Triangle Interior Angles Worksheet(pdf) and Answer Key. Scaffolded ... - Free Printable

Let’s solve each problem step by step.

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Problem 5:

We have two triangles that share a common vertex where angles w° and x° are located. These two angles are vertical angles, which means they are equal:
→ w = x

Also, in any triangle, the sum of the interior angles is always 180°.

Left triangle:
Angles given: 42°, 85°, and w°
So:
42 + 85 + w = 180
127 + w = 180
w = 180 - 127 = 53°

Since w = x (vertical angles), then:
→ x = 53°

Right triangle:
Angles given: 88°, y°, and x° (which we now know is 53°)
So:
88 + y + 53 = 180
141 + y = 180
y = 180 - 141 = 39°

✔ So for Problem 5:
w = 53°, x = 53°, y = 39°

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Problem 6:

We have a large triangle with angles 73°, 75°, and x° at the left tip. But there’s also a smaller triangle inside it with angles y°, z°, and 26°.

First, let’s find x° using the big outer triangle.

Sum of angles in a triangle = 180°
So:
x + 73 + 75 = 180
x + 148 = 180
x = 180 - 148 = 32°

Now look at the small inner triangle. It has angles: y°, z°, and 26°.
But notice: angle y° and angle z° are on a straight line with other parts? Wait — actually, looking carefully:

The small triangle shares side with the big one. The angle labeled z° is adjacent to the 73° angle? Not exactly.

Wait — better approach: Look at the point where the small triangle meets the top side of the big triangle.

Actually, here’s a clearer way:

In the small triangle: angles are y°, z°, and 26° → so:
y + z + 26 = 180 → y + z = 154 ...(Equation 1)

Now, look at the top part: the angle next to z° along the top edge — since the whole top angle of the big triangle is 73°, and if z° is part of that, then the remaining angle on that side would be 73° - something? Hmm, maybe not.

Wait — another idea: use the fact that the small triangle is inside, and its base is along the bottom of the big triangle.

Actually, let’s consider the quadrilateral or use exterior angles? Maybe overcomplicating.

Alternative plan:

Look at the big triangle again. We found x = 32°.

Now, focus on the small triangle. One of its angles is 26°, and the other two are y° and z°.

But also, notice that angle y° and the 26° angle are part of a larger triangle? Or perhaps use the fact that the line dividing the big triangle creates two triangles.

Wait — here's a better observation:

The small triangle has angles: y°, z°, 26°.

Also, angle z° and the 73° angle are on the same straight line? No, they’re not.

Wait — actually, look at the vertex where z° is. That vertex is on the top side of the big triangle. The entire angle at that corner of the big triangle is 73°. But z° is an angle *inside* the small triangle at that same vertex? That can’t be unless z° is part of the 73°.

Ah! Yes — likely, the 73° angle is split into two parts: one is z°, and the other is... wait, no, because the small triangle is drawn from the bottom up.

Actually, re-examining the diagram mentally:

The big triangle has vertices: left (x°), top-right (73°), bottom-right (75°).

Inside, there’s a line from the bottom side (near the 75° angle) going up to the top side, forming a small triangle with angles y°, z°, and 26°.

That 26° angle is at the bottom, near the 75° angle.

So, at the bottom-right corner of the big triangle, the total angle is 75°. Part of that is taken by the 26° angle of the small triangle, and the rest is another angle — let’s call it A.

So: 26° + A = 75° → A = 75 - 26 = 49°

Now, this angle A (49°) is actually an angle in the quadrilateral or in the other triangle? Actually, it’s part of the remaining shape.

But more directly: now consider the triangle that includes angles x°, y°, and A (49°)? Let’s see.

Actually, the figure is divided into two triangles by the internal line:

- Small triangle: angles y°, z°, 26°
- Larger remaining polygon? Not quite.

Wait — perhaps the internal line creates two triangles:

1. The small one: with angles y°, z°, 26°
2. The other one: which includes angles x°, 73°, and the angle adjacent to z°?

This is getting messy. Let me try a different strategy.

Use the fact that the sum of all angles around a point is 360°, but maybe not needed.

Another idea: use the exterior angle theorem.

Notice that in the small triangle, the angle y° is an exterior angle to some other triangle? Not obviously.

Wait — let’s go back to basics.

We already have x = 32° from the big triangle.

Now, look at the triangle formed by the left part: it has angles x° (32°), y°, and the angle at the bottom which is part of the 75°.

Earlier I said that at the bottom-right, 75° is split into 26° and 49°. So the 49° is adjacent to the small triangle.

Now, consider the triangle that has angles: x° (32°), y°, and 49°? Is that a triangle?

Yes! Because the internal line connects from the bottom (creating the 26° and 49° split) to the top side, so the left portion is a triangle with angles:

- At left: x° = 32°
- At bottom: 49° (since 75° - 26° = 49°)
- At top: y°

So in that triangle: 32 + 49 + y = 180
81 + y = 180
y = 180 - 81 = 99°

Now, go back to the small triangle: angles y°, z°, 26°
We just found y = 99°, so:
99 + z + 26 = 180
125 + z = 180
z = 180 - 125 = 55°

Let me verify if this makes sense with the top angle.

At the top-right corner of the big triangle, the angle is 73°. In our calculation, z° = 55°, which is part of that? But 55° is less than 73°, so the remaining part would be 73 - 55 = 18°, which should be the angle in the other triangle at that vertex.

Check the other triangle (the right one): it has angles 73° (but wait, we used only part of it?), no.

Actually, the big triangle’s top angle is 73°, and we have z° = 55° as part of the small triangle at that vertex. Then the remaining angle at that vertex for the other triangle is 73° - 55° = 18°.

Now, what are the angles of the right-side triangle? It should have:

- Top: 18° (as above)
- Bottom-right: 75°? But we already used part of it.

Wait, no — the bottom-right angle of the big triangle is 75°, which we split into 26° (for small triangle) and 49° (for left triangle). So the right-side triangle doesn't include the bottom-right corner directly.

Perhaps the right-side "triangle" isn't a triangle — actually, the figure is divided into two triangles by the internal line:

Triangle 1 (left): angles x=32°, y=99°, and 49° → sum 32+99+49=180 ✓

Triangle 2 (small): angles y=99°? Wait no — y is shared? That can’t be.

I think I made a mistake here.

If y° is in both triangles, that would mean it's the same angle, but in reality, y° is only in the small triangle.

Let me redefine.

Label the points:

Call the big triangle ABC:

- A: left vertex, angle x°
- B: top-right vertex, angle 73°
- C: bottom-right vertex, angle 75°

Internal line from D on AC to E on AB, such that triangle DEC has angles: at D: 26°, at E: y°, at C: z°? Not matching.

Perhaps better to accept my earlier calculation was wrong.

Let me start over for Problem 6.

Big triangle: angles x°, 73°, 75° → x = 180 - 73 - 75 = 32° ✓

Now, the internal line creates a small triangle with angles 26°, y°, z°.

Also, this internal line creates another triangle or quadrilateral.

Notice that the angle adjacent to the 26° angle at the bottom is part of the 75° angle.

So, at vertex C (bottom-right), the total angle is 75°. The small triangle takes 26° of it, so the remaining angle between the internal line and side BC is 75° - 26° = 49°.

This 49° angle is now an angle in the quadrilateral or in the other triangle.

Consider the triangle that includes vertex A (x=32°), vertex B (73°), and the point where the internal line meets AB.

Actually, the internal line meets AB at some point, say P, and meets AC at Q.

Then triangle APQ or something.

Perhaps use the fact that the sum of angles in the small triangle is 180°, and relate to the big triangle.

Another approach: the angle y° is an exterior angle to the triangle that has the 26° and the 49° angles.

Recall the exterior angle theorem: the exterior angle is equal to the sum of the two opposite interior angles.

In this case, at the point where the internal line meets the bottom side, the angle outside the small triangle is 49° (as calculated), and the small triangle has 26° at that vertex.

But y° is at the top of the small triangle.

Notice that y° and the angle in the big triangle at the top are related.

Let's calculate the third angle of the big triangle first — we did, x=32°.

Now, the internal line divides the big triangle into two parts: a small triangle and a quadrilateral? No, it should divide it into two triangles if it goes from one side to another.

Assume the internal line goes from the bottom side to the top side, so it divides the big triangle into two smaller triangles.

Triangle 1: left part, with angles: x°=32°, and two others.

Triangle 2: right part, with angles: 73°, 75°, but that can't be because 73+75=148, plus x=32 is 180, so if we draw a line, it must create two triangles sharing the internal line.

Suppose the internal line is from a point on the bottom side to a point on the top side.

Then, the small triangle mentioned has angles 26°, y°, z°.

The 26° is at the bottom, so at the bottom side, near the 75° vertex.

So, at the bottom-right vertex C, angle is 75°. The small triangle has an angle of 26° at C, so the other part of the 75° is 49°, which is the angle between the internal line and side BC.

This 49° angle is now an angle in the other triangle (the one that includes vertex B and A).

Specifically, the other triangle has vertices: A, B, and the point where the internal line meets AB or AC.

Assume the internal line meets AC at D and AB at E.

Then triangle CDE has angles: at C: 26°, at D: z°, at E: y°.

Then, the remaining part is quadrilateral or triangle ADEB? Messy.

Perhaps the key is that the angle y° is supplementary to something.

Let's use the following:

In the small triangle: y + z + 26 = 180 → y + z = 154 ...(1)

Now, at the top vertex B, the angle is 73°. This angle is composed of z° and another angle, say α, so z + α = 73 ...(2)

Similarly, at the bottom vertex C, 75° = 26° + β, so β = 49° ...(3)

Now, consider the triangle that has angles x°, β, and α. That would be the triangle formed by A, B, and the intersection point.

Vertices: A (x=32°), B (α), and the point on AC or AB.

Actually, the triangle that includes A, B, and the point where the internal line meets AB — but it's complicated.

Notice that the three angles at the internal points must add up.

Another idea: the sum of all angles in the two small triangles should equal the sum of the big triangle plus the angles on the internal line, but since the internal line is straight, the angles on it sum to 180°.

When you draw a line inside a triangle from one side to another, it creates two triangles, and the sum of their angles is 180° + 180° = 360°, while the original triangle is 180°, so the extra 180° comes from the two angles on the internal line, which are adjacent and form a straight line, so they sum to 180°, which checks out.

In this case, the two triangles are:

- Triangle 1: the small one with angles 26°, y°, z°
- Triangle 2: the other one, which has angles: x°=32°, the angle at B which is 73° minus z°, and the angle at C which is 75° minus 26° = 49°.

Is that correct?

Let's define triangle 2 as having vertices: A, B, and the point D on AC.

Then angles in triangle 2:

- At A: x° = 32°
- At B: the angle between AB and BD. Since the full angle at B is 73°, and if z° is the angle in the small triangle at B, then the angle in triangle 2 at B is 73° - z°
- At D: the angle between AD and BD. But D is on AC, so at D, the angle in triangle 2 is the supplement of z° if z° is on the other side, but it's messy.

Perhaps assume that the small triangle is CED, with C being the bottom-right vertex, E on AB, D on AC.

Then in triangle CED: angle at C is 26°, angle at E is y°, angle at D is z°.

Then, in the remaining part, we have triangle AED or something.

The angle at D in the small triangle is z°, and since D is on AC, the angle between ED and DA is 180° - z°, because they are on a straight line.

Similarly, at E on AB, the angle between CE and EA is 180° - y°.

Now, consider triangle AED: it has angles:

- At A: x° = 32°
- At E: 180° - y°
- At D: 180° - z°

Sum of angles in triangle AED: 32 + (180 - y) + (180 - z) = 180

Calculate: 32 + 180 - y + 180 - z = 180
392 - y - z = 180
- y - z = 180 - 392 = -212
So y + z = 212

But from the small triangle, y + z = 154

Contradiction! 212 ≠ 154, so my assumption is wrong.

What's wrong? Ah, I see — triangle AED may not be a triangle; the points may not form a triangle that way.

Perhaps the internal line is from C to a point on AB, but the 26° is at C, so it's not.

Let's look for a standard solution.

I recall that in such problems, often the angle y° is an exterior angle.

Notice that the 26° angle and the 49° angle (75-26) are in the same triangle as y°? No.

Another thought: the angle y° is equal to the sum of the two remote interior angles in some triangle.

For example, in the big triangle, if we consider the small triangle, then y° might be an exterior angle to the triangle that has the 32° and 49° angles.

Let's try that.

Suppose we have a triangle with angles 32° and 49°, then the exterior angle at the third vertex would be 32 + 49 = 81°.

And if y° is that exterior angle, then y = 81°.

Then from small triangle: y + z + 26 = 180 → 81 + z + 26 = 180 → z = 180 - 107 = 73°.

But then at the top, z=73°, and the big triangle has 73° at that vertex, so it matches if z is the entire angle, but in the diagram, z is part of it, so probably not.

If z=73°, and the big triangle's angle is 73°, then it works, but then what about the other part.

Let's check the sum.

If y=81°, z=73°, then in small triangle: 81+73+26=180 ✓

Now, at the top vertex, if z=73°, and the big triangle has 73° there, then the internal line must be along the side, which is not the case.

Perhaps z is not at the top vertex.

I think I need to accept that in many such diagrams, the angle y° is the exterior angle for the triangle with angles x° and the 49° angle.

So y = x + 49 = 32 + 49 = 81°

Then from small triangle: 81 + z + 26 = 180 → z = 73°

Then at the top, the angle is 73°, and z=73°, so perhaps the internal line is such that z is the entire angle at B, which means the small triangle includes the whole top angle, which might be possible if the internal line starts from C and goes to B, but then it wouldn't create a small triangle with 26° at C.

Unless the 26° is not at C.

Perhaps the 26° is at the bottom, but not at the vertex.

Let's read the diagram description again.

In problem 6, the figure has a large triangle with angles x° at left, 73° at top-right, 75° at bottom-right.

Inside, there is a line from the bottom side (not at the vertex) to the top side, creating a small triangle with angles y° at the top of the small triangle, z° at the top-right of the small triangle, and 26° at the bottom of the small triangle.

So, the 26° is at the bottom, on the bottom side, not at the vertex.

So, at the bottom side, the 75° angle is at the vertex, and the 26° is somewhere on the side, so it's not part of the 75°.

That changes everything.

So, the 26° is an angle in the small triangle at a point on the bottom side, not at the vertex.

So, the small triangle has vertices: let's say P on bottom side, Q on top side, R on left side or something.

Typically, in such diagrams, the small triangle is formed by drawing a line from a point on the bottom side to a point on the top side, and the 26° is the angle at the bottom point of the small triangle.

Then, the angle at the bottom vertex of the big triangle is 75°, which is separate.

To solve, we can use the fact that the sum of angles in the small triangle is 180°, and also use the big triangle.

But we need more relations.

Notice that the line inside creates two triangles and a quadrilateral, but perhaps use the formula for the sum.

Another idea: the angle y° and the 73° are on the same side, but not directly related.

Let's consider the triangle that includes the 26° angle and the 75° angle.

At the bottom, the big triangle has angle 75° at the vertex. The small triangle has an angle of 26° at a point on the bottom side. So, the angle between the internal line and the bottom side is 26°, but that's not helpful.

Perhaps the 26° is the angle between the internal line and the bottom side.

In that case, then at the bottom, the angle between the internal line and the side BC is 26°, but the vertex angle is 75°, so the angle between BA and the internal line would be 75° - 26° = 49°, but only if the internal line is from the vertex, which it's not.

I think I found a reliable method.

Let me denote the points:

Let the big triangle be ABC, with A at left, B at top-right, C at bottom-right.

Angle at A: x°, at B: 73°, at C: 75°.

Draw a line from a point D on AC to a point E on AB, such that triangle CDE has angles: at D: 26°, at E: y°, at C: z°.

But then at C, the angle is 75°, which is split into z° and another angle.

So, at C, angle BCA = 75° = angle BCD + angle DCA, but if D is on AC, then it's not.

Assume D is on BC, E on AB.

Then triangle CDE with C at bottom-right, D on BC, E on AB.

Then angle at C in triangle CDE is the angle between CD and CE, but CD is part of BC, so if D is on BC, then CD is along BC, so the angle at C in the small triangle is the angle between BC and CE, which is part of the 75°.

So, if the small triangle has angle 26° at C, then that means the angle between BC and CE is 26°, so the remaining angle between CE and CA is 75° - 26° = 49°.

Then, in the small triangle CDE, angles are: at C: 26°, at E: y°, at D: z°.

Sum: 26 + y + z = 180 → y + z = 154 ...(1)

Now, consider triangle ACE or something.

Consider triangle BCE or the triangle that includes A, C, E.

Note that at point E on AB, the angle in the small triangle is y°, and the angle in the big triangle at B is 73°, but E is on AB, so the angle at E for the small triangle is between CE and BE.

Then, the angle between CE and AE is 180° - y°, since AB is a straight line.

Now, consider triangle ACE: it has vertices A, C, E.

Angles in triangle ACE:

- At A: x° = 32°
- At C: the angle between AC and CE, which is 49° (as calculated above, since 75° - 26° = 49°)
- At E: the angle between AE and CE, which is 180° - y° (because at E, on line AB, the angles on one side sum to 180°)

So, sum of angles in triangle ACE: 32 + 49 + (180 - y) = 180

Calculate: 32 + 49 = 81, so 81 + 180 - y = 180

261 - y = 180

y = 261 - 180 = 81°

Then from (1): 81 + z = 154 → z = 154 - 81 = 73°

Now, check if this makes sense.

At B, the angle is 73°, and in the small triangle, at E, y=81°, which is on AB, so the angle between CE and BE is 81°, but the angle at B is 73°, which is between AB and CB, so it might be consistent if we consider the directions.

In triangle BCE, for example, but we don't need to.

So, x = 32°, y = 81°, z = 73°

And at the top, z=73°, which matches the big triangle's angle at B, so perhaps the internal line is such that it doesn't affect the top angle, or in this case, z is not at B.

In our setup, z is at D on BC, not at B.

So, it should be fine.

Verify the sums.

Small triangle: 26 + 81 + 73 = 180 ✓

Triangle ACE: 32 + 49 + (180 - 81) = 32 + 49 + 99 = 180 ✓

Perfect.

So for Problem 6: x = 32°, y = 81°, z = 73°

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Final Answers:

Problem 5: w = 53°, x = 53°, y = 39°

Problem 6: x = 32°, y = 81°, z = 73°

Final Answer:
For problem 5: w = 53°, x = 53°, y = 39°
For problem 6: x = 32°, y = 81°, z = 73°