Students can practice finding missing angles labeled a through l in Section A of this Angles on Parallel Lines worksheet.

Let's solve each of the missing angles in Section A of the worksheet, using properties of parallel lines, transversals, corresponding angles, alternate angles, co-interior (consecutive) angles, and triangle angle sums.

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🔷 Problem 1: Find a°

- Given: One angle is 51°.
- The lines are parallel, and the transversal forms an angle a° on the opposite side of the transversal but in the same relative position — this is a corresponding angle.
- Corresponding angles are equal when lines are parallel.

✔ So,
$$
a = 51^\circ
$$

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🔷 Problem 2: Find b° and c°

We have two parallel lines cut by two transversals.

Given:
- Top right angle = 104°
- Bottom left angle = 117°

#### Step 1: Find b°
The angle marked b° is vertically opposite to the 104° angle at the top.
Vertically opposite angles are equal.

So,
$$
b = 104^\circ
$$

#### Step 2: Find c°
Now look at the bottom right angle c°. It’s on the same side as the 117° angle and forms a co-interior (same-side interior) angle with it.

But wait — the two lines are parallel, so co-interior angles add up to 180°.

So:
$$
c + 117^\circ = 180^\circ \Rightarrow c = 180 - 117 = 63^\circ
$$

✔ So,
$$
b = 104^\circ,\quad c = 63^\circ
$$

---

🔷 Problem 3: Find d° and e°

We have multiple parallel lines and transversals.

Given:
- One angle = 67°
- Another = 75°
- We need d° and e°

Let’s analyze:

#### Step 1: Find e°
Look at the 67° angle. It appears to be corresponding to e°, since they are both on the same side of the transversal and in the same relative position between parallel lines.

So,
$$
e = 67^\circ
$$

#### Step 2: Find d°
Now look at the 75° angle. It’s adjacent to d° along a straight line? Wait — actually, let's trace carefully.

There’s a triangle formed or just a transversal cutting through.

Alternatively, notice that d° and 75° appear to be alternate interior angles if we consider the lower transversal.

Wait — more accurately, d° and 75° are vertically opposite or corresponding?

Actually, from the diagram:
The 75° angle is on one line, and d° is on a different transversal.

But note: There’s a transversal crossing the lines forming 75° and d°.

Wait — better idea: Look at the angle between the two transversals.

Actually, the key is that d° is corresponding to the 75° angle, because they’re in the same position relative to the parallel lines and transversal.

Yes — they are corresponding angles.

So,
$$
d = 75^\circ
$$

✔ So,
$$
d = 75^\circ,\quad e = 67^\circ
$$

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🔷 Problem 4: Find f° and g°

This is an isosceles triangle with two equal sides (marked with ticks), and the base is on a horizontal line, with another horizontal line above.

The triangle has:
- Two equal sides → base angles are equal.
- The top angle is g°, and the bottom-left angle is f°.

But also, the base of the triangle is on a straight line, and there’s a parallel line above.

Let’s use properties of parallel lines and isosceles triangles.

Since the triangle is isosceles with two equal sides, the base angles are equal.

Let’s denote:
- The two base angles = f° (left) and g° (right)? Wait — no.

Wait: The equal sides are the two legs from the apex down to the base.

So the two base angles are equal.

But the top angle is labeled g°, and the bottom-left angle is f°.

Wait — no: looking at the diagram:

- The triangle has two equal sides: from the top vertex to the base.
- So the two base angles are equal.
- But only f° is labeled at the bottom-left.
- The top angle is g°.

But f° is not the base angle — it's the exterior angle at the bottom-left vertex.

Ah! That’s important.

So the interior angle at the bottom-left is adjacent to f°, and they form a straight line.

But the triangle is drawn such that the base is on the lower line, and the top vertex touches the upper line.

Also, the upper line is parallel to the lower line.

So, the angle at the top (g°) is between the upper line and one side of the triangle.

And the side of the triangle is a transversal.

Let’s re-analyze.

Let’s suppose:
- The triangle has two equal sides (marked with ticks).
- The top vertex is on the upper line.
- The base is on the lower line.
- The two sides go from the top to the base.

Then, the base angles (at the bottom) are equal.

But the angle f° is outside the triangle, on the lower line, and is adjacent to the interior base angle.

So, the interior base angle = $180^\circ - f^\circ$, but that can't be.

Wait — no: f° is the exterior angle at the base.

But actually, f° is on the straight line, and the triangle is below it.

So the interior angle of the triangle at the base is supplementary to f°.

But the triangle is isosceles, so the two base angles are equal.

Let’s call the interior base angles both $ x $. Then:
$$
\text{Sum of angles in triangle} = 180^\circ
\Rightarrow x + x + g = 180^\circ \Rightarrow 2x + g = 180
$$

But also, f° is the angle between the extension of the base and the side — that is, f° is the exterior angle at the base.

So:
$$
f = 180^\circ - x
\Rightarrow x = 180^\circ - f
$$

Substitute into triangle sum:
$$
2(180 - f) + g = 180 \\
360 - 2f + g = 180 \\
g = 180 - 360 + 2f = 2f - 180
$$

But we don’t know f yet.

Wait — perhaps we can use parallel lines.

Note: The upper line and lower line are parallel, and the side of the triangle is a transversal.

The angle at the top (g°) is between the upper line and the transversal.

The angle at the bottom-left inside the triangle is between the lower line and the transversal.

Because the lines are parallel, these two angles are alternate interior angles.

So:
$$
\text{Interior base angle} = g^\circ
$$

But earlier we said the two base angles are equal, so both are $ g^\circ $.

Then:
$$
\text{Sum of angles in triangle} = g + g + g = 3g = 180^\circ \Rightarrow g = 60^\circ
$$

Wait — that would mean all angles are 60°, but then the triangle is equilateral.

But the top angle is g°, and the base angles are also g°, so yes.

So:
$$
g = 60^\circ
$$

Then, the interior base angle = $ 60^\circ $

Now, f° is the exterior angle at the base, so:
$$
f = 180^\circ - 60^\circ = 120^\circ
$$

✔ So,
$$
f = 120^\circ,\quad g = 60^\circ
$$

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🔷 Problem 5: Find h° and i°

We have an isosceles triangle again, with two equal sides (marked with ticks), and the base angles are equal.

Given:
- An angle outside the triangle at the bottom-left is 124°.
- This is adjacent to the interior base angle.

So, the interior base angle = $180^\circ - 124^\circ = 56^\circ$

Since the triangle is isosceles, both base angles are $56^\circ$.

So, the top angle (h°) is:
$$
h = 180^\circ - 56^\circ - 56^\circ = 68^\circ
$$

Now, i° is the angle at the top, between the upper line and the side of the triangle.

But the upper line is parallel to the lower line.

So, the angle i° and the interior base angle (56°) are alternate interior angles.

Thus:
$$
i = 56^\circ
$$

✔ So,
$$
h = 68^\circ,\quad i = 56^\circ
$$

---

🔷 Problem 6: Find j°, k°, l°

We have a triangle with:
- Top angle = 41°
- Right-bottom angle = 119°
- A horizontal line cuts across the triangle.

We need to find j°, k°, l°

Let’s label the triangle:

- Vertex A: top (41°)
- Vertex B: bottom-right (119°)
- Vertex C: bottom-left (unknown)

But we can find angle at C:
$$
\text{Sum of angles in triangle} = 180^\circ \\
\Rightarrow \text{Angle C} = 180 - 41 - 119 = 20^\circ
$$

So, the bottom-left interior angle = $20^\circ$

Now, k° is the angle at the bottom-left, but it's labeled on the lower line, and it's adjacent to the triangle’s angle.

Wait — no: k° is inside the triangle, at the bottom-left corner.

Wait — the diagram shows:

- A triangle with angles: top = 41°, bottom-right = 119°, so bottom-left = 20°
- A horizontal line crosses the triangle, forming angles j° and l°

Let’s examine:

- j° is the angle between the horizontal line and the left side of the triangle.
- k° is the interior angle at the bottom-left, which we found is $20^\circ$
- l° is the angle between the horizontal line and the right side of the triangle.

But the horizontal line is a transversal cutting through the triangle.

We can use parallel lines? Are the top and bottom lines parallel?

No — not necessarily. But the horizontal line is straight, and it intersects the triangle.

But here’s the key: The horizontal line is a transversal, and we can use angle relationships.

Let’s look at j°:

At the bottom-left vertex:
- The interior angle is $20^\circ$
- The horizontal line splits the angle?
- No — actually, j° is on the other side of the triangle.

Wait — the horizontal line passes through the left side of the triangle, forming j°.

But j° is alternate interior or corresponding?

Wait — perhaps j° is equal to the top angle?

Let’s think.

The horizontal line is cutting across the triangle.

But j° is the angle between the horizontal line and the left side of the triangle.

Similarly, l° is the angle between the horizontal line and the right side.

But we know:
- The interior angle at the bottom-left is $20^\circ$
- The interior angle at the bottom-right is $119^\circ$

Now, k° is the interior angle at the bottom-left, so:
$$
k = 20^\circ
$$

Now, j°: At the bottom-left vertex, the horizontal line is above the triangle.

So, the angle between the horizontal line and the left side is j°, and it's adjacent to the interior angle $k = 20^\circ$.

But wait — the horizontal line is not the base — the base is the bottom edge.

Wait — the diagram shows a horizontal line cutting across the triangle, not the base.

So the base is the bottom edge, and the horizontal line is above it, cutting through the triangle.

So, at the left side, the horizontal line intersects the left side of the triangle, forming angle j°.

Similarly, on the right side, it forms angle l°.

Now, the top angle is $41^\circ$, and the horizontal line is parallel to the base? Not stated.

Wait — but the horizontal line is shown with arrows — suggesting it's parallel to the base?

Yes — both lines have arrows pointing right, so likely parallel.

So assume: The horizontal line is parallel to the base.

Then, the left side of the triangle is a transversal.

So, j° and the interior angle at the bottom-left (k° = 20°) are alternate interior angles.

Therefore:
$$
j = 20^\circ
$$

Similarly, on the right side, l° and the interior angle at the bottom-right (119°) are alternate interior angles?

Wait — but l° is on the same side of the transversal.

Wait — let’s see:

- The horizontal line is above the base.
- The right side of the triangle goes from the top to the bottom-right.
- The angle l° is between the horizontal line and the right side, on the outer side.

But the interior angle at bottom-right is 119°, and the horizontal line is above.

So, the angle between the horizontal line and the right side — l° — is corresponding to the interior angle at the bottom-right?

Wait — no.

Actually, the angle l° and the interior angle at the bottom-right are on the same side of the transversal, and the lines are parallel, so they are co-interior angles?

Wait — no.

Let’s clarify.

The transversal is the right side of the triangle.

It intersects:
- The horizontal line (upper)
- The base (lower)

So, the angle l° is on the upper line, and the interior angle at bottom-right is on the lower line.

They are on the same side of the transversal → so they are co-interior angles.

But co-interior angles add to $180^\circ$ if the lines are parallel.

So:
$$
l + 119^\circ = 180^\circ \Rightarrow l = 61^\circ
$$

Alternatively, maybe l° is equal to the top angle?

Wait — no.

Wait — the top angle is 41°, and it's not adjacent.

But here’s a better idea:

Use triangle angles and parallel lines.

We already have:
- k = 20° (interior bottom-left angle)
- j = 20° (alternate interior to k, since lines are parallel)

Now for l°:

The interior angle at bottom-right is $119^\circ$. Since the horizontal line is parallel to the base, and the right side is a transversal, then:

- The angle between the horizontal line and the right side (l°) and the interior angle at bottom-right are co-interior angles → sum to $180^\circ$

So:
$$
l + 119^\circ = 180^\circ \Rightarrow l = 61^\circ
$$

Alternatively, you might think of l° as corresponding to some angle.

But l° is on the upper line, and the interior angle at bottom-right is on the lower line — same side of transversal → co-interior.

So yes, $ l = 61^\circ $

Now, what about j°?

Earlier I said $ j = 20^\circ $, because it’s alternate interior to $ k = 20^\circ $

But wait — j° is on the upper line, and k° is on the lower line, same side of transversal?

No — j° and k° are on opposite sides of the transversal.

Let’s define:

- Transversal: left side of triangle
- Upper line: horizontal
- Lower line: base

Then:
- j° is the angle between upper line and transversal, on the left side
- k° is the angle between lower line and transversal, on the left side

So they are on the same side of the transversal → co-interior angles

Wait — that means:
$$
j + k = 180^\circ \Rightarrow j + 20^\circ = 180^\circ \Rightarrow j = 160^\circ
$$

Wait — that contradicts earlier.

So which is correct?

Let’s look at the diagram carefully.

The horizontal line is above the triangle.

At the left side, the triangle has a side going from top to bottom-left.

The horizontal line cuts this side.

So, the angle j° is between the horizontal line and the left side, and it's outside the triangle.

Similarly, the angle k° is inside the triangle at the bottom-left.

Now, the horizontal line and the base are parallel.

The left side is a transversal.

Then, the angle j° and the angle k° are on the same side of the transversal.

But j° is above the transversal, k° is below?

No — both are on the left side.

Wait — j° is between the horizontal line and the left side, and k° is between the base and the left side.

So, if you follow the transversal (left side), the angle j° is on the upper side, and k° is on the lower side.

So they are on opposite sides of the transversal → alternate interior angles?

But alternate interior angles are between the two parallel lines.

So yes — j° and k° are alternate interior angles → equal.

So:
$$
j = k = 20^\circ
$$

But wait — j° is not between the lines — it's on the upper line, and k° is on the lower line.

But both are between the parallel lines?

Yes — the horizontal line and the base are parallel.

The left side of the triangle crosses both.

So, the angle j° is between the upper line and the transversal, and k° is between the lower line and the transversal.

And they are on opposite sides of the transversal → alternate interior angles → equal.

So:
$$
j = k = 20^\circ
$$

Now, for l°:

On the right side, the angle l° is between the horizontal line and the right side.

The interior angle at bottom-right is $119^\circ$, which is between the base and the right side.

These two angles are on opposite sides of the transversal → alternate interior angles?

But l° and $119^\circ$ are both on the right side of the transversal.

Wait — l° is above, $119^\circ$ is below.

So they are on opposite sides → alternate interior angles → should be equal?

But $119^\circ$ is large, and $l°$ is small.

Wait — no.

Actually, l° and the interior angle at bottom-right are on the same side of the transversal.

So they are co-interior angles → sum to $180^\circ$

So:
$$
l + 119^\circ = 180^\circ \Rightarrow l = 61^\circ
$$

Alternatively, l° could be corresponding to some angle.

But the top angle is $41^\circ$, and l° is not corresponding.

Wait — but l° is adjacent to the top angle?

No.

Another way: The angle at the top is $41^\circ$, and the horizontal line is parallel to the base.

Then, the angle between the horizontal line and the right side (l°) and the angle between the base and the right side (119°) are co-interior → sum to $180^\circ$ → $l = 61^\circ$

✔ So,
$$
j = 20^\circ,\quad k = 20^\circ,\quad l = 61^\circ
$$

Wait — but k° is the interior angle at bottom-left, which we calculated as $20^\circ$, so that’s fine.

But j° is alternate interior to k°, so $j = 20^\circ$

And l° is co-interior with $119^\circ$, so $l = 61^\circ$

Yes.

But wait — is there a contradiction?

Let’s double-check the top angle.

We had:
- Top = $41^\circ$
- Bottom-right = $119^\circ$
- Bottom-left = $180 - 41 - 119 = 20^\circ$ → correct.

So k = 20^\circ

Now, j° is alternate interior to k° → $j = 20^\circ$

l° is co-interior with $119^\circ$ → $l = 61^\circ$

✔ All good.

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✔ Final Answers:

| Variable | Value |
|--------|-------|
| a | $51^\circ$ |
| b | $104^\circ$ |
| c | $63^\circ$ |
| d | $75^\circ$ |
| e | $67^\circ$ |
| f | $120^\circ$ |
| g | $60^\circ$ |
| h | $68^\circ$ |
| i | $56^\circ$ |
| j | $20^\circ$ |
| k | $20^\circ$ |
| l | $61^\circ$ |

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📌 Summary of Key Concepts Used:

- Corresponding angles = equal (when lines are parallel)
- Alternate interior angles = equal
- Co-interior (consecutive) angles = sum to $180^\circ$
- Vertically opposite angles = equal
- Triangle angle sum = $180^\circ$
- Isosceles triangle → base angles equal
- Exterior angles = supplementary to interior angles

Let me know if you'd like a visual explanation or diagram annotations!