Corresponding Angles and Alternate Angles Worksheets and revision ... - Free Printable

We are given a geometric figure involving several lines and angles, and we are asked to find the value of $ x^\circ $ in triangle $ \triangle KHL $, where angle $ \angle KHL = x^\circ $. Let's analyze the diagram step by step.

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Step 1: Understand the Diagram

- Lines $ CF $ and $ GJ $ are horizontal lines (parallel to each other).
- Line $ AD $ intersects $ CF $ at point $ D $, forming a $ 76^\circ $ angle.
- Line $ BE $ intersects $ CF $ at point $ E $, forming a $ 39^\circ $ angle.
- Lines $ AD $ and $ BE $ intersect at point $ H $, which lies on line $ GJ $.
- Lines $ KH $ and $ LH $ form angle $ x^\circ $ at point $ H $, with $ K $ and $ L $ below $ H $.

We need to find $ x $, the measure of $ \angle KHL $.

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Step 2: Use Parallel Lines and Transversals

Since $ CF \parallel GJ $, and both are cut by transversal lines $ AD $ and $ BE $, we can use properties of corresponding and alternate interior angles.

#### Angle at D:
At point $ D $, the angle between $ AD $ and $ CF $ is $ 76^\circ $. This means that the angle between $ AD $ and the horizontal line $ CF $ is $ 76^\circ $. Since $ CF \parallel GJ $, the same angle will be formed at $ H $ on the opposite side.

But note: The angle between $ AD $ and $ CF $ at $ D $ is $ 76^\circ $. That means the interior angle on the upper side of $ CF $ is $ 76^\circ $. Then, the alternate interior angle at $ H $ (on line $ GJ $) would also be $ 76^\circ $, but on the lower side.

Wait — let’s clarify:

Actually, since $ AD $ crosses $ CF $ at $ D $, forming a $ 76^\circ $ angle above $ CF $, then when it continues down to $ H $, the angle between $ AD $ and $ GJ $ at $ H $ should be equal to $ 76^\circ $ if we consider the alternate interior angles.

Similarly, at $ E $, the angle between $ BE $ and $ CF $ is $ 39^\circ $. So the angle between $ BE $ and $ CF $ is $ 39^\circ $, meaning the alternate interior angle at $ H $ (between $ BE $ and $ GJ $) is also $ 39^\circ $.

Let’s define:

- At point $ H $, line $ AD $ forms an angle with $ GJ $: this angle is equal to the angle at D, because $ CF \parallel GJ $, and $ AD $ is a transversal. So, the alternate interior angle at $ H $ between $ AD $ and $ GJ $ is $ 76^\circ $.
- Similarly, the angle between $ BE $ and $ GJ $ at $ H $ is $ 39^\circ $, due to alternate interior angles.

But these two angles are on opposite sides of $ H $, so they form angles around point $ H $.

Let’s look more carefully.

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Step 3: Focus on Point H

At point $ H $, two lines cross:
- Line $ AD $, going from $ A $ through $ D $ to $ H $
- Line $ BE $, going from $ B $ through $ E $ to $ H $

These two lines intersect at $ H $, and form four angles there.

We are interested in angle $ \angle KHL $, which is formed by extending lines from $ H $ downward to points $ K $ and $ L $. But from the diagram, $ K $ and $ L $ lie on extensions of $ AD $ and $ BE $, respectively, below $ H $.

So:
- $ HK $ is part of line $ AD $ extended beyond $ H $
- $ HL $ is part of line $ BE $ extended beyond $ H $

Thus, $ \angle KHL $ is the angle between the extensions of $ AD $ and $ BE $ below point $ H $.

This is the vertical angle opposite to the angle formed by $ AD $ and $ BE $ above $ H $.

But we don't know that directly. Instead, we can compute the angle between $ AD $ and $ BE $ at $ H $ using the angles formed with the parallel lines.

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Step 4: Use Angles Formed with Parallel Lines

Let’s define:

- At point $ D $: angle between $ AD $ and $ CF $ is $ 76^\circ $. Since $ CF \parallel GJ $, the angle between $ AD $ and $ GJ $ at $ H $ is also $ 76^\circ $, but depending on direction.

Actually, consider the angles formed by transversal $ AD $ with parallel lines $ CF $ and $ GJ $:

- The angle at $ D $ between $ AD $ and $ CF $ is $ 76^\circ $. Since $ CF \parallel GJ $, the corresponding angle at $ H $ (on $ GJ $) is also $ 76^\circ $, on the same side of the transversal.

But we need to be careful about orientation.

Let’s suppose:

- The angle at $ D $ is $ 76^\circ $ above line $ CF $, on the left side of $ AD $.
- Then, the corresponding angle at $ H $ (on $ GJ $) would be $ 76^\circ $ above $ GJ $, on the left side of $ AD $.

Similarly, for line $ BE $:

- At $ E $, the angle between $ BE $ and $ CF $ is $ 39^\circ $. It's shown as the angle inside triangle $ BEF $, so it's the angle between $ BE $ and $ CF $, measured upward from $ BE $ to $ CF $.

So, again, since $ CF \parallel GJ $, the corresponding angle at $ H $ (on $ GJ $) would be $ 39^\circ $ above $ GJ $, on the right side of $ BE $.

Now, at point $ H $, we have:

- From line $ AD $: angle between $ AD $ and $ GJ $ is $ 76^\circ $, on the left side.
- From line $ BE $: angle between $ BE $ and $ GJ $ is $ 39^\circ $, on the right side.

But both angles are measured from the horizontal line $ GJ $ upward.

So, the total angle between the two lines $ AD $ and $ BE $ above $ H $ is the sum of these two angles?

Wait — not necessarily. Because they are on opposite sides of the intersection point.

Let’s think differently.

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Alternative Approach: Use Triangle or Angle Sum

Let’s consider triangle $ \triangle DHE $, but wait — $ D $, $ H $, $ E $ are not necessarily connected.

Better idea: Consider triangle $ \triangle HDE $, but we don’t have enough info.

Instead, focus on the angles at $ H $.

Let’s define:

- Let $ \angle DHG = 76^\circ $ — this is the angle between $ AD $ and $ GJ $ at $ H $, since $ AD $ makes $ 76^\circ $ with $ CF $, and $ CF \parallel GJ $, so alternate interior angles give $ \angle DHG = 76^\circ $.
- Similarly, $ \angle EHG = 39^\circ $ — angle between $ BE $ and $ GJ $, because $ BE $ makes $ 39^\circ $ with $ CF $, so alternate interior angle at $ H $ is $ 39^\circ $.

Wait — but are these on the same side?

Let’s suppose:

- $ AD $ comes from above, hits $ CF $ at $ D $, makes $ 76^\circ $ with $ CF $, then goes down to $ H $.
- So, the angle between $ AD $ and $ CF $ at $ D $ is $ 76^\circ $, so the interior angle on the upper side is $ 76^\circ $.
- Therefore, the alternate interior angle at $ H $ (on $ GJ $) is also $ 76^\circ $, meaning the angle between $ AD $ and $ GJ $ at $ H $ is $ 76^\circ $, on the lower side.

Wait — perhaps better to draw mentally:

Let’s assume $ CF $ and $ GJ $ are horizontal lines, $ CF $ above $ GJ $.

- $ AD $ starts at $ A $, goes down through $ D $ on $ CF $, then to $ H $ on $ GJ $.
- The angle at $ D $ between $ AD $ and $ CF $ is $ 76^\circ $. If $ AD $ is coming from above and going down to the right, then the angle between $ AD $ and $ CF $ is $ 76^\circ $, measured clockwise from $ CF $ to $ AD $, say.

Then, since $ CF \parallel GJ $, the angle between $ AD $ and $ GJ $ at $ H $ is also $ 76^\circ $, on the same side — i.e., the angle between $ AD $ and $ GJ $ is $ 76^\circ $, measured clockwise from $ GJ $ to $ AD $.

Similarly, for $ BE $:

- $ BE $ goes from $ B $, down to $ E $ on $ CF $, then to $ H $ on $ GJ $.
- At $ E $, the angle between $ BE $ and $ CF $ is $ 39^\circ $. Suppose it's measured from $ CF $ to $ BE $, going counterclockwise.

Then, at $ H $, the angle between $ BE $ and $ GJ $ is also $ 39^\circ $, on the same side.

So now, at point $ H $, we have:

- $ AD $ makes $ 76^\circ $ with $ GJ $ (say, to the right)
- $ BE $ makes $ 39^\circ $ with $ GJ $ (say, to the left)

Wait — directions matter.

Let’s assign directions.

Assume:

- $ CF $ and $ GJ $ are horizontal, $ CF $ above $ GJ $.
- $ AD $ goes from $ A $ (top-left) to $ D $ on $ CF $, then down to $ H $ on $ GJ $, continuing to $ K $.
- So $ AD $ is going downward to the right.
- At $ D $, the angle between $ AD $ and $ CF $ is $ 76^\circ $, and since $ AD $ is going down-right, the angle between $ AD $ and $ CF $ is $ 76^\circ $ below $ CF $, on the right side.

Then, the alternate interior angle at $ H $ would be $ 76^\circ $ above $ GJ $, on the right side? No.

Wait — alternate interior angles are on opposite sides of the transversal.

Better to use the fact that the interior angles on the same side of the transversal add up to $ 180^\circ $ if the lines are parallel.

But here’s a better idea:

Let’s consider the triangle formed by the two transversals and the parallel lines.

Alternatively, consider the angles at $ H $.

Let’s define:

- The angle between $ AD $ and $ GJ $ at $ H $ is $ 76^\circ $, because $ AD $ makes $ 76^\circ $ with $ CF $, and $ CF \parallel GJ $, so the corresponding angle at $ H $ is $ 76^\circ $.
- Similarly, the angle between $ BE $ and $ GJ $ at $ H $ is $ 39^\circ $.

Now, both angles are measured from the horizontal line $ GJ $ to the respective lines $ AD $ and $ BE $.

But are they on the same side?

From the diagram:

- $ AD $ is sloping down to the right.
- $ BE $ is sloping down to the left.

So at $ H $:

- $ AD $ makes an angle of $ 76^\circ $ with $ GJ $, measured downward to the right.
- $ BE $ makes an angle of $ 39^\circ $ with $ GJ $, measured downward to the left.

Therefore, the total angle between $ AD $ and $ BE $ at $ H $, below $ GJ $, is the sum of these two angles: $ 76^\circ + 39^\circ = 115^\circ $.

But wait — no. Because the angle between them is the angle between the two lines.

Let’s think: the angle between $ AD $ and $ BE $ at $ H $ is the angle between the two lines.

If $ AD $ is going down to the right at $ 76^\circ $ from horizontal, and $ BE $ is going down to the left at $ 39^\circ $ from horizontal, then the angle between them is:

$$
76^\circ + 39^\circ = 115^\circ
$$

Because one is $ 76^\circ $ below horizontal to the right, and the other is $ 39^\circ $ below horizontal to the left, so the angle between them is the sum.

So the angle between $ AD $ and $ BE $ at $ H $ is $ 115^\circ $.

Now, $ \angle KHL $ is the angle formed by extending $ AD $ and $ BE $ beyond $ H $ to $ K $ and $ L $, so it's the angle between the two lines below $ H $.

But the angle between two lines is the same regardless of direction — it's the smaller angle between them.

But in this case, the angle between the two lines at $ H $ is $ 115^\circ $, so the adjacent angle (the other one) is $ 180^\circ - 115^\circ = 65^\circ $.

But which one is $ \angle KHL $?

Look at the diagram: $ K $ is on the extension of $ AD $ below $ H $, $ L $ is on the extension of $ BE $ below $ H $. So $ \angle KHL $ is the angle between those two extensions, which is the same as the angle between $ AD $ and $ BE $ at $ H $, but on the lower side.

But the angle between the two lines is $ 115^\circ $, so the angle between the extensions is also $ 115^\circ $? Or is it the supplementary angle?

No — the angle between two lines is defined as the smaller one, but here, since the two lines form two pairs of vertical angles: one pair of $ 115^\circ $, and one pair of $ 65^\circ $.

Now, looking at the diagram: $ K $ and $ L $ are below $ H $, and the angle $ \angle KHL $ is shown as opening downward, so it's the angle between the two lines on the bottom side.

But which one is it?

Let’s see: if $ AD $ goes down to the right, and $ BE $ goes down to the left, then the angle between them below $ H $ is the acute angle? Or obtuse?

Actually, if you imagine:

- From $ H $, $ HK $ goes down to the right (extension of $ AD $)
- $ HL $ goes down to the left (extension of $ BE $)
- Then the angle between them at $ H $ is the angle between the two downward rays.

The total angle between the two lines is $ 76^\circ + 39^\circ = 115^\circ $, so the angle between $ HK $ and $ HL $ is $ 115^\circ $.

But wait — is that correct?

Let’s suppose:

- The angle between $ AD $ and the horizontal $ GJ $ is $ 76^\circ $, so the angle from $ GJ $ to $ AD $ is $ 76^\circ $ downward to the right.
- The angle between $ BE $ and $ GJ $ is $ 39^\circ $, downward to the left.

So, the total angle between $ AD $ and $ BE $, measured from $ AD $ to $ BE $ across the bottom, is $ 76^\circ + 39^\circ = 115^\circ $.

Yes.

So the angle between $ HK $ and $ HL $ is $ 115^\circ $.

But wait — the problem asks for $ x^\circ $, and it looks like $ x $ is the angle at $ H $ between $ K $ and $ L $, which is this angle.

But let’s double-check.

Alternatively, consider triangle $ \triangle KHL $. But we don’t have triangle info.

Another way: use the fact that the sum of angles around point $ H $ is $ 360^\circ $.

But we can use the following:

The angle between $ AD $ and $ GJ $ is $ 76^\circ $, so the angle between $ AD $ and $ GJ $ on the upper side is $ 76^\circ $, and on the lower side is $ 180^\circ - 76^\circ = 104^\circ $? No.

Wait — actually, the angle between $ AD $ and $ GJ $ is $ 76^\circ $, so the adjacent angle is $ 104^\circ $, but that’s not helpful.

Let’s try a different approach.

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Correct Approach: Use Triangle and Exterior Angles

Consider triangle $ \triangle DEH $, but we don’t have $ DE $.

Wait — better idea:

Lines $ AD $ and $ BE $ intersect at $ H $, and we know the angles they make with the parallel lines.

Let’s consider the angles at $ H $ with respect to the horizontal.

- $ AD $ makes $ 76^\circ $ with $ CF $, so it makes $ 76^\circ $ with $ GJ $, since $ CF \parallel GJ $. So the angle between $ AD $ and $ GJ $ is $ 76^\circ $.
- $ BE $ makes $ 39^\circ $ with $ CF $, so it makes $ 39^\circ $ with $ GJ $.

Now, since $ AD $ and $ BE $ are not parallel, they intersect at $ H $.

The angle between $ AD $ and $ BE $ can be found by considering the difference in their inclinations.

Suppose we measure the angles from the horizontal:

- $ AD $ has a slope such that it makes $ 76^\circ $ with the horizontal (from $ CF $ to $ AD $), so its angle with horizontal is $ 76^\circ $.
- $ BE $ makes $ 39^\circ $ with the horizontal.

But are they on the same side?

From the diagram:

- $ AD $ is going from $ A $ to $ D $ to $ H $, so it’s descending to the right, making $ 76^\circ $ with $ CF $, so the angle between $ AD $ and $ CF $ is $ 76^\circ $, so the acute angle between $ AD $ and the horizontal is $ 76^\circ $.
- $ BE $ is going from $ B $ to $ E $ to $ H $, so it’s descending to the left, making $ 39^\circ $ with $ CF $, so its angle with the horizontal is $ 39^\circ $.

So at point $ H $, the two lines $ AD $ and $ BE $ are both below the horizontal, but on opposite sides.

So the angle between them is the sum of their angles with the horizontal:

$$
\angle KHL = 76^\circ + 39^\circ = 115^\circ
$$

Therefore, $ x = 115^\circ $.

But wait — let’s verify with a known property.

Final Check Using Triangle Angles

Let’s consider the quadrilateral or the triangle formed.

Alternatively, consider the triangle $ \triangle HDE $, but we don’t have it.

Wait — here’s a better way.

Let’s consider the angles at $ H $.

Let’s define:

- Let $ \angle DHG = 76^\circ $ — angle between $ AD $ and $ GJ $ at $ H $
- Let $ \angle EHG = 39^\circ $ — angle between $ BE $ and $ GJ $ at $ H $

But are these on the same side?

If $ AD $ is on the left and $ BE $ on the right, then the angle between $ AD $ and $ BE $ at $ H $ is the sum of the angles from $ AD $ to $ GJ $ and from $ GJ $ to $ BE $, but only if they are on opposite sides.

In this case, since $ AD $ is on the left side of $ GJ $, and $ BE $ is on the right side, then the angle between $ AD $ and $ BE $ at $ H $ is the sum of the two angles with the horizontal.

So yes, $ x = 76^\circ + 39^\circ = 115^\circ $.

But is $ \angle KHL $ the angle between $ AD $ and $ BE $ at $ H $?

Yes — because $ K $ is on the extension of $ AD $ beyond $ H $, and $ L $ is on the extension of $ BE $ beyond $ H $, so $ \angle KHL $ is the angle between the two lines at $ H $, on the lower side.

And since both lines are going downward, and $ AD $ is to the right, $ BE $ to the left, the angle between them is $ 76^\circ + 39^\circ = 115^\circ $.

Therefore, $ x = 115^\circ $.

But let’s confirm with a common geometry principle.

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Answer:

$$
x = 76^\circ + 39^\circ = \boxed{115^\circ}
$$

✔ Final Answer: $ \boxed{115^\circ} $