Find the missing angles in various polygons using this printable worksheet.

Let’s solve each problem one by one. We’ll use the rule:
The sum of interior angles in a polygon = (n - 2) × 180°, where n is the number of sides.

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Problem 1: Quadrilateral (4 sides)
Sum of angles = (4 - 2) × 180 = 360°
Given angles: 135°, 60°, 50°
Add them: 135 + 60 + 50 = 245°
Missing angle x = 360 - 245 = 115°

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Problem 2: Quadrilateral with two right angles (90° each)
Sum = 360°
Given: 90°, 90°, 50°
Add: 90 + 90 + 50 = 230°
x = 360 - 230 = 130°

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Problem 3: Triangle with two equal sides → Isosceles triangle
Two sides are marked equal → base angles are equal.
One angle is 60°, and the other two are equal? Wait — actually, look again:
It shows one angle as 60°, and another as x°, and the third also looks like it might be x°? But wait — the marks are on the *sides*, not the angles.
Actually, if two sides are equal, then the angles opposite them are equal.
So if the two sides with tick marks are equal, then the angles opposite them are equal.
Looking at the diagram: the 60° angle is between the two equal sides? Or opposite?
Wait — better to think: total angles in triangle = 180°
If two angles are equal (because two sides are equal), and one is 60°, then:
Case 1: If 60° is the vertex angle, then base angles = (180 - 60)/2 = 60° → equilateral → all 60° → so x = 60°
But in the diagram, only one angle is labeled 60°, and another is x°, and the third is unlabeled but has no mark.
Wait — actually, looking closely: the two sides with tick marks are adjacent to the 60° angle? No — let me re-express.

Actually, standard interpretation: when two sides have tick marks, they are equal → angles opposite them are equal.
In this triangle, the side opposite the 60° angle does NOT have a tick. The two sides that DO have ticks are the ones forming the 60° angle? That would mean the two base angles are equal.

Wait — perhaps simpler: assume the triangle has angles: 60°, x°, and y°. But since two sides are equal, two angles must be equal.
If the 60° is one of the base angles, then the other base angle is also 60°, so top angle = 60° → equilateral → x=60°.
But in the diagram, x is shown at the top, and 60° at bottom left. And the two sides from the top vertex down to the base have ticks — meaning those two sides are equal → so the base angles are equal.
Therefore, the two bottom angles are equal. One is given as 60°, so the other bottom angle is also 60°. Then top angle x = 180 - 60 - 60 = 60°

Wait — but that makes it equilateral. Maybe that’s correct. Alternatively, maybe the 60° is the apex? Let me check the drawing mentally.

Actually, in many such problems, if two sides are equal and one angle is 60°, it often turns out to be equilateral. But let's calculate properly.

Assume: triangle with two equal sides → isosceles → two equal angles.
Total = 180°
Suppose the two equal angles are both x, and the third is 60° → then 2x + 60 = 180 → 2x=120 → x=60 → again equilateral.

Alternatively, suppose the 60° is one of the equal angles → then the other equal angle is 60°, and the third is x → 60+60+x=180 → x=60.

So either way, x=60°. So answer is 60°

But wait — in the diagram, x is at the top, and 60° at bottom left, and the two sides from top to bottom corners have ticks — meaning those two sides are equal → so the angles at the bottom are equal. Since one bottom angle is 60°, the other is 60°, so top angle x = 180 - 60 - 60 = 60°. Yes.

So x = 60°

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Problem 4: Heptagon (7 sides)
Sum = (7-2)×180 = 5×180 = 900°
Given angles: 130°, 122°, 130°, 130°, 128°, 138°, and x°
Add known angles:
130 + 122 = 252
252 + 130 = 382
382 + 130 = 512
512 + 128 = 640
640 + 138 = 778
So x = 900 - 778 = 122°

Wait — let me add again carefully:

List: 130, 122, 130, 130, 128, 138

Step-by-step:

130 + 122 = 252
252 + 130 = 382
382 + 130 = 512
512 + 128 = 640
640 + 138 = 778 → yes
900 - 778 = 122 → so x = 122°

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Problem 5: Pentagon (5 sides)
Sum = (5-2)×180 = 3×180 = 540°
Given angles: 120°, 107°, and three right angles? Wait — look: there are three squares indicating 90° angles.
So angles: 120°, 107°, 90°, 90°, 90°
Add: 120 + 107 = 227; 227 + 90 = 317; 317 + 90 = 407; 407 + 90 = 497
x = 540 - 497 = 43°

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Problem 6: Quadrilateral (4 sides)
Sum = 360°
Given: 140°, 70°, 50°, and x°
Add: 140 + 70 = 210; 210 + 50 = 260
x = 360 - 260 = 100°

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Problem 7: Quadrilateral (parallelogram?)
Opposite angles equal? In parallelogram, opposite angles are equal, consecutive add to 180.
Given: 60°, 120°, 60°, and x°
If it’s a parallelogram, then opposite angles equal: so if one pair is 60° and 60°, the other pair should be 120° and 120° → so x should be 120°? But wait, 120° is already given.
Angles: let’s list them: say A=60°, B=x°, C=120°, D=60°
In quadrilateral, sum=360°
So 60 + x + 120 + 60 = 360 → 240 + x = 360 → x=120°
Yes, so x=120°

Even without assuming parallelogram, just add: 60+120+60=240, 360-240=120.

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Problem 8: Kite or quadrilateral?
Angles: 140°, 40°, 140°, and x°
Sum = 360°
Add: 140 + 40 = 180; 180 + 140 = 320
x = 360 - 320 = 40°

Note: This looks like a kite, with two pairs of equal adjacent angles. Here, 140° appears twice, and 40° and x° — so if symmetric, x should be 40°, which matches.

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Now, compiling all answers:

1. 115
2. 130
3. 60
4. 122
5. 43
6. 100
7. 120
8. 40

Final Answer:
1. 115
2. 130
3. 60
4. 122
5. 43
6. 100
7. 120
8. 40