Geometry worksheet on angles formed by parallel lines and transversals, including alternate interior and exterior angles.

Let’s solve each angle step by step, using what we know about parallel lines and transversals.

We’ll go diagram by diagram. Remember:

- When two parallel lines are cut by a transversal:
- Alternate interior angles are equal.
- Alternate exterior angles are equal.
- Corresponding angles are equal.
- Consecutive interior angles (same side) add to 180°.
- Vertical angles are equal.
- Angles on a straight line add to 180°.
- Triangle angles add to 180°.
- Quadrilateral angles add to 360°.

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First Diagram (top left): Parallelogram with diagonal

Given: One angle is 86°, another is 38°. We need ∠1 and 2.

In a parallelogram, opposite angles are equal, consecutive angles add to 180°.

But here, there’s a diagonal splitting it into two triangles.

Look at the triangle that has the 38° angle and the 86° angle — wait, actually, let’s label carefully.

Actually, looking at the figure: The 86° is at top right, 38° is at bottom left. The diagonal connects them? Not exactly.

Wait — better approach: In the lower triangle, we have angle 38° and angle 2. Also, since it's a parallelogram, opposite sides are parallel.

Actually, notice: The 86° angle and angle 1 are alternate interior angles if we consider the top and bottom sides as parallel and the diagonal as transversal? Let me think.

Alternatively, in the upper triangle: angles are ∠1, 86°, and the third angle which is same as ∠2 because of parallel lines? Hmm.

Better: Since it’s a parallelogram, consecutive angles sum to 180°.

So, the angle adjacent to 86° should be 180° - 86° = 94°. But that’s not directly helpful.

Wait — look at the triangle formed by the diagonal. It splits the parallelogram into two congruent triangles.

In the lower triangle: we have angle 38°, angle 2, and the angle between them which is part of the parallelogram.

Actually, perhaps easier: The angle marked 86° and angle 1 are on opposite sides of the diagonal. If the top and bottom are parallel, then angle 1 and the 38° angle might be related.

Wait — I think I’m overcomplicating.

Let me use this: In the triangle that includes angle 1 and the 86° angle — they share a vertex. Actually, no.

Alternative idea: The 86° angle and the angle next to it (on the same side) must add to 180° if they’re consecutive in the parallelogram. So the angle below the 86° is 94°. Then in the lower triangle, we have angles: 38°, 94°, and angle 2? No, that doesn’t make sense.

Wait — let’s assume the diagonal creates two triangles. In the upper triangle: angles are ∠1, 86°, and the angle at the left-top corner. That left-top corner angle is equal to the bottom-right angle due to parallelogram properties? This is messy.

Perhaps the 86° and 38° are not in the same triangle.

Looking again: The figure shows a parallelogram with a diagonal from top-left to bottom-right. At top-right, angle is 86°. At bottom-left, angle is 38°. We need ∠1 (at top-left, between top side and diagonal) and ∠2 (at bottom-left, between bottom side and diagonal).

Since top and bottom are parallel, and diagonal is transversal, then ∠1 and the 38° angle are alternate interior angles → so ∠1 = 38°.

Similarly, the 86° angle and ∠2 are alternate interior angles → so ∠2 = 86°.

Is that correct? Let’s verify.

If top || bottom, and diagonal crosses them, then yes: ∠1 (above diagonal, left side) and 38° (below diagonal, left side) — wait, no, 38° is at bottom-left, which is on the same side as ∠1? Actually, depending on orientation.

Standard rule: Alternate interior angles are on opposite sides of the transversal and inside the parallel lines.

So if diagonal is transversal, and top and bottom are parallel, then:

- Angle between top side and diagonal (∠1) and angle between bottom side and diagonal on the opposite side — that would be the angle at bottom-right, not bottom-left.

I think I got it wrong.

Let me draw mentally: Parallelogram ABCD, A top-left, B top-right, C bottom-right, D bottom-left. Diagonal AC.

Angle at B is 86°, angle at D is 38°.

We want ∠1 = angle at A between AB and AC.

∠2 = angle at D between CD and AC? Or between AD and AC? The diagram says "m∠2" near the 38°, so probably ∠2 is part of the 38° or adjacent.

Actually, re-examining: The 38° is labeled at vertex D, and ∠2 is also at D, but between the diagonal and the side. Similarly, 86° at B, ∠1 at A.

Perhaps in triangle ADC or something.

Another approach: In triangle ABC or ADC.

Assume diagonal is from A to C.

Then in triangle ABC: angles at B is 86°, at A is ∠1, at C is some angle.

But we don't know.

Note that in parallelogram, angle at A + angle at B = 180°, so angle at A = 180° - 86° = 94°.

This 94° is split into ∠1 and another angle by the diagonal.

Similarly, angle at D is 38°, and it's split into ∠2 and another angle.

But without more info, hard.

Wait — perhaps the 38° and 86° are the full angles at those vertices, and the diagonal creates smaller angles.

But in that case, for triangle formed by diagonal, say triangle ABD or something.

I recall that in such problems, often the given angles are the ones in the triangles.

Let me try this: In the lower triangle (vertices D, C, and the intersection), but it's not clear.

Perhaps the 38° is angle at D in the lower triangle, and 86° is angle at B in the upper triangle, and since the diagonal is common, and sides are parallel, then the other angles can be found.

Assume that the diagonal makes two triangles: triangle ABD and CBD, but usually it's AC or BD.

To save time, let's look for standard solution or logical deduction.

Notice that if we consider the diagonal as transversal, and top and bottom parallel, then the alternate interior angles should be equal.

Specifically, the angle between the top side and the diagonal at A (∠1) should equal the angle between the bottom side and the diagonal at C, but we don't have that.

At D, the angle between the bottom side and the diagonal is ∠2, and at B, the angle between the top side and the diagonal is part of the 86°.

This is confusing.

Let me calculate based on triangle sum.

Suppose in the upper triangle (say triangle ABC if diagonal is AC), but let's define.

Assume the parallelogram has points: let's call P Q R S, with P top-left, Q top-right, R bottom-right, S bottom-left. Diagonal PR.

Angle at Q is 86°, angle at S is 38°.

We want m∠1 = angle at P between PQ and PR.

m∠2 = angle at S between SR and PR? Or between SP and PR? The diagram likely has ∠2 at S between the diagonal and the side towards R or P.

Typically, in such diagrams, ∠2 is the angle in the triangle at S.

So in triangle PSR or PQR.

Consider triangle PQR: but Q is 86°, P is ∠1, R is unknown.

Not helpful.

Consider that the diagonal divides the parallelogram into two congruent triangles only if it's a rhombus or rectangle, but not necessarily.

Another idea: The sum of angles around point P is 360°, but too vague.

Perhaps the 86° and 38° are not the vertex angles but the angles in the triangles.

Let's read the diagram description: "m∠1 = ___", "m∠2 = ___", and there's a 86° at top-right, 38° at bottom-left, and the diagonal.

I recall that in many textbooks, for a parallelogram with diagonal, if you have an angle at one end, the alternate interior angle gives you the other.

Let's assume that the line from top-left to bottom-right is the diagonal.

Then, the angle at top-left between the top side and diagonal is ∠1.

The angle at bottom-left between the bottom side and diagonal is ∠2.

Now, since top and bottom are parallel, and the diagonal is transversal, then ∠1 and the angle at bottom-right between the bottom side and diagonal are alternate interior angles, so equal.

Similarly, ∠2 and the angle at top-right between the top side and diagonal are alternate interior angles.

At top-right, the full angle is 86°, which is composed of the angle between top side and diagonal and the angle between diagonal and right side.

But we don't know how it's split.

Unless the 86° is the angle in the triangle.

Perhaps the 86° is the angle at Q in triangle PQR, but then we need more.

Let's look at the second diagram for clue, but better to move on and come back.

Perhaps for this diagram, since it's a parallelogram, and diagonal, then the triangle containing the 38° has angles 38°, ∠2, and the angle at the other end.

But let's calculate the missing angle in the triangle that has the 38°.

Assume that in the lower triangle, we have angles: at S (bottom-left) is 38°, at R (bottom-right) is some angle, at P (top-left) is ∠1 or something.

I think I found a better way: in the parallelogram, opposite angles are equal, so angle at P = angle at R, angle at Q = angle at S.

But here, angle at Q is 86°, angle at S is 38°, but 86° ≠ 38°, so they are not opposite; probably Q and S are adjacent or something.

In standard labeling, if P,Q,R,S in order, then P and R are opposite, Q and S are opposite.

So if angle at Q is 86°, then angle at S should also be 86° if opposite, but it's given as 38°, contradiction.

Unless the 86° and 38° are not the vertex angles of the parallelogram, but the angles in the triangles formed by the diagonal.

That must be it. The 86° is the angle at Q in the upper triangle, and 38° is the angle at S in the lower triangle.

So, for the upper triangle (say P-Q-R, but with diagonal P-R, so triangle P-Q-R is not a triangle; the diagonal is P-R, so triangles are P-Q-R and P-S-R? No.

If diagonal is from P to R, then triangles are P-Q-R and P-S-R, but P-Q-R includes points P,Q,R, which is not a triangle if Q and R are connected.

Standard: in quadrilateral PQRS, diagonal PR divides it into triangle PQR and triangle PSR? No, triangle PQR would include side QR, but diagonal is PR, so the two triangles are triangle PQR and triangle PSR only if S is connected, but typically it's triangle PQR and triangle RSP or something.

Let's define: vertices A,B,C,D in order, A-B-C-D-A. Diagonal A-C. Then triangles are ABC and ADC.

So in triangle ABC, angles at A,B,C.

In triangle ADC, angles at A,D,C.

In this case, for our diagram, suppose A top-left, B top-right, C bottom-right, D bottom-left. Diagonal A-C.

Then angle at B is given as 86° — this is angle of the parallelogram at B, which is angle ABC.

Similarly, angle at D is 38°, angle ADC.

In parallelogram, angle ABC + angle BAD = 180°, etc.

But for the triangles, in triangle ABC, we have angle at B = 86°, angle at A is part of angle BAD, angle at C is part of angle BCD.

This is still complicated.

Perhaps the 86° is the angle in triangle ABC at B, and 38° is the angle in triangle ADC at D.

Then in triangle ABC, we have angle at B = 86°, and since AB || CD, and BC is transversal, then angle at B and angle at C are consecutive interior, so angle at C in the parallelogram is 180° - 86° = 94°, but in triangle ABC, angle at C is the same as the parallelogram's angle at C if no diagonal issue, but with diagonal, in triangle ABC, angle at C is the angle between BC and AC, which is not the full angle.

I think I need to accept that for this diagram, using alternate interior angles with the diagonal as transversal.

Let me search for a different strategy.

Notice that in the first diagram, the two triangles share the diagonal, and the sum of angles in each triangle is 180°.

Also, the angles at the ends of the diagonal are related.

For example, at vertex A, the angle of the parallelogram is split into ∠1 and another angle, say x.

At vertex C, the angle is split into y and z.

But too many variables.

Perhaps the 86° and 38° are the angles that are not at the diagonal vertices.

Let's assume that in the upper triangle, the angles are ∠1, 86°, and the angle at the bottom-right of the upper triangle.

But the bottom-right of the upper triangle is the same as the top-right of the lower triangle.

And since the sides are parallel, the alternate interior angles are equal.

Specifically, the angle between the diagonal and the top side at A (∠1) should equal the angle between the diagonal and the bottom side at C, because top || bottom, diagonal transversal.

Similarly, the angle between the diagonal and the left side at A should equal the angle between the diagonal and the right side at C, but we don't have that.

For the lower triangle, at D, we have 38°, which is the angle between the bottom side and the left side, but with diagonal, it's split.

I recall that in some sources, for this exact problem, m∠1 = 38°, m∠2 = 86°, by alternate interior angles.

Let me verify with logic.

Suppose we have two parallel lines (top and bottom), and a transversal (diagonal).

Then, the alternate interior angles are equal.

So, the angle that the diagonal makes with the top line at A should equal the angle it makes with the bottom line at C, on the opposite side.

But at D, the 38° is given, which is at the bottom-left, so if we consider the angle between the bottom line and the diagonal at D, that would be ∠2, and it should be equal to the angle between the top line and the diagonal at B, on the opposite side.

At B, the full angle is 86°, which is between the top line and the right side, not the diagonal.

Unless the diagonal is such that at B, the angle between top line and diagonal is part of it.

Perhaps the 86° is the angle in the triangle at B, so in triangle ABC, angle at B is 86°, and since AB || CD, and BC is transversal, then angle at B and angle at C are supplementary, so angle at C in the parallelogram is 94°, but in triangle ABC, angle at C is the angle between BC and AC, which is not 94°.

I think I have to guess that m∠1 = 38°, m∠2 = 86°, as it's a common result.

Or perhaps calculate.

Let's do this: in the lower triangle, assume it has angles: at D: 38°, at C: let's call it c, at A: let's call it a.

Sum 38 + c + a = 180.

In the upper triangle, at B: 86°, at C: d, at A: b.

Sum 86 + d + b = 180.

At vertex A, the total angle is a + b.

At vertex C, c + d.

In parallelogram, a + b + 86 + 38 = 360? No, the sum of all four vertex angles is 360°, so (a+b) + 86 + (c+d) + 38 = 360, so a+b+c+d = 360 - 86 - 38 = 236.

From the two triangles: from lower: a + c = 180 - 38 = 142

From upper: b + d = 180 - 86 = 94

Then a+b+c+d = (a+c) + (b+d) = 142 + 94 = 236, which matches, good.

But we need individual values.

Now, because of parallel lines, at vertex A, the angle a + b is the angle of the parallelogram, and at C, c + d is the opposite angle, so a+b = c+d, because opposite angles in parallelogram are equal.

Is that true? Yes, in parallelogram, opposite angles are equal, so angle at A = angle at C, so a+b = c+d.

But from above, a+b + c+d = 236, and a+b = c+d, so 2(a+b) = 236, so a+b = 118, c+d = 118.

From lower triangle: a + c = 142

From upper: b + d = 94

And a+b = 118

c+d = 118

Now, from a+b = 118, and a+c = 142, subtract: (a+c) - (a+b) = 142 - 118 => c - b = 24

From b+d = 94, and c+d = 118, subtract: (c+d) - (b+d) = 118 - 94 => c - b = 24, same.

Now, we have c = b + 24

From b+d = 94, d = 94 - b

From c+d = 118, (b+24) + (94 - b) = 118 => 118 = 118, always true.

So we have infinite solutions? That can't be.

We need another relation.

Ah, yes! Because the lines are parallel, the alternate interior angles are equal.

Specifically, for the diagonal AC as transversal, cutting parallel lines AB and CD.

AB and CD are parallel (top and bottom).

Transversal AC.

Then, the alternate interior angles are: angle between AB and AC at A, and angle between CD and AC at C.

Angle at A between AB and AC is ∠1 = b (in my notation, since at A, in upper triangle, angle between AB and AC is b).

At C, between CD and AC, in the lower triangle, angle between CD and AC is c.

And since AB || CD, and AC transversal, then alternate interior angles are equal, so b = c.

Oh! So b = c.

From earlier, c = b + 24, so b = b + 24, which implies 0=24, contradiction.

What's wrong?

Perhaps the alternate interior angles are different.

Let's define carefully.

Line AB (top), line CD (bottom), parallel.

Transversal AC.

At point A, the angle between AB and AC. Depending on direction.

If we go from A to C, then at A, the angle between the transversal and the line AB.

The alternate interior angle would be at C, between the transversal and the line CD, on the opposite side.

In the diagram, at A, the angle between AB and AC is ∠1, which is inside the parallelogram.

At C, the angle between CD and AC: if CD is the bottom line, and AC is coming from A to C, then at C, the angle between CD and AC could be the angle in the lower triangle or upper.

In the lower triangle, at C, the angle is between BC and AC or between CD and AC.

In triangle ADC, at C, the angle is between DC and AC, which is c in my earlier notation.

And since AB || CD, and AC transversal, then the alternate interior angles are: the angle at A between AB and AC, and the angle at C between CD and AC, and they should be equal if they are on opposite sides of the transversal.

In this case, both are on the same side or opposite?

Typically, for transversal AC, at A, the angle below AB and above AC, and at C, the angle above CD and below AC, which would be the alternate interior.

In the parallelogram, if we assume it's convex, then at A, ∠1 is the angle inside, between AB and AC.

At C, in the lower triangle, the angle between CD and AC is also inside, and they are on opposite sides of the transversal AC, so yes, they should be alternate interior angles, so equal.

So b = c.

But from earlier, from the sums, we had c = b + 24, which is impossible unless 24=0.

So where is the mistake?

I think I misidentified the angles.

In the lower triangle, at C, the angle is between BC and AC or between DC and AC?

In triangle ADC, vertices A,D,C, so at C, the angle is between sides DC and AC.

DC is the bottom side, from D to C.

AC is the diagonal.

So yes, angle between DC and AC.

At A, in triangle ABC, angle between AB and AC.

AB is top side, from A to B.

AC is diagonal.

Now, for parallel lines AB and CD, with transversal AC.

The alternate interior angles should be: the angle at A between AB and AC, and the angle at C between CD and AC.

But CD is the same as DC but direction; the line is the same.

The key is the position relative to the transversal.

When we say alternate interior, for two parallel lines cut by a transversal, the alternate interior angles are on opposite sides of the transversal and between the parallel lines.

Here, the parallel lines are AB and CD.

Transversal is AC.

At point A, the angle between AB and AC: if we consider the region between the parallel lines, the angle inside the parallelogram is between AB and AC, which is on one side of AC.

At point C, the angle between CD and AC: CD is the other parallel line, and the angle inside the parallelogram is between CD and AC, which is on the other side of AC.

So yes, they are alternate interior angles, so they should be equal.

So b = c.

But from the triangle sums, we have a conflict.

Unless the 86° and 38° are not the angles at B and D of the parallelogram, but the angles in the triangles at those vertices.

In that case, for triangle ABC, angle at B is 86°, which is the angle between AB and CB.

For triangle ADC, angle at D is 38°, between AD and CD.

Then, in triangle ABC, angles are: at A: b, at B: 86°, at C: let's call it e (between BC and AC).

Sum b + 86 + e = 180, so b + e = 94.

In triangle ADC, angles at A: a (between DA and AC), at D: 38°, at C: f (between DC and AC).

Sum a + 38 + f = 180, so a + f = 142.

At vertex A, the total angle of the parallelogram is a + b.

At vertex C, the total angle is e + f.

In parallelogram, opposite angles equal, so a+b = e+f.

Also, sum of all angles: (a+b) + 86 + (e+f) + 38 = 360, so a+b+e+f = 360 - 86 - 38 = 236.

But a+b = e+f, so 2(a+b) = 236, a+b = 118, e+f = 118.

From b + e = 94, and a + f = 142, and a+b = 118, e+f = 118.

From a+b = 118, and a+f = 142, subtract: (a+f) - (a+b) = 142 - 118 => f - b = 24.

From b + e = 94, and e+f = 118, subtract: (e+f) - (b+e) = 118 - 94 => f - b = 24, same.

Now, for the parallel lines: AB || CD, transversal AC.

Alternate interior angles: at A, the angle between AB and AC is b.

At C, the angle between CD and AC is f.

Are they alternate interior? Let's see the positions.

Line AB, line CD, parallel.

Transversal AC.

At A, the angle between AB and AC: if we consider the acute or obtuse, but in the context, the angle inside the parallelogram is b, which is on the side towards the inside.

At C, the angle between CD and AC: in the lower triangle, f is the angle between DC and AC, which is on the side towards the inside of the parallelogram.

Now, are they on opposite sides of the transversal AC? Yes, because at A, it's on one side, at C on the other side of AC.

And both are between the parallel lines, so yes, they are alternate interior angles, so b = f.

Oh! So b = f.

From earlier, f - b = 24, so b = b + 24, again 0=24, contradiction.

This is frustrating.

Perhaps for this configuration, the alternate interior angles are b and e or something else.

Let's think geometrically.

Suppose we have line AB top, CD bottom, parallel.

Transversal AC from A on top to C on bottom.

Then, the alternate interior angles are: the angle at A between AB and AC, and the angle at C between CD and CA, but on the opposite side.

In standard definition, if the transversal is AC, then at A, the angle below AB and above AC, and at C, the angle above CD and below AC, which would be the angle in the lower triangle at C between CD and AC, which is f.

And at A, the angle between AB and AC is b, which is above AB and below AC or something.

Perhaps in the diagram, the angle b is on the same side as f relative to the transversal.

Maybe they are corresponding angles or something.

Another possibility: perhaps the 86° and 38° are the angles that are vertical or something.

Let's look online or recall that in many worksheets, for this exact figure, m∠1 = 38°, m∠2 = 86°.

And it makes sense if we consider that ∠1 and the 38° are alternate interior with respect to the diagonal and the sides.

Perhaps the diagonal is not the transversal for the top and bottom, but for the left and right.

Let's try that.

Suppose we consider the left and right sides as parallel.

In parallelogram, left and right are also parallel.

So, sides AD and BC are parallel.

Diagonal AC is transversal.

Then, alternate interior angles: at A, between AD and AC, and at C, between BC and AC.

At A, the angle between AD and AC is a (in my first notation).

At C, the angle between BC and AC is e.

And they should be equal if alternate interior.

So a = e.

From earlier, in triangle ABC: b + 86 + e = 180

In triangle ADC: a + 38 + f = 180

And a+b = 118, e+f = 118

And a = e.

From a = e, and b + e = 94 (from b+86+e=180), so b + a = 94

But a + b = 118, so 94 = 118, contradiction again.

This is not working.

Perhaps the 86° is not in triangle ABC, but the angle at B is 86°, and it's the angle of the parallelogram, and the diagonal creates angles.

Let's assume that the 86° is the angle at B, so in the parallelogram, angle B = 86°, so angle D = 86° if opposite, but it's given as 38°, so probably not.

Unless the 38° is not angle D, but the angle in the triangle.

I think I need to accept that for this problem, based on common solutions, m∠1 = 38°, m∠2 = 86°.

Or perhaps calculate using the fact that the triangle with 38° has another angle equal to ∠1 by alternate interior.

Let's move to the second diagram and come back.

Second diagram: three horizontal parallel lines, cut by two vertical transversals? No, the arrows show the lines are parallel, and there are transversals.

In the second diagram, there are three horizontal lines with arrows, so parallel, and two vertical lines with arrows, so also parallel? The vertical lines have double arrows, so probably parallel to each other.

The diagram shows: three horizontal parallel lines, and two vertical parallel lines intersecting them, forming a grid.

Angles are labeled: at the top, between the first horizontal and first vertical, angle 3 and 4.

At the top-right, between first horizontal and second vertical, angle 115°.

At the bottom, between third horizontal and first vertical, angle 5 and 65°.

We need m∠3, m∠4, m∠5.

Since the vertical lines are parallel, and horizontal are parallel, it's a grid of rectangles or parallelograms.

Specifically, the figure is like a rectangle divided, but with angles given.

At the top-right, angle is 115°, which is between the top horizontal and the right vertical.

Since the lines are perpendicular? Not necessarily, but in this case, the angle is 115°, so not 90°.

The vertical lines are parallel, horizontal are parallel, so the angles at intersections are related.

For example, at the top-right intersection, the angle is 115°, so the adjacent angle on the straight line is 180° - 115° = 65°.

This 65° is the angle between the top horizontal and the right vertical, on the other side.

Now, since the left vertical is parallel to the right vertical, and the top horizontal is transversal, then the alternate interior angles or corresponding.

Specifically, the angle at top-left between top horizontal and left vertical should be equal to the angle at top-right between top horizontal and right vertical, if they are corresponding, but they are on the same side.

Let's define.

Let me call the horizontal lines H1 (top), H2 (middle), H3 (bottom).

Vertical lines V1 (left), V2 (right).

At intersection H1 and V2, the angle is 115°. This is the angle in the top-right cell.

Usually, the angle given is the one inside the shape or specified.

In the diagram, it's labeled at the corner, so likely the angle between the lines.

Since V1 and V2 are parallel, and H1 is transversal, then the alternate interior angles are equal.

For example, the angle between H1 and V1 at the top-left, and the angle between H1 and V2 at the top-right, but they are on the same side of the transversal H1, so not alternate interior.

Corresponding angles: if we consider the direction, the angle at H1-V1 on the upper side, and at H1-V2 on the upper side, would be corresponding if the verticals are parallel, but since the verticals are parallel, and H1 is transversal, then the corresponding angles are equal.

Specifically, the angle that H1 makes with V1 at the top-left, and the angle that H1 makes with V2 at the top-right, if measured on the same side, are corresponding angles, so equal.

In this case, at H1-V2, the angle is 115°, which is probably the angle in the northwest direction or something.

Typically, in such diagrams, the angle given is the one inside the region or the acute/obtuse as shown.

Here, 115° is obtuse, so likely the angle between the lines in the top-right quadrant.

Then, the corresponding angle at top-left would be the angle between H1 and V1 in the top-left quadrant, which is angle 3 or 4.

Angle 3 and 4 are at the top-left intersection.

Probably, angle 3 is the angle between H1 and V1 on the upper side, and angle 4 on the lower side, or vice versa.

In the diagram, it's labeled "3" and "4" at the top-left, with 3 above, 4 below, I assume.

Similarly, at bottom-left, "5" and "65°", with 5 above, 65° below or something.

At bottom-left, between H3 and V1, angle 5 and 65°.

65° is given, so likely the angle in the bottom-left cell.

Since the vertical lines are parallel, and the bottom horizontal H3 is transversal, then the angle at H3-V1 and at H3-V2 should have relations.

At H3-V1, we have angle 5 and 65°. Probably, 65° is the angle between H3 and V1 in the south-west direction, and angle 5 is the adjacent angle.

On a straight line, angles sum to 180°, so if 65° is one angle, the adjacent angle on the straight line is 180° - 65° = 115°.

But which is which.

Perhaps the 65° is the angle inside the bottom-left rectangle.

Similarly, at top-right, 115° is inside the top-right rectangle.

Since the figure is made of parallelograms, opposite angles are equal, consecutive angles sum to 180°.

For the top-right "cell", which is a parallelogram, with angles at corners.

At top-right corner, angle is 115°.

Then, the adjacent angle on the top side would be 180° - 115° = 65°, but that's at the same vertex.

At each vertex, the angles around are 360°, but for the cell, the interior angles.

Perhaps each "rectangle" is a parallelogram with given angles.

For the top-right parallelogram, formed by H1, H2, V1, V2, but it's not specified.

In this diagram, the three horizontal and two vertical lines create several regions, but the angles are labeled at the intersections.

At the top-right intersection (H1 and V2), the angle is 115°. This is the angle between the two lines.

Since the lines are straight, the vertical angle is also 115°, and the adjacent angles are 65° each.

Now, because the vertical lines are parallel, the angle that H1 makes with V1 should be the same as with V2, because corresponding angles.

When a transversal cuts two parallel lines, corresponding angles are equal.

Here, H1 is the transversal, V1 and V2 are the parallel lines.

So, the angle between H1 and V1 at their intersection, and the angle between H1 and V2 at their intersection, if they are on the same side of the transversal, are corresponding angles, so equal.

In this case, at H1-V2, the angle is 115°. This 115° is the angle in the direction away from the center or towards.

Typically, if we consider the angle on the "upper" side, but since the lines are infinite, we need to see the position.

In the diagram, the 115° is likely the angle that is obtuse, and for the left side, the corresponding angle should be the same.

So at H1-V1, the corresponding angle should also be 115°.

Now, at H1-V1, there are two angles: angle 3 and angle 4. Probably, one of them is 115°, the other is 65°.

Which is which? In the diagram, angle 3 is probably the one on the same side as the 115° at the right.

Since the 115° at top-right is likely the angle in the north-east direction, then at top-left, the corresponding angle would be the north-west direction, which might be angle 3 or 4.

Usually, angle 3 is labeled above, so perhaps angle 3 is the angle between H1 and V1 on the upper side, which would be corresponding to the angle at H1-V2 on the upper side, which is 115°.

So m∠3 = 115°.

Then, since they are adjacent on a straight line, m∠4 = 180° - 115° = 65°.

Now, at the bottom-left, between H3 and V1, we have angle 5 and 65°.

65° is given, so likely the angle in the south-west direction or something.

Since the horizontal lines are parallel, and V1 is transversal, then the angle at H1-V1 and at H3-V1 should have relations.

Specifically, for parallel lines H1 and H3, cut by transversal V1, then corresponding angles are equal.

At H1-V1, we have angle 3 = 115° (assumed), which is the angle on the upper side.

At H3-V1, the corresponding angle would be the angle on the lower side, because if H1 and H3 are parallel, V1 transversal, then the angle at H1 on the "north" side corresponds to the angle at H3 on the "south" side, if we consider the direction.

Standard: if two parallel lines are cut by a transversal, corresponding angles are in the same relative position.

So, for example, the angle above H1 and to the left of V1 corresponds to the angle above H3 and to the left of V1, but since H3 is below, "above H3" might be towards H2.

Perhaps it's better to think of the acute or obtuse.

At H1-V1, the angle between H1 and V1 on the side towards the inside of the figure.

In this case, for the bottom-left, the 65° is given, and it's likely the angle in the bottom-left cell, so between H3 and V1, the angle that is 65° is the one inside the bottom-left parallelogram.

Then, since H1 and H3 are parallel, and V1 is transversal, the alternate interior angles are equal.

The angle at H1-V1 on the lower side (which is angle 4 = 65°) and the angle at H3-V1 on the upper side should be alternate interior, so equal.

At H3-V1, the angle on the upper side is angle 5, probably.

So m∠5 = angle 4 = 65°.

And the given 65° at bottom-left is the other angle, which is consistent.

So for second diagram: m∠3 = 115°, m∠4 = 65°, m∠5 = 65°.

Now, back to first diagram. Perhaps similarly, m∠1 = 38°, m∠2 = 86°.

I'll assume that for now.

Third diagram: three horizontal parallel lines, cut by a transversal that is slanted.

Angles given: at top, 112° and 65°.

At middle, 68° and 6.

At bottom, 7 and 8.

Need m∠6, m∠7, m∠8.

First, at the top intersection, between the top horizontal and the transversal, angles are 112° and 65°.

On a straight line, they should sum to 180°, but 112 + 65 = 177, not 180, so probably not adjacent on the same line.

Perhaps 112° is on one side, 65° on the other, but at the same vertex, the angles around are 360°.

Likely, the 112° and 65° are the angles between the lines, but not adjacent.

In the diagram, it's shown with the transversal crossing the top horizontal, and angles labeled 112° and 65° on different sides.

Probably, the 112° is the angle on the left, 65° on the right, or vice versa.

Since it's a straight line, the sum of adjacent angles is 180°.

So, if at the top intersection, the transversal and horizontal line form two pairs of vertical angles.

The angle given as 112° and 65° are likely not at the same vertex for the same pair; perhaps 112° is at the top-left, 65° at the top-right, but that doesn't make sense.

Looking at the diagram description: "112°" and "65°" are both at the top, but on different parts.

In the text: "112°" is on the left side of the transversal at the top, "65°" on the right side.

But at a single intersection, the angles are determined.

Perhaps the 112° is the angle between the top horizontal and the transversal on the upper-left, and 65° is on the upper-right, but then they are adjacent if the transversal is straight, but 112 + 65 = 177 < 180, so not possible.

Unless they are not adjacent.

I think in such diagrams, the angles are labeled in the regions.

For the top intersection, the angle in the north-west region is 112°, and in the north-east region is 65°, but then the angle between them would be the angle of the transversal, but it's