Find the angle x in each geometric figure.

Let’s solve each problem one by one. We’ll use basic angle rules:

- Angles on a straight line add to 180°.
- Angles around a point add to 360°.
- In any polygon, the sum of interior angles = (n–2) × 180°, where n is number of sides.
- For triangles: sum of interior angles = 180°.
- Exterior angle = 180° – interior angle at that vertex.

We’ll go step by step for each question.

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Problem 1)

We have a triangle with two known exterior angles: 130° and 70°, and one interior angle given as 80°. Wait — let’s look carefully.

Actually, the diagram shows:

- One angle outside the triangle is 130° → so the adjacent interior angle is 180° – 130° = 50°.
- Another angle inside the triangle is 80°.
- The third interior angle is next to x, and we’re told another exterior angle is 70° → so the interior angle there is 180° – 70° = 110°? That can’t be right because 50 + 80 + 110 = 240 > 180.

Wait — maybe I misread. Let me re-express.

Looking again: It seems like we have a triangle with:

- One interior angle = 80°
- One exterior angle = 130° → so its adjacent interior angle = 50°
- Then the third interior angle must be 180 – 80 – 50 = 50°
- But then x is an exterior angle to that 50° interior angle → so x = 180 – 50 = 130°? But wait, there's also a 70° marked.

Alternatively, perhaps the 70° is part of a different shape.

Wait — actually, looking at the figure: it looks like a triangle with one side extended, forming exterior angles.

Let me label:

Assume triangle ABC.

At vertex A: exterior angle = 130° → interior angle A = 50°

At vertex B: interior angle = 80°

Then interior angle C = 180 – 50 – 80 = 50°

Now, at vertex C, there’s an exterior angle labeled x, but also another angle marked 70° — which might be adjacent?

Wait — perhaps the 70° is not part of the triangle. Maybe it’s a separate angle.

Actually, looking more carefully: the figure has a triangle, and from one vertex, two lines extend, making angles 70° and x with the side.

Perhaps it’s better to think: the three angles around the point where x is located should add to 360°? No.

Alternative approach: Use the fact that in any triangle, the exterior angle equals the sum of the two opposite interior angles.

But let’s try this:

From the diagram:

- One exterior angle is 130° → so interior angle = 50°
- Another interior angle is 80°
- So third interior angle = 50°
- Now, at that vertex, the exterior angle would be 130°, but we see 70° and x.

Wait — perhaps the 70° and x are on a straight line? If so, 70 + x = 180 → x = 110? But that doesn't fit.

I think I need to reinterpret.

Let me consider the entire figure.

Actually, upon closer inspection, Problem 1 appears to show a triangle with:

- One exterior angle = 130° → interior = 50°
- One interior angle = 80°
- So third interior angle = 50°
- Now, at that third vertex, the exterior angle is split into two parts: 70° and x? Or perhaps x is the exterior angle, and 70° is something else.

Wait — maybe the 70° is an exterior angle at another vertex? But we already used 130°.

Another idea: perhaps the 70° and x are adjacent angles forming a straight line with the interior angle.

If interior angle is 50°, then the two exterior angles on either side would add to 130°, but that doesn’t help.

Let’s try using the rule: sum of exterior angles of any polygon is 360°.

For a triangle, sum of exterior angles (one per vertex) is 360°.

So if we have exterior angles: 130°, 70°, and x — then 130 + 70 + x = 360 → x = 160°? But that seems too big, and we have an 80° interior angle which might conflict.

Wait — the 80° is interior, so its exterior is 100°.

Ah! That’s it!

In the triangle:

- At one vertex, exterior angle = 130° → interior = 50°
- At another vertex, interior angle = 80° → exterior = 100°
- At the third vertex, exterior angle = ? Let’s call it y
- Sum of exterior angles = 360° → 130 + 100 + y = 360 → y = 130°

But in the diagram, at that third vertex, we see two angles: 70° and x. Perhaps they are parts of the exterior angle?

Maybe x is the exterior angle, and 70° is adjacent to it on a straight line? Then x + 70 = 180 → x = 110°, but that contradicts.

I think I found the issue: in Problem 1, the 70° is likely the exterior angle at the third vertex, and x is something else.

Let’s look at the diagram description again.

Upon second thought, perhaps the figure is not a simple triangle. It might be a quadrilateral or have intersecting lines.

To save time, let’s move to other problems and come back.

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Problem 2)

We have a right angle (90°), and angles 75°, 60°, 25°, and x.

It looks like a polygon or a set of angles around a point.

Let’s assume it’s a pentagon or something, but probably it’s angles around a point or on lines.

Notice: there’s a right angle symbol, so 90°.

Also, 75°, 60°, 25°, and x.

Perhaps these are all angles around a single point? Sum should be 360°.

So 90 + 75 + 60 + 25 + x = 360?

Calculate: 90+75=165, +60=225, +25=250, so x = 360 - 250 = 110°.

Is that it? Let me check if all are around one point.

The diagram shows several lines meeting, so likely yes.

So for Problem 2, x = 110°.

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Problem 3)

We have a quadrilateral with some exterior angles.

Given: 60°, 70°, 75°, 35°, and x.

These appear to be exterior angles.

Sum of exterior angles of any polygon is 360°.

So 60 + 70 + 75 + 35 + x = 360?

Add: 60+70=130, +75=205, +35=240, so x = 360 - 240 = 120°.

But is it a pentagon? Five exterior angles, so yes, pentagon.

Sum of exterior angles is always 360°, regardless of number of sides.

So x = 120°.

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Problem 4)

Angles: 140°, 120°, 60°, and x.

Again, likely exterior angles of a quadrilateral.

Sum = 360°.

140 + 120 + 60 + x = 360

140+120=260, +60=320, so x = 40°.

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Problem 5)

Pentagon? Angles: 65°, 100°, 50°, right angle (90°), and x.

These seem to be exterior angles.

Sum = 360°.

65 + 100 + 50 + 90 + x = 360

65+100=165, +50=215, +90=305, so x = 360 - 305 = 55°.

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Problem 6)

Quadrilateral? Angles: 125°, 50°, 60°, and x.

Exterior angles? Sum 360°.

125 + 50 + 60 + x = 360

125+50=175, +60=235, so x = 125°.

But let me confirm: 360 - 235 = 125, yes.

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Problem 7)

Angles around a point: 55°, 110°, 70°, 65°, and x.

Sum should be 360°.

55 + 110 = 165, +70=235, +65=300, so x = 60°.

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Now back to Problem 1).

Let me try again.

Diagram: triangle with:

- One exterior angle = 130° → interior = 50°
- One interior angle = 80°
- So third interior angle = 50°
- At that third vertex, there is an exterior angle composed of two parts: 70° and x? Or perhaps x is the exterior angle, and 70° is adjacent.

Notice that at the vertex with interior angle 50°, the exterior angle is 130°, but in the diagram, it's shown as two angles: 70° and x, which might be on a straight line with the interior angle.

Actually, the interior angle is 50°, so the two adjacent exterior angles on the extensions should add to 130°, but typically we take one exterior angle per vertex.

Perhaps the 70° is not an exterior angle of the triangle, but part of another line.

Another idea: use the fact that the sum of angles in the triangle is 180°, and then use linear pairs.

Let me define:

Let the triangle have vertices A, B, C.

At A: exterior angle 130° → interior A = 50°

At B: interior angle 80°

At C: interior angle = 50°

Now, at vertex C, the side is extended, and we have two angles marked: 70° and x, which are on the straight line.

Since the interior angle at C is 50°, the exterior angle on one side is 130°, but if it's split into 70° and x, then 70 + x = 130? So x = 60°.

That makes sense!

Because the exterior angle is 130°, and it's divided into two parts: 70° and x, so x = 130 - 70 = 60°.

Yes, that fits.

So for Problem 1, x = 60°.

Let me verify with another method.

The three interior angles: 50°, 80°, 50° — sum 180°, good.

At vertex C, the exterior angle is 130°, which is equal to the sum of the two opposite interior angles: 50° + 80° = 130°, yes.

And if that exterior angle is split into 70° and x, then x = 60°.

Perfect.

So now we have all answers.

Summary:

1) x = 60°

2) x = 110°

3) x = 120°

4) x = 40°

5) x = 55°

6) x = 125°

7) x = 60°

Let me double-check Problem 2.

Problem 2: angles 90° (right angle), 75°, 60°, 25°, and x.

If they are around a point, sum 360°.

90 + 75 = 165

165 + 60 = 225

225 + 25 = 250

360 - 250 = 110°, yes.

Problem 3: exterior angles 60,70,75,35,x sum to 360.

60+70=130, 130+75=205, 205+35=240, 360-240=120, yes.

Problem 4: 140+120+60=320, 360-320=40, yes.

Problem 5: 65+100+50+90=305, 360-305=55, yes.

Problem 6: 125+50+60=235, 360-235=125, yes.

Problem 7: 55+110+70+65=300, 360-300=60, yes.

All good.

Final Answer:
1) 60°
2) 110°
3) 120°
4) 40°
5) 55°
6) 125°
7) 60°