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

I'll solve each problem step by step, finding the angle marked x in each question.

Problem 1:
- We have a triangle with angles 80° and 70°, plus an exterior angle of 130°
- The interior angle adjacent to 130° is: 180° - 130° = 50°
- In the triangle: 80° + 70° + 50° = 200° (this doesn't work for a triangle)
- Let me reconsider: The 130° is an exterior angle at one vertex
- Using exterior angle theorem: exterior angle = sum of two opposite interior angles
- So 130° = 80° + (angle at top)
- Angle at top = 130° - 80° = 50°
- Now in the triangle with angles 50°, 70°, and x:
- 50° + 70° + x = 180°
- x = 180° - 120° = 60°

Problem 2:
- We have a right angle (90°), 75°, 60°, 25°, and need to find x
- Looking at the quadrilateral formed: angles are 90°, 75°, 60°, and an angle adjacent to x
- Sum of angles in quadrilateral = 360°
- 90° + 75° + 60° + angle = 360°
- angle = 360° - 225° = 135°
- x and this angle form a linear pair: x + 135° = 180°
- x = 45°

Wait, let me reconsider problem 2 more carefully:
- The figure shows angles around a point or in connected triangles
- Let me use the fact that angles on a straight line = 180°
- Starting from bottom: 25° + some angle = 180° on that line
- Actually, looking at the structure, I need to find the angle in the triangle
- Triangle has angles: (90° - 75° = 15°), 60°, and x
- Wait, that's not right either
- Let me trace through: The vertical line makes 90° with horizontal
- 75° is measured from vertical, so from horizontal it's 90° - 75° = 15°
- In the triangle: 15° + 60° + x = 180°
- x = 180° - 75° = 105°

Actually, let me be more careful. Looking at the diagram structure:
- There's a right angle at bottom left
- 75° angle from the vertical
- 60° angle shown
- 25° angle at bottom
- Need to find x

Using the polygon/angle relationships:
The angles going around should account for the geometry. Let me use exterior angles:
x = 75° + 60° - 25° = 110°... no that doesn't seem right.

Let me try: In the quadrilateral, if we consider the angles properly:
The angle supplementary to x plus 75° + 60° + (90° - 25°) = 360°
Angle supplementary to x + 75° + 60° + 65° = 360°
Angle supplementary to x = 360° - 200° = 160°
x = 180° - 160° = 20°

Hmm, let me try another approach for problem 2:
Using the fact that the sum of exterior angles of any polygon = 360°
Or using parallel lines and transversals...

Actually, looking more carefully at the structure:
- The angle marked x appears to be an exterior angle
- Using the relationship: x = 75° + 60° + 25° - 90° = 70°

Let me verify: If we go around the figure, tracking the turns...
Starting from horizontal, going up 90°, then turning 75°, then 60°, then down 25°, and back to horizontal.
The total turn should be 360°.
90° + 75° + 60° + 25° + x = 360°
x = 360° - 250° = 110°

Actually I think x = 110° for problem 2.

Problem 3:
- Quadrilateral with exterior angles 60°, 70°, 35°, and x
- Sum of exterior angles of any polygon = 360°
- 60° + 70° + 35° + x = 360°
- 165° + x = 360°
- x = 195°

Wait, that can't be right for an exterior angle. Let me reconsider.
Looking at the diagram, these appear to be the exterior angles at each vertex.
But 75° is shown inside... Let me look again.

Actually, I see: 60°, 70°, 35° are exterior angles, and there's a 75° interior angle shown.
If 75° is interior, its exterior angle = 180° - 75° = 105°
Sum of exterior angles = 60° + 70° + 35° + 105° = 270° ≠ 360°

That means my interpretation is wrong. Let me reconsider the diagram.
Perhaps all four angles shown (60°, 70°, 35°, and the one at the fourth vertex) are exterior angles.
Then: 60° + 70° + 35° + x = 360°
x = 195°

But wait, there's also a 75° shown. Let me think about this differently.
Maybe the 75° is used to help find one of the exterior angles?
If an interior angle is 75°, the exterior angle at that vertex = 180° - 75° = 105°
So: 60° + 70° + 35° + 105° = 270°... still not 360°

I think I need to reconsider which angles are exterior vs interior.
Looking at the positions: 60°, 70°, and 35° appear to be exterior angles.
The 75° appears to be an interior angle.
And x is an exterior angle.

For a quadrilateral: sum of interior angles = 360°
Interior angles would be: (180°-60°), (180°-70°), 75°, (180°-x)
= 120° + 110° + 75° + (180°-x) = 360°
485° - x = 360°
x = 125°

Let me verify: Interior angles: 120°, 110°, 75°, 55°
Sum = 120° + 110° + 75° + 55° = 360° ✓

Problem 4:
- Exterior angles: 140°, 120°, 60°, and x
- Sum of exterior angles = 360°
- 140° + 120° + 60° + x = 360°
- 320° + x = 360°
- x = 40°

Problem 5:
- Pentagon with exterior angles: 65°, 100°, 50°, 90° (right angle), and x
- Sum of exterior angles of pentagon = 360°
- 65° + 100° + 50° + 90° + x = 360°
- 305° + x = 360°
- x = 55°

Problem 6:
- Pentagon with exterior angles: 125°, 50°, x, and two others
- I see 125°, 50°, 60°, and x marked
- That's only 4 angles for a pentagon... let me recount
- Looking at the vertices: 125°, 50°, x, 60°, and one more unmarked
- Wait, I count 5 vertices but only 4 exterior angles shown explicitly
- Let me assume the unmarked one needs to be found from the geometry
- Actually, looking more carefully, I see: 125°, 50°, x, 60°
- If this is a pentagon, sum of exterior angles = 360°
- But I only have 4 values including x
- Let me check if there's a fifth angle I'm missing...
- Oh, I see now - there are 5 exterior angles but one might be calculated
- 125° + 50° + 60° + x + (some angle) = 360°

Actually, re-examining: I count 5 vertices with exterior angles at 4 of them shown, plus x.
Let me assume all 5 are: 125°, 50°, x, 60°, and one more.
But I don't see a fifth value...

Let me try: Maybe it's a quadrilateral?
If quadrilateral: 125° + 50° + 60° + x = 360°
235° + x = 360°
x = 125°

Or maybe one angle is interior? If 60° is interior, exterior = 120°
Then: 125° + 50° + 120° + x = 360° (for quadrilateral)
295° + x = 360°
x = 65°

Given the shape looks like a pentagon but I may be miscounting, let me go with quadrilateral interpretation:
x = 125° (if all shown are exterior) or x = 65° (if 60° is interior)

Looking at the marking style, 60° appears to be exterior like the others.
So for a quadrilateral: x = 360° - 125° - 50° - 60° = 125°

Problem 7:
- Angles around a point: 55°, 110°, 70°, 65°, and x
- Angles around a point sum to 360°
- 55° + 110° + 70° + 65° + x = 360°
- 300° + x = 360°
- x = 60°

Let me double-check my answers:

1) x = 60°
2) x = 110° (after careful consideration)
3) x = 125°
4) x = 40°
5) x = 55°
6) x = 125° (assuming quadrilateral with all exterior angles)
7) x = 60°

Final Answer:
1) x = 60°
2) x = 110°
3) x = 125°
4) x = 40°
5) x = 55°
6) x = 125°
7) x = 60°