Find the missing angles in triangles with this math worksheet.

Let’s solve each triangle one by one. Remember: The sum of the interior angles in any triangle is always 180°.

---

Problem 1:
Angles given: 70°, 35°, and a°
So, a = 180 - 70 - 35 = 75°

---

Problem 2:
Angles given: 82°, 44°, and a°
a = 180 - 82 - 44 = 54°

---

Problem 3:
Angles given: 117°, 17°, and a°
a = 180 - 117 - 17 = 46°

---

Problem 4:
This is an isosceles triangle (two sides marked equal → two base angles are equal).
One angle is 68°, so the other base angle is also 68°.
Then top angle a = 180 - 68 - 68 = 44°

Wait — let me double-check: The markings show the two *sides* are equal, which means the angles *opposite* those sides are equal. In this diagram, the two equal sides are the legs, so the base angles (at the bottom) are equal. One base angle is labeled 68°, so the other is also 68°. Then the top angle a = 180 - 68 - 68 = 44° ✔

---

Problem 5:
Again, isosceles triangle — two sides marked equal → two angles equal.
Given angle: 42° at the left vertex.
The side opposite that angle is NOT one of the equal sides? Wait — look carefully: the two equal sides are the ones with tick marks — they meet at the right-angle-looking corner? Actually, no — the tick marks are on the two sides forming the angle labeled “a°” and the 42° angle? Let me re-express:

Actually, in Problem 5:
- The triangle has two sides with single tick marks → those two sides are equal → so the angles opposite them are equal.
- The angle between the two equal sides is NOT labeled — wait, actually, looking again:
The 42° angle is at the left vertex.
The side opposite to it is the right side (which has a tick mark).
The side adjacent to 42° going down has a tick mark too.
So the two sides with tick marks are:
- from top vertex to bottom-right vertex
- from bottom-left vertex to bottom-right vertex
That would mean the two equal sides share the bottom-right vertex → so the two base angles are at top and bottom-left? This is confusing without seeing orientation.

Better approach: Since two sides are equal, two angles are equal. We’re told one angle is 42°, and another is a°. If the 42° is NOT one of the equal angles, then the other two must be equal → but we have only one unknown. Alternatively, maybe 42° is one of the equal angles?

Wait — standard convention: when two sides are marked equal, the angles opposite them are equal. So if the two sides with ticks are the ones meeting at the vertex where angle “a°” is, then the two base angles (opposite the equal sides) are equal. But here, the 42° is at a different vertex.

Actually, let's assume the triangle has vertices A, B, C. Suppose side AB and AC have tick marks → then angles at B and C are equal. But in the drawing, the 42° is at one end, and a° is at the other end, and the third angle is unlabeled.

Perhaps simpler: since two sides are equal, two angles are equal. The only way this works is if the 42° is one of the equal angles, and a° is the other equal angle? But then both would be 42°, and the third angle would be 96° — but a° is labeled at a different spot.

Wait — looking again: in Problem 5, the angle labeled “a°” is at the bottom-right vertex, and the 42° is at the left vertex. The two sides with tick marks are:
- from top vertex to bottom-right vertex (side opposite 42°)
- from bottom-left vertex to bottom-right vertex (side opposite the top angle)

That doesn’t help. Maybe the two equal sides are the ones forming the angle at the bottom-right? No.

Alternative idea: perhaps the two equal sides are the ones that do NOT include the 42° angle? Then the two base angles would be the ones at the ends of the unequal side — which would be the 42° and the a°? That can't be unless they are equal.

I think I made a mistake earlier. Let me reinterpret:

In many textbooks, when two sides are marked equal, and one angle is given not at the apex, you need to see which angles are opposite the equal sides.

But perhaps for simplicity — let’s count: total angles = 180°. Two angles are equal because two sides are equal. One angle is 42°. So either:
Case 1: 42° is one of the equal angles → then the other equal angle is also 42°, so a° = 180 - 42 - 42 = 96° — but a° is labeled at a specific vertex.

Case 2: 42° is the unique angle → then the other two are equal → so each is (180 - 42)/2 = 69°. Now, which one is a°? In the diagram, a° is at the vertex opposite the side that is NOT equal? Or adjacent?

Looking back at the original problem layout (even though I shouldn’t describe image), based on common worksheet patterns, in Problem 5, the angle labeled “a°” is likely the vertex angle (between the two equal sides), or one of the base angles.

Wait — there’s a better way: in Problem 5, the two sides with tick marks are the ones that form the angle at the bottom-right? No — typically, tick marks on sides indicate those sides are equal, so the angles opposite them are equal.

Assume:
- Side opposite 42° has a tick mark.
- Side opposite the top angle has a tick mark.
→ Then 42° equals the top angle? But top angle isn’t labeled.

This is messy. Let me try a different strategy: perhaps in Problem 5, the two equal sides are the ones that include the angle “a°”, meaning a° is the vertex angle, and the two base angles are equal. But one base angle is 42°, so the other is also 42°, then a° = 180 - 42 - 42 = 96°.

Yes, that makes sense. Because if the two sides with tick marks are the legs (meeting at the top), then the base angles are equal. But in the diagram, the 42° is at the left base, and a° is at the right base? No — a° is labeled at the bottom-right, which might be a base angle.

I recall now: in some diagrams, if two sides are equal and the angle between them is not labeled, but one base angle is given, then the other base angle is equal, and the vertex angle is calculated.

But in Problem 5, the angle labeled “a°” is at the vertex where the two equal sides meet? Or not?

Let’s look at Problem 6 for comparison: it’s equilateral? All three sides have tick marks → all angles equal → 60° each. So a° = 60°.

Back to Problem 5: perhaps the two equal sides are the ones that do NOT include the 42° angle. So the 42° is the vertex angle, and the two base angles are equal. Then each base angle = (180 - 42)/2 = 69°. And if a° is one of the base angles, then a° = 69°.

That seems most logical. In many worksheets, when they label "a°" at a base angle in an isosceles triangle with vertex angle given, a° is the base angle.

Moreover, in the diagram description (though I won't say it), typically for Problem 5, the 42° is at the apex, and a° is at the base. But wait — the user's image shows 42° at the left, a° at the bottom-right, and the two equal sides are probably the ones from top to bottom-left and top to bottom-right? I'm overcomplicating.

Let me use logic: if two sides are equal, two angles are equal. The sum is 180. Given one angle is 42, the other two add to 138. If they are equal, each is 69. Now, which one is a°? In the absence of more info, and since a° is labeled at a vertex that is likely a base angle (not the apex), and 42° is at another vertex, probably 42° is the apex angle, so a° = 69°.

I think that's standard. Let's go with that.

So Problem 5: a = 69°

---

Problem 6:
All three sides have tick marks → equilateral triangle → all angles equal → 180 / 3 = 60°

So a° = 60°

---

Problem 7:
This is an exterior angle problem. The angle labeled “a°” is outside the triangle, adjacent to the top vertex.

First, find the interior angle at the top vertex:
Interior angles given: 40° and 17° at the base.
So top interior angle = 180 - 40 - 17 = 123°

Now, a° is the exterior angle adjacent to that 123° angle.
Exterior angle + interior angle = 180° (linear pair)
So a° = 180 - 123 = 57°

Alternatively, exterior angle theorem: exterior angle = sum of two remote interior angles = 40 + 17 = 57° → same answer.

✔ So a° = 57°

---

Problem 8:
Again, exterior angle. Angle “a°” is outside, adjacent to the bottom-left vertex.

First, find the interior angle at that vertex.
Given angles inside: 88° at top, 39° at bottom-right.
So bottom-left interior angle = 180 - 88 - 39 = 53°

Then a° is the exterior angle adjacent to it → a° = 180 - 53 = 127°

Or by exterior angle theorem: a° = sum of two remote interior angles = 88 + 39 = 127° → same.

✔ So a° = 127°

---

Now, let’s list all answers:

1. 75°
2. 54°
3. 46°
4. 44°
5. 69°
6. 60°
7. 57°
8. 127°

Double-check Problem 5: I assumed 42° is the vertex angle, so base angles are 69° each, and a° is a base angle → 69°. Is there any chance a° is the vertex angle? If a° were the vertex angle, and the two base angles are equal, but one is 42°, then the other is 42°, so a° = 96°. But in the diagram, the 42° is not at the base? Without seeing, but based on typical labeling, and since in Problem 4 the equal angles were at the base, likely in Problem 5 the 42° is at the apex. Also, 69° is a nicer number than 96° for a worksheet? Not necessarily.

Wait — let's think differently. In Problem 5, the two sides with tick marks are the ones that are equal. The angle between them is the vertex angle. If the 42° is at a vertex that is not between the two equal sides, then it's a base angle. So if 42° is a base angle, and the triangle is isosceles, then the other base angle is also 42°, and the vertex angle a° = 180 - 42 - 42 = 96°.

Which is it? I think I had it backwards initially.

Standard rule: in an isosceles triangle, the two equal sides are called legs, and the angle between them is the vertex angle. The other two angles are base angles and are equal.

In Problem 5, if the two sides with tick marks are the legs (meeting at the top), then the base angles are at the bottom. If one base angle is 42°, then the other is 42°, and the top angle a° = 96°.

But in the diagram, a° is labeled at the bottom-right, which would be a base angle, not the top. Contradiction.

Perhaps the two equal sides are the ones from bottom-left to bottom-right and from bottom-left to top? Then the equal angles would be at top and bottom-right.

If side from bottom-left to bottom-right has a tick, and side from bottom-left to top has a tick, then the angles opposite them are equal: angle at top (opposite bottom side) and angle at bottom-right (opposite left side).

So if angle at bottom-left is 42°, then the other two angles are equal: let's call them x. Then 42 + x + x = 180 → 2x = 138 → x = 69°. So a° = 69° if it's at bottom-right or top.

In the diagram, a° is at bottom-right, so yes, a° = 69°.

I think 69° is correct.

To confirm, let's see Problem 4: two sides equal, one base angle 68°, so other base angle 68°, top angle 44° — that matches.

In Problem 5, if 42° is at bottom-left, and a° at bottom-right, and the two equal sides are bottom-left to top and bottom-left to bottom-right, then the angles at top and bottom-right are equal, so a° = (180 - 42)/2 = 69°.

Yes.

Final Answers:

1. 75
2. 54
3. 46
4. 44
5. 69
6. 60
7. 57
8. 127

──────────────────────────────────────

Final Answer:
1. 75°
2. 54°
3. 46°
4. 44°
5. 69°
6. 60°
7. 57°
8. 127°