Worksheet on calculating angles in various quadrilaterals and triangles.

Let’s solve each problem one by one. Remember: the sum of all interior angles in any quadrilateral is always 360°. We’ll use that rule for every shape here.

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First Problem (Y = ?)
We have a quadrilateral with angles: 135°, 140°, 60°, and y.
Add the known angles:
135 + 140 = 275
275 + 60 = 335
Now subtract from 360:
360 - 335 = 25
So, y = 25

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Second Problem (p = ?, q = ?)
This shape has two parts — it looks like a trapezoid with an extra triangle on the left. But actually, we can treat the whole figure as a pentagon? Wait — no, let’s look again.

Actually, this is a quadrilateral with some external angles marked. Let’s focus on the inner quadrilateral first.

The inner four angles are:
- Top-left: 125°
- Top-right: 140°
- Bottom-right: q (unknown)
- Bottom-left: p (unknown)

But wait — there’s also a 30° angle outside at the bottom-left corner. That means the *interior* angle at that corner is 90° - 30° = 60°? No — actually, looking at the diagram, the 30° is part of a right angle split into two parts: one is 30°, the other is p. So if they form a right angle together, then:

p + 30° = 90° → p = 60°

Now, for the quadrilateral inside: angles are 125°, 140°, q, and the angle adjacent to p — which is NOT p itself, but the full interior angle at that vertex.

Wait — let me re-express.

Actually, the shape shown is a quadrilateral where:
- One angle is 125°
- Another is 140°
- At the bottom right, angle is q
- At the bottom left, the interior angle is made up of p plus something? Hmm.

Looking more carefully: The diagram shows a vertical line on the left forming a right angle (90°), split into 30° and p. So yes — p = 90° - 30° = 60°

Now, the quadrilateral has four interior angles:
- Top-left: 125°
- Top-right: 140°
- Bottom-right: q
- Bottom-left: the angle next to p — but since p is part of the 90° corner, and the quadrilateral’s interior angle at that corner is actually the supplement? Wait — no.

Actually, the quadrilateral’s bottom-left interior angle is the same as the angle labeled “p” — because the 30° is outside the quadrilateral. So the quadrilateral’s four angles are: 125°, 140°, q, and p.

And we just found p = 60°.

So add them: 125 + 140 + 60 = 325
Then q = 360 - 325 = 35°

Wait — but let’s double-check. Is p really an interior angle of the quadrilateral?

Looking at the diagram description: the 30° is between the vertical side and the slanted side, and p is between the horizontal base and the slanted side. So together they make the corner angle of the quadrilateral? Actually, no — the quadrilateral’s corner at bottom-left is composed of both 30° and p? That would mean the interior angle is 30° + p.

But earlier I assumed p was the interior angle — that might be wrong.

Let me rethink.

If the vertical line and horizontal line meet at 90°, and the slanted line splits that 90° into 30° and p, then the interior angle of the quadrilateral at that corner is actually the angle between the horizontal base and the slanted side — which is p. And the 30° is between the vertical side and the slanted side — so it's outside the quadrilateral? Or inside?

Actually, in standard diagrams like this, when you see a right angle split into two parts by a diagonal, and one part is labeled 30°, the other is p, and the quadrilateral includes the area below the slanted line — then the interior angle of the quadrilateral at that corner is p.

Moreover, the top-left angle is 125°, which is clearly inside the quadrilateral.

So assuming the four interior angles of the quadrilateral are: 125°, 140°, q, and p.

And since p + 30° = 90° (because they form a right angle), then p = 60°.

Then total of three known angles: 125 + 140 + 60 = 325
q = 360 - 325 = 35°

Yes, that makes sense.

So:
p = 60
q = 35

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Third Problem (b = ?)
Quadrilateral with angles: 131°, 79°, 83°, and b.
Add knowns: 131 + 79 = 210; 210 + 83 = 293
b = 360 - 293 = 67

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Fourth Problem (a = ?, b = ?, c = ?)
This is a complex shape — looks like a large triangle divided into smaller triangles.

Let’s break it down.

First, notice the big shape is a triangle with base angles 30° and... well, we need to find angles inside.

There’s a point inside where several lines meet.

Label the points mentally.

At the top, there’s an angle of 130° — that’s part of a triangle on the left.

In the left small triangle: angles are 30° (bottom-left), 130° (top), and a — so:

a = 180 - 30 - 130 = 20°

Now, moving to the middle section.

There’s a straight line at the bottom — so angles on a straight line add to 180°.

At the bottom center, we have 50° and b next to it — and they are on a straight line? Looking at the diagram: the 50° and b are adjacent angles forming a straight line? Actually, no — they are parts of different triangles.

Wait — let’s look at the triangle that contains angle b.

There’s a triangle with angles: 30° (at top-right), b (at bottom-right), and another angle at the center.

Also, at the center point, several angles meet.

Note that around a point, angles add to 360°.

But perhaps easier: consider the triangle that has angles 50°, b, and the angle above b.

Wait — let’s use the fact that the big outer shape is a triangle.

Big triangle: left angle 30°, right angle c, top angle... what is the top angle?

The top angle is split into two parts: one is part of the left triangle (which had 130° — but that’s not the top angle of the big triangle).

Actually, the 130° is an interior angle of the left small triangle, not the big triangle.

Let me try a different approach.

Focus on the triangle that contains angle a: we already did — a = 20°.

Now, look at the triangle that shares the side with angle a — the middle triangle.

It has angles: a = 20°, 50°, and the angle at the top-center.

Call that angle d.

Then d = 180 - 20 - 50 = 110°

Now, at the top-center point, we have angle d = 110°, and next to it is the 30° angle (from the right small triangle), and also the angle from the left part — but wait, the 130° was in the left triangle, which is separate.

Actually, at the very top vertex of the big triangle, the angle is composed of the 130°? No — that doesn’t make sense because 130° is already used in the left triangle.

I think I made a mistake.

Let me redraw mentally.

The big triangle has:

- Left base angle: 30°
- Right base angle: c (unknown)
- Top angle: let’s call it T

Inside, there are lines dividing it.

From the left, a line goes up to a point, making a triangle with angles 30°, 130°, and a — so a = 20°, as before.

That 130° is the angle at the top of that small left triangle — which is not the same as the big triangle’s top angle.

Actually, the 130° is an obtuse angle inside, so likely it’s the angle between the left side and the internal line.

Perhaps better to use the fact that the sum of angles in each small triangle is 180°.

List all small triangles:

1. Left triangle: angles 30°, 130°, a → a = 20° (as before)

2. Middle triangle: angles a=20°, 50°, and let’s say x (at the top-center) → x = 180 - 20 - 50 = 110°

3. Right triangle: angles 30° (given at top-right), b (at bottom-right), and y (at center-top) → but y is adjacent to x.

At the center-top point, the angles from the middle and right triangles meet. Since they are on a straight line? Or around a point?

Actually, the two triangles share a common vertex at the top-center, and their angles at that vertex are adjacent and form a straight line? Looking at the diagram description, it seems that the 110° (x) and the angle from the right triangle at that point are on a straight line — because the top edge is straight.

So if x = 110°, then the angle in the right triangle at that vertex is 180° - 110° = 70°

Then in the right triangle: angles are 30° (top-right), 70° (top-center), and b (bottom-right)

So b = 180 - 30 - 70 = 80°

Now, for c: c is the bottom-right angle of the big triangle.

The big triangle has angles: left 30°, right c, and top angle.

What is the top angle of the big triangle? It is composed of the 130°? No — the 130° is inside the left small triangle.

Actually, the top angle of the big triangle is the angle between the two sides — which is split by the internal lines.

From the left, the internal line creates a 130° angle with the left side, but that’s not helpful.

Note that at the bottom, the big triangle’s base is straight, so the angles at the bottom should add appropriately.

At the bottom-left: 30° (part of big triangle)

At the bottom-center: we have 50° and b=80°, but these are not on the base — wait, the 50° and b are angles in the small triangles, not necessarily on the base.

Actually, looking back: the 50° is at the bottom-center, between the base and the internal line, and b is at the bottom-right, between the base and the right side.

So the entire bottom angle of the big triangle is composed of the 30° (left), plus the angle between the left internal line and the right internal line, plus b? This is messy.

Alternative approach: the big triangle’s three angles are:

- Left: 30°
- Right: c
- Top: let’s calculate from the internal angles.

At the top vertex of the big triangle, the angle is formed by the two sides. From the left small triangle, the angle at the top is 130°, but that’s the angle inside the small triangle, which is actually the supplement of the big triangle’s top angle? No.

I think I need to consider that the 130° is an exterior angle or something.

Let’s use the fact that the sum of angles around the internal points must work.

Another idea: the big triangle’s top angle can be found by noting that the internal lines create angles that add up.

From the left small triangle: we have angle a=20° at the internal point on the left.

At the bottom-center point, we have angles: 50° (in the middle triangle), and also the angle from the left triangle? The left triangle has angles 30°, 130°, 20° — so at the bottom-left vertex, it’s 30°, which is shared with the big triangle.

At the bottom-center vertex, the middle triangle has 50°, and there might be another angle from the left or right.

Actually, at the bottom-center point, the angles on the straight line should add to 180°.

The bottom side is straight, so the angles below the internal lines should add to 180°.

Specifically, at the bottom-center point, the angle between the left internal line and the right internal line is part of the middle triangle, which is 50°, but that’s above the base.

Below the base, it’s straight, so no issue.

Perhaps for angle c, we can look at the right small triangle.

We have b=80°, and the right small triangle has angles 30°, 70°, 80° — sum 180°, good.

Now, the big triangle’s right angle c is the same as b? No, because b is inside the small triangle, while c is the angle of the big triangle at the bottom-right.

In the diagram, c is labeled at the bottom-right corner, between the base and the right side — which is exactly the same as angle b in the small triangle? Let’s see.

In the right small triangle, angle b is at the bottom-right vertex, between the base and the right side — so yes, that is the same as angle c of the big triangle.

Is that correct? If the small triangle is part of the big triangle, and they share that corner, then yes, c = b.

But we calculated b = 80°, so c = 80°?

But let’s verify with the big triangle’s angle sum.

Big triangle angles: left 30°, right c, top T.

Sum must be 180°.

What is T? The top angle of the big triangle.

From the internal structure, the top angle T is composed of the angle from the left small triangle and the angle from the right small triangle at the top.

In the left small triangle, the top angle is 130° — but that can't be, because 130° is already larger than 180° for the whole triangle.

I think I have a fundamental mistake.

Let me start over for this problem.

Look at the diagram description: there is a large triangle. Inside, there are two lines from the top vertex to the base, dividing it into three parts.

Angles given:

- At bottom-left: 30° (of big triangle)
- At top-left part: 130° — this must be the angle in the left small triangle at the top vertex.
- Then a is the angle at the internal point on the left.
- In the middle, at the bottom, 50° is given.
- At the top-right part, 30° is given.
- b is at the bottom-right of the middle section.
- c is at the bottom-right of the big triangle.

Perhaps the 130° is not an angle of the small triangle, but an angle between the left side and the first internal line.

Assume that the 130° is the angle at the top-left vertex between the left side and the first internal line.

Then in the left small triangle, the angles are:

- At bottom-left: 30°
- At top-left: 130° — but that would make the third angle negative, impossible.

130° + 30° = 160°, so a = 20°, which is fine, but then the top angle of the big triangle would include this 130°, which is too big.

Unless the 130° is the reflex angle or something, but that's unlikely.

Another possibility: the 130° is the angle inside the quadrilateral or something, but the shape is a triangle.

Let's read the diagram again: "130°" is written in the left part, near the top, and "a" is at the internal point.

Perhaps the 130° is the angle of the big triangle at the top-left, but that doesn't make sense.

I recall that in some diagrams, the angle marked 130° might be the exterior angle, but let's think differently.

Use the fact that the sum of angles in the big triangle is 180°.

Let me denote the big triangle's angles as A, B, C at vertices A (top), B (bottom-left), C (bottom-right).

Given: angle at B is 30°.

Angle at C is c.

Angle at A is unknown.

Now, there is a point D on AB, and E on AC, and DE is drawn, but from the description, it seems there are two lines from A to the base, creating three triangles.

Assume that from the top vertex A, two lines are drawn to the base BC, meeting at points D and E, with D closer to B, E closer to C.

Then we have three triangles: ABD, ADE, AEC.

But in the diagram, it's described as having angles 130°, a, 50°, etc.

Perhaps the 130° is in triangle ABD.

In triangle ABD: angles at B is 30°, at A is part of the top angle, at D is a.

But 130° is given, so perhaps angle at D is 130°? But then 30° + 130° = 160°, so a = 20°, same as before.

Then in triangle ADE or whatever.

At point D on the base, the angle in triangle ABD is 130°, but since the base is straight, the adjacent angle in the next triangle would be 180° - 130° = 50°, which matches the given 50° in the middle.

Oh! That's it!

So at the bottom-center point (let's call it D), the angle in the left triangle is 130°, but since it's on a straight line, the angle in the middle triangle at D is 180° - 130° = 50°, which is given.

Perfect.

So in the left triangle: angles are 30° (at B), 130° (at D), and a (at A) — so a = 180 - 30 - 130 = 20°.

Now, in the middle triangle: it has vertices at A, D, and say E (another point on the base).

Angles: at D is 50°, at A is part of the top angle, and at E is b? Or what.

The middle triangle has angles: at D: 50°, at A: let's call it f, and at the other base point, say E, angle g.

But we have a given 50° at D, and also in the diagram, there is a 50° mentioned, which is at D for the middle triangle.

Additionally, there is a 30° at the top-right, which is probably in the right triangle.

Also, b is at the bottom-right of the middle section, which might be at E.

Assume the middle triangle has angles: at D: 50°, at A: f, at E: b.

But we don't know f yet.

At vertex A, the total angle is split into three parts: one for left triangle (a=20°), one for middle triangle (f), and one for right triangle (say h).

And we know that in the right triangle, there is a 30° angle at the top, which is h=30°.

So total top angle of big triangle is a + f + h = 20° + f + 30° = 50° + f.

Now, in the middle triangle, angles sum to 180°: 50° (at D) + f (at A) + b (at E) = 180°.

So f + b = 130°. [equation 1]

Now, at point E on the base, the angle in the middle triangle is b, and since the base is straight, the angle in the right triangle at E is 180° - b.

In the right triangle: vertices A, E, C.

Angles: at A: h=30°, at E: 180° - b, at C: c.

Sum: 30° + (180° - b) + c = 180°

Simplify: 30 + 180 - b + c = 180 → 210 - b + c = 180 → c - b = -30 → b - c = 30. [equation 2]

Also, for the big triangle: angles at B: 30°, at C: c, at A: 50° + f.

Sum: 30 + c + 50 + f = 180 → c + f = 100. [equation 3]

Now we have:

From eq1: f + b = 130

From eq2: b - c = 30

From eq3: c + f = 100

Let me solve these equations.

From eq2: b = c + 30

Plug into eq1: f + (c + 30) = 130 → f + c = 100

But from eq3: c + f = 100 — same thing.

So we have dependency.

We need another relation.

In the right triangle, we have angles: at A: 30°, at E: 180° - b, at C: c.

Sum is 180°, which gave us eq2.

But we also know that in the right triangle, the angle at E is 180° - b, and it must be positive, etc.

Perhaps we can use the fact that the middle triangle's angle at E is b, and it's on the straight line, so the right triangle's angle at E is 180° - b, as we have.

Now, from eq3: c + f = 100

From eq1: f + b = 130

Subtract eq3 from eq1: (f + b) - (c + f) = 130 - 100 → b - c = 30, which is eq2.

So we have only two independent equations.

We need to find numerical values.

Perhaps I missed that in the middle triangle, there is another given angle.

In the diagram, it says "50°" and "b", and also "30°" at top-right.

But in the middle triangle, is there a specific angle given besides 50°?

Looking back at the user's description: "50°" is at the bottom-center, which we assigned to the middle triangle at D.

And "b" is at the bottom-right of the middle section, which is at E.

Also, in the right triangle, "30°" is at the top, which is at A.

But we have three variables.

Perhaps the "50°" is not at D for the middle triangle, but let's check the initial calculation.

Another thought: in the middle triangle, the angle at A is not f, but perhaps it's given or can be found.

Notice that at vertex A, the angle for the middle triangle is between the two internal lines.

But we have a = 20° for the left part, h = 30° for the right part, so the middle part f = total top angle - 20° - 30°.

But we don't know total top angle.

Perhaps from the big triangle.

Let's assume that the big triangle's top angle is T.

Then T = a + f + h = 20 + f + 30 = 50 + f.

Big triangle sum: 30 + c + T = 180 → 30 + c + 50 + f = 180 → c + f = 100, as before.

Now, in the middle triangle: angles are at D: 50°, at A: f, at E: b, sum 180, so f + b = 130.

In the right triangle: angles at A: 30°, at E: 180 - b (since straight line), at C: c, sum 180: 30 + (180 - b) + c = 180 → 210 - b + c = 180 → c - b = -30.

So we have:

1) c + f = 100

2) f + b = 130

3) c - b = -30

From 3) c = b - 30

Plug into 1): (b - 30) + f = 100 → b + f = 130, which is the same as 2).

So indeed, we have infinite solutions? That can't be.

I must have misidentified the angles.

Let's look back at the user's input: "50°" is written, and "b", and "30°" at top-right.

Perhaps the 50° is not at D, but at another place.

Another idea: perhaps the "50°" is the angle at the bottom-center for the middle triangle, but in the context, it might be the angle between the base and the internal line, which is what we have.

Perhaps for angle b, it is in the right triangle.

Let's try to calculate c directly.

Notice that the big triangle's right angle c is the same as the angle in the right small triangle at C, which is c.

In the right small triangle, we have angles: at A: 30°, at C: c, and at E: let's call it e.

Sum: 30 + c + e = 180 → c + e = 150.

At point E on the base, the angle e in the right triangle and the angle b in the middle triangle are adjacent on a straight line, so b + e = 180°.

So e = 180 - b.

Then c + (180 - b) = 150 → c - b = -30, same as before.

Still stuck.

Perhaps the "50°" is not in the middle triangle at D, but at A or something.

Let's read the user's description carefully: "50°" is listed, and in the diagram, it's likely at the bottom-center, between the base and the internal line for the middle triangle.

Another thought: in the middle triangle, the angle at A is not f, but perhaps it's given as part of the 130° or something.

Perhaps the 130° is the angle of the big triangle at the top, but that can't be because then with 30° at bottom, the third angle would be 20°, but we have more.

Let's calculate the top angle of the big triangle from the internal angles.

From the left, the internal line makes an angle of 130° with the left side, but that 130° is the angle in the small triangle, which is at the top vertex for that small triangle.

Perhaps for the big triangle, the top angle is the sum of the angles from the three small triangles at A.

We have a = 20° for left, h = 30° for right, and for middle, let's say m.

Then top angle = 20 + m + 30 = 50 + m.

In the middle triangle, angles are: at A: m, at D: 50°, at E: b, sum 180, so m + 50 + b = 180 → m + b = 130.

In the right triangle: at A: 30°, at E: 180 - b (since straight line), at C: c, sum 180: 30 + 180 - b + c = 180 → c - b = -30.

Big triangle: 30 + c + (50 + m) = 180 → c + m = 100.

Same as before.

Now, from m + b = 130 and c + m = 100, subtract: (m + b) - (c + m) = 130 - 100 → b - c = 30.

From c - b = -30, which is the same as b - c = 30.

So consistent, but we need another equation.

Perhaps the "50°" is not the angle at D for the middle triangle, but the angle at A for the middle triangle or something.

Let's look at the user's text: "50°" is written, and in the context, it might be the angle between the two internal lines at the bottom or something.

Another idea: perhaps the 50° is the angle at the bottom-center for the big triangle's base, but that doesn't help.

Let's assume that in the middle triangle, the angle at A is 50°, but the user said "50°" is at the bottom.

Perhaps "50°" is the angle b, but no, b is separate.

Let's try to guess from typical problems.

Often in such diagrams, the 50° is the angle in the middle triangle at the bottom, and b is at the other bottom, and we can find.

Perhaps for angle b, it is in the right triangle.

Let's calculate the angle at E for the middle triangle.

From the left, at D, the angle is 130° for the left triangle, so for the middle triangle at D, it is 50°, as supplementary.

Then in the middle triangle, if we knew another angle, but we don't.

Unless the 30° at top-right is for the middle triangle, but it's specified as top-right.

Perhaps the "30°" at top-right is the angle in the right triangle at A, which we have.

Let's use the fact that the sum of angles around the internal points.

At vertex A, the angles add to the top angle.

At point D on the base, the angles from left and middle triangles add to 180°, which we have: 130° + 50° = 180°, good.

At point E on the base, the angles from middle and right triangles add to 180°: b + e = 180°, where e is the angle in the right triangle at E.

In the right triangle, angles are 30° at A, e at E, c at C, sum 180: 30 + e + c = 180 → e + c = 150.

But e = 180 - b, so 180 - b + c = 150 → c - b = -30, same as before.

Now, for the middle triangle: angles at D: 50°, at A: m, at E: b, sum 180: 50 + m + b = 180 → m + b = 130.

For the big triangle: angles at B: 30°, at C: c, at A: a + m + h = 20 + m + 30 = 50 + m, sum 30 + c + 50 + m = 180 → c + m = 100.

So we have:

1) m + b = 130

2) c + m = 100

3) c - b = -30

From 2) and 3): from 3) c = b - 30, plug into 2): (b - 30) + m = 100 → b + m = 130, which is 1).

So the system is dependent, and we have one free variable.

That means we need another constraint.

Perhaps in the diagram, the "50°" is not the angle at D for the middle triangle, but the angle at A for the middle triangle.

Let me try that.

Suppose that in the middle triangle, the angle at A is 50°.

Then from left triangle: a = 20° (as before).

Then at A, for middle triangle: 50°.

For right triangle: 30° at A.

So total top angle = 20 + 50 + 30 = 100°.

Then big triangle: angles 30° at B, c at C, 100° at A, sum 30 + c + 100 = 180 → c = 50°.

Now, in the middle triangle: angles at A: 50°, at D: ? , at E: b.

Sum 180, so angle at D + b = 130°.

At D, on the base, the angle from left triangle is 130°, so for middle triangle at D, it is 180° - 130° = 50°.

So angle at D for middle triangle is 50°.

Then 50° + b = 130° → b = 80°.

Then in the right triangle: angles at A: 30°, at C: c = 50°, so at E: 180 - 30 - 50 = 100°.

At E on the base, the angle from middle triangle is b = 80°, and from right triangle is 100°, and 80° + 100° = 180°, perfect, on a straight line.

Also, the 50° given in the diagram is the angle at A for the middle triangle, which matches our assumption.

In the user's description, "50°" is listed, and in many diagrams, it might be placed at the top for the middle section.

So likely, the 50° is the angle at the top for the middle triangle, not at the bottom.

That makes sense.

So to confirm:

- Left triangle: angles 30° (B), 130° (D), a=20° (A)

- Middle triangle: angles 50° (A), 50° (D), b=80° (E) — sum 180°, good.

- Right triangle: angles 30° (A), 100° (E), c=50° (C) — sum 180°, good.

Big triangle: A: 20+50+30=100°, B:30°, C:50°, sum 180°, good.

And at D: left triangle 130°, middle triangle 50°, sum 180°.

At E: middle triangle 80°, right triangle 100°, sum 180°.

Perfect.

So a = 20°, b = 80°, c = 50°.

In the user's notation, a is in the left triangle, b is in the middle triangle at E, c is at C.

Yes.

So:

a = 20
b = 80
c = 50

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Fifth Problem (x = ?)
Quadrilateral with angles: 92°, 110°, 128°, and x.
Add knowns: 92 + 110 = 202; 202 + 128 = 330
x = 360 - 330 = 30

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Now, compiling all answers:

First: Y = 25
Second: p = 60, q = 35
Third: b = 67
Fourth: a = 20, b = 80, c = 50
Fifth: x = 30

Note: in the fourth problem, there are two 'b's — one in the third problem and one in the fourth. But they are different problems, so it's ok. In the fourth problem, b is 80, in the third, b is 67.

In the user's output, for the fourth problem, it's a,b,c, and for third, it's b, so we'll keep them separate.

Final Answer:

Y = 25
p = 60
q = 35
b = 67 (for the third problem)
a = 20
b = 80 (for the fourth problem)
c = 50
x = 30

To avoid confusion, since the user has separate sections, we'll list them as per the order.

In the user's message, the problems are:

1. Y =
2. p = , q =
3. b =
4. a = , b = , c =
5. x =

So for problem 3, b=67; for problem 4, b=80.

In the final answer, we'll write them as requested.

Final Answer:
Y = 25
p = 60
q = 35
b = 67
a = 20
b = 80
c = 50
x = 30