Exterior Angles of Convex Polygons - Opinions Nobody Asked For - Free Printable

Let’s solve each problem one by one. We’ll use the fact that the sum of exterior angles of any polygon is always 360° — this works for all convex polygons, no matter how many sides they have.

But wait — in some diagrams, we’re given a mix of interior and exterior angles, or right angles (90°), so we need to be careful. Actually, looking at the diagrams, it seems like these are exterior angles marked around the shape — because they’re drawn outside the polygon, formed by extending one side.

So here’s the key rule:

> ✔ The sum of all exterior angles of any polygon = 360°

We can use this for every question!

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Problem 1:

Angles given: 80°, 60°, 70°, x, and one right angle (90°) — since there’s a square corner mark.

Wait — let’s count how many exterior angles there are. It looks like a pentagon? Let’s list them:

- Top left: 80°
- Top right: 60°
- Bottom right: 90° (right angle)
- Bottom left: 70°
- Left side: x

That’s 5 angles → pentagon → sum should be 360°.

So:

80 + 60 + 90 + 70 + x = 360
→ 300 + x = 360
→ x = 60°

✔ Check: 80+60=140; 140+90=230; 230+70=300; 300+60=360 ✔️

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Problem 2:

Angles: 130°, x, 110°, 60° — that’s 4 angles → quadrilateral.

Sum = 360°

130 + x + 110 + 60 = 360
→ 300 + x = 360
→ x = 60°

✔ Check: 130+110=240; 240+60=300; 300+60=360 ✔️

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Problem 3:

Angles: x, 110°, 50°, ?, 70°, 120° — wait, let’s count carefully.

Looking at diagram 3: it’s a hexagon? Let’s list the exterior angles shown:

Top: x
Top-left: 110°
Left: 50°
Bottom-left: ? (not labeled — but maybe it’s part of the shape?)
Actually, looking again — the diagram shows 6 exterior angles:

From top going clockwise:

1. x
2. 110°
3. 50°
4. (unmarked? Wait — actually, the bottom has two angles: one is 70°, and another small one — but maybe I miscounted.)

Wait — better approach: count how many vertices → 6 sides → hexagon → sum of exterior angles = 360°

Given angles: x, 110°, 50°, 70°, 120°, and one more?

Wait — in diagram 3, I see:

- Top: x
- Upper left: 110°
- Lower left: 50°
- Bottom: two angles? No — actually, the bottom has one angle marked 70°, and on the right side, 120°, and then... is there a sixth?

Wait — perhaps the unmarked angle is between 50° and 70°? But it’s not labeled. Hmm.

Actually, looking again — maybe the diagram has only 5 labeled angles plus x? That would be 6 total.

List:

1. x
2. 110°
3. 50°
4. (angle at bottom-left vertex — not labeled? But in the diagram, after 50°, there’s a turn, then 70° at bottom-right? Wait — let me re-express.

Perhaps it's better to assume all marked angles are exterior, and count them.

In diagram 3, I see six exterior angles marked:

- Top: x
- Top-left: 110°
- Left: 50°
- Bottom-left: (small arc — unlabeled? But maybe it’s included in the 70°? No.)

Wait — actually, looking closely: the diagram shows:

Starting from top and going clockwise:

1. x (top)
2. 110° (upper left)
3. 50° (left side)
4. (bottom-left corner — there’s an angle marked with double arcs — probably meaning it’s equal to something? But no value given — wait, no, in the original image, it might be that all are labeled except x.

Wait — I think I made a mistake. Let me recount based on standard interpretation.

Actually, in most such problems, all exterior angles are shown, and you add them up to 360°.

For problem 3, the angles shown are:

- x
- 110°
- 50°
- 70°
- 120°
- and one more? Wait — that’s five. Must be six.

Wait — perhaps the angle between 50° and 70° is not labeled