make a C (or U) shape.
These rules work only for parallel lines. If the lines are not parallel, the two crossings are different and the angles do not match. The rules also work backward: if a pair of corresponding angles is equal, the lines must be parallel.
Take any triangle $ABC$. Through $C$, draw the line parallel to side $AB$.
Side $CA$ crosses both parallel lines, so it is a transversal. The angle between $CA$ and the new line is alternate interior to $\angle A$, so it equals $\angle A$. In the same way, side $CB$ is a transversal, and the angle between $CB$ and the new line equals $\angle B$.
Now look along the new line at $C$. There are three angles side by side: the copy of $\angle A$, then $\angle C$ itself, then the copy of $\angle B$. Together they fill the straight line, so
$$\angle A + \angle B + \angle C = 180^\circ.$$
Nothing in this argument used the size or shape of the triangle, so it works for every triangle. That is what makes it a proof and not just a pattern.
The exterior angle theorem follows in one line. Extend side $BC$ past $C$. The exterior angle at $C$ and $\angle C$ make a straight line, so the exterior angle is $180^\circ - \angle C$. But $180^\circ - \angle C = \angle A + \angle B$. So the exterior angle equals the two remote interior angles added together.
Harder problems chain two or three facts. Suppose two parallel lines have a triangle drawn between them, with one corner on the top line and two on the bottom line. A side of the triangle makes $40^\circ$ with the top line. That side is a transversal, so the alternate interior angle at the bottom line is also $40^\circ$: one angle of the triangle. If another angle of the triangle is $65^\circ$, the angle sum gives the third: $180 - 40 - 65 = 75^\circ$.
The angle sum also reaches shapes with more sides. Any four-sided shape can be cut along a diagonal into two triangles, so its angles total $2 \times 180^\circ = 360^\circ$. A five-sided shape splits into three triangles, $540^\circ$ in all. Each extra side adds one more triangle and one more $180^\circ$.
Almost every angle problem in this lesson is solved with the same routine.
Why each move is allowed. Each fact in step 2 has a reason from this lesson: straight lines and turns, the translation that copies one crossing onto another, and the parallel line through a corner. Step 4 uses the rule that doing the same thing to both sides keeps an equation true.
A useful check for parallel lines: at any crossing, there are only two sizes of angle, and they add to $180^\circ$. If you ever get three different sizes, one pair was named wrongly.
Many parking lots use angled spaces, often at $60^\circ$ to the curb, because cars can pull in without a sharp turn. The painted lines between spaces are parallel, and the curb is a transversal crossing all of them. So every line meets the curb at the same angle: if the first stripe makes $60^\circ$ with the curb on one side, every stripe does, and the angle on the other side of each stripe is $180^\circ - 60^\circ = 120^\circ$. The painter only needs to set the angle once and slide the template along the curb.
The same idea shows up where a road crosses railroad tracks. The two rails are parallel, so the road meets both rails at the same angle. If a road crosses at $70^\circ$ instead of $90^\circ$, the gap between road and rail on one side is only $70^\circ$, and a bicycle wheel can slip into the gap along the rail. That is why road signs warn cyclists to cross tracks as close to $90^\circ$ as they can.
Builders describe how steep a roof is by its pitch: how many inches it rises for every $12$ inches across. A common "6-in-12" roof slopes at about $26.6^\circ$ from level. A simple gable roof, seen from the end, is an isosceles triangle: the two sloping sides match, so the two angles at the bottom are equal.
The angle at the peak comes from the angle sum: $180^\circ - 26.6^\circ - 26.6^\circ = 126.8^\circ$. A steeper "12-in-12" roof slopes at $45^\circ$, so its peak is $180^\circ - 45^\circ - 45^\circ = 90^\circ$, a right angle. Carpenters cutting the wooden rafters use these angles to mark where each board is cut so the two sides meet flush at the top.
Using parallel-line rules on lines that are not parallel. The F, Z and C rules need parallel lines. Look for the arrow marks that show lines are parallel, or a statement that they are.
Mixing up equal and supplementary. Corresponding and alternate angles are equal, but co-interior angles add to $180^\circ$. If your answer makes an obtuse angle equal an acute one, you used the wrong rule.
Using $360^\circ$ for a triangle. $360^\circ$ is a full turn and the total for a four-sided shape. A triangle's angles total $180^\circ$.
Adding the wrong angles for an exterior angle. The exterior angle equals the two interior angles that are not next to it. The one next to it is its straight-line partner, and together those two make $180^\circ$.
Stopping at $x$. When angles are given as expressions like $(3x + 15)^\circ$, the question may ask for the angle, not $x$. Put $x$ back in before you answer.
Triangle $ABC$ has $\angle A = 52^\circ$ and $\angle B = 71^\circ$. Draw a line through $C$ parallel to $AB$.
$\ell \parallel AB, \quad C \text{ on } \ell$
This line makes $CA$ and $CB$ into transversals, so the parallel-line rules apply.
Copy $\angle A$ to $C$ using side $CA$.
$\text{left angle at } C = \angle A = 52^\circ$
They are alternate interior angles, which are equal for parallel lines.
Copy $\angle B$ to $C$ using side $CB$.
$\text{right angle at } C = \angle B = 71^\circ$
Again alternate interior angles.
The three angles at $C$ fill the straight line $\ell$.
$52^\circ + \angle C + 71^\circ = 180^\circ$
Angles along a straight line add to $180^\circ$.
Add the known angles, then subtract from $180$.
$\angle C = 180^\circ - 123^\circ = 57^\circ$
$52 + 71 = 123$, and subtracting it from both sides leaves $\angle C$.
Triangle $PQR$ has $\angle P = 48^\circ$ and $\angle Q = 67^\circ$. Side $QR$ is extended past $R$. Find $\angle PRQ$ first.
$\angle PRQ = 180^\circ - 48^\circ - 67^\circ$
The three interior angles add to $180^\circ$.
Do the subtraction.
$\angle PRQ = 180^\circ - 115^\circ = 65^\circ$
$48 + 67 = 115$.
The exterior angle at $R$ and $\angle PRQ$ lie on the straight line $QR$ extended.
$\text{ext} + 65^\circ = 180^\circ$
Angles along a straight line add to $180^\circ$.
Solve for the exterior angle.
$\text{ext} = 180^\circ - 65^\circ = 115^\circ$
Subtract $65^\circ$ from both sides.
Now add the two remote interior angles.
$\angle P + \angle Q = 48^\circ + 67^\circ = 115^\circ$
The exterior angle theorem says this sum should match.
Compare the two answers.
$\text{ext} = \angle P + \angle Q = 115^\circ$
Both routes agree. Next time the theorem gives the exterior angle in one step.
Lines $\ell$ and $m$ are parallel. A transversal makes corresponding angles of $(3x + 15)^\circ$ at $\ell$ and $(5x - 25)^\circ$ at $m$. Set them equal.
$3x + 15 = 5x - 25$
Corresponding angles are equal when the lines are parallel.
Subtract $3x$ from both sides.
$15 = 2x - 25$
This puts all the $x$ terms on one side.
Add $25$ to both sides.
$40 = 2x$
This puts all the numbers on the other side.
Divide both sides by $2$.
$x = 20$
Dividing undoes the multiplication by $2$.
Find the angle by putting $x = 20$ into both expressions.
$3(20) + 15 = 75, \qquad 5(20) - 25 = 75$
Both give $75^\circ$, which checks that they are equal.
Find the co-interior partner of the $75^\circ$ angle at $m$.
$180^\circ - 75^\circ = 105^\circ$
Co-interior angles add to $180^\circ$.
List the angles at each crossing and check.
$75^\circ, \ 105^\circ, \ 75^\circ, \ 105^\circ \quad\text{and}\quad 75 + 105 = 180$
At every crossing there are only two sizes of angle, and they add to $180^\circ$.
Write the angle sum.
$x + 2x + 3x = 180$
The three angles of a triangle add to $180^\circ$.
Combine the $x$ terms.
$6x = 180$
$1 + 2 + 3 = 6$.
Divide both sides by $6$.
Find each angle and check.
Triangle $ABC$ has $\angle A = 40^\circ$ and $\angle B = 53^\circ$. A line is drawn through $C$ parallel to $AB$. Which reason shows that $\angle A + \angle B + \angle C = 180^\circ$?
Lines $\ell$ and $m$ are parallel, and a transversal crosses both. Between the lines on the left of the transversal, the angle at $\ell$ is $49^\circ$. Complete the worked solution to find the angle at $m$ on the same side, between the lines, and then its vertical angle.
Name the pair: both angles are between the lines, on the same side of the transversal.
$\text{co-interior angles}$
Same side and between the parallel lines is what co-interior means.
Write the rule for co-interior angles, with $y$ for the unknown angle.
$49 + y = 180$
Co-interior angles add to $180^\circ$: one of them is the straight-line partner of an angle equal to the other.
Subtract $49$ from both sides.
$y = 180 - 49 =$ y
Subtracting undoes the addition.
Find the vertical angle across the crossing point at $m$.
$\text{vertical angle} = y =$ v
Vertical angles are equal, because each is $180^\circ$ minus the same neighbor.
A transversal crosses two parallel lines. At the upper crossing, the angle above the top line and to the right of the transversal is $39^\circ$. What is the angle above the lower line and to the right of the transversal?
The angle is answer degrees.
Two angles of a triangle measure $42^\circ$ and $53^\circ$. What is the third angle?
The third angle is answer degrees.
One side of a triangle is extended to make an exterior angle of $87^\circ$. One of the two interior angles not next to it is $30^\circ$. What is the other one?
The other angle is answer degrees.
Two parallel lines are cut by a transversal. A pair of corresponding angles measure $(6x - 26)^\circ$ and $(3x + 10)^\circ$. Find $x$.
Answer:
A stepladder's two legs are the same length and meet at the top at an angle of $30^\circ$. The legs and the floor make a triangle. What angle does each leg make with the floor?
Each leg makes an angle of answer degrees with the floor.
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Two parallel lines are cut by a transversal. A pair of corresponding angles measure $(5x - 30)^\circ$ and $(3x + 10)^\circ$. Find $x$.
Answer:
You can find missing angles in a triangle and around a transversal, including angles written with $x$. Without looking: why do a triangle's angles add to a straight line, and which angles at two parallel crossings are equal?
15. Your turn: a triangle's angles are $x^\circ$, $2x^\circ$ and $3x^\circ$. Find each angle., step 3
$x = 30$
Dividing undoes the multiplication by $6$.
15. Your turn: a triangle's angles are $x^\circ$, $2x^\circ$ and $3x^\circ$. Find each angle., step 4
$30^\circ, \ 60^\circ, \ 90^\circ; \qquad 30 + 60 + 90 = 180$
Put $x = 30$ into each expression. The largest is a right angle.