Okay, let's try to figure out the smallest angle in triangle PQS. Hmm, first, I need to visualize the diagram. There are four circles with radius 1, centers at P, Q, R, and S. They're tangent to each other and the sides of triangle ABC. The Asymptote code shows that R is at (0,0), Q is at (-2,0), S is at (2,0), and P is at (1, approximately 1.732). So, P is above the x-axis, and Q, R, S are on the x-axis. The triangle ABC is drawn with points A, B, and C, but the exact coordinates are given in the Asymptote code. Wait, maybe I can use those coordinates to help me.

Since the centers are given as P(1, 1.732), Q(-2,0), R(0,0), S(2,0). So triangle PQS is formed by connecting these three points. Let me note down their coordinates:

- P: (1, 1.732)
- Q: (-2, 0)
- S: (2, 0)

Wait, 1.732 is approximately √3, right? Because √3 is about 1.732. So maybe the coordinates are exact. Let me confirm:

√3 ≈ 1.732, so yes. So, point P is (1, √3). That might be exact. Let me check the Asymptote code again. The coordinates are given as (1,1.732), but in the code, the exact value might be (1, sqrt(3)). So maybe the coordinates are exact. Let's assume that.

So, triangle PQS has vertices at P(1, √3), Q(-2, 0), and S(2, 0). To find the smallest angle in this triangle, I need to compute the angles at each vertex and determine which is the smallest. The smallest angle will be opposite the shortest side. Alternatively, I can use the Law of Cosines to compute each angle.

First, let's compute the lengths of the sides of triangle PQS.

Let's label the triangle:

- Side PQ: between P and Q
- Side QS: between Q and S
- Side PS: between P and S

Compute each length using distance formula.

First, PQ:

P(1, √3) and Q(-2, 0). The distance is sqrt[(1 - (-2))² + (√3 - 0)²] = sqrt[(3)² + (√3)²] = sqrt[9 + 3] = sqrt[12] = 2*sqrt(3).

QS:

Q(-2, 0) and S(2, 0). The distance is sqrt[(2 - (-2))² + (0 - 0)²] = sqrt[(4)² + 0] = sqrt[16] = 4.

PS:

P(1, √3) and S(2, 0). The distance is sqrt[(2 - 1)² + (0 - √3)²] = sqrt[1 + 3] = sqrt[4] = 2.

So the sides are:

PQ: 2√3

QS: 4

PS: 2

So the sides are 2, 2√3, and 4. Let's confirm:

Wait, PQ is 2√3, QS is 4, PS is 2. So the sides opposite the vertices:

- Angle at P: opposite side QS (4)
- Angle at Q: opposite side PS (2)
- Angle at S: opposite side PQ (2√3)

Since the smallest side is PS (2), the angle opposite to it is the smallest angle. So the angle at Q is the smallest. Let me verify:

Wait, sides:

Opposite angle P is QS (4)

Opposite angle Q is PS (2)

Opposite angle S is PQ (2√3 ≈ 3.464)

So yes, the smallest side is PS (2), so angle at Q is the smallest.

Alternatively, maybe I can compute all angles to confirm.

Let me use the Law of Cosines to compute each angle.

First, angle at Q:

Law of Cosines: cos(angle Q) = (PQ² + QS² - PS²)/(2*PQ*QS)

Wait, no. Wait, Law of Cosines is c² = a² + b² - 2ab cos(C), where angle C is opposite side c.

So, for angle at Q, the sides adjacent to Q are PQ and QS, and the side opposite is PS.

So, angle Q is between sides PQ and QS, and opposite side PS.

Therefore, Law of Cosines:

PS² = PQ² + QS² - 2*PQ*QS*cos(angle Q)

So:

2² = (2√3)² + 4² - 2*(2√3)*4*cos(angle Q)

Compute:

4 = 12 + 16 - 16√3 cos(angle Q)

4 = 28 - 16√3 cos(angle Q)

Then, 16√3 cos(angle Q) = 28 - 4 = 24

So, cos(angle Q) = 24 / (16√3) = (24)/(16√3) = (3)/(2√3) = (√3)/2

Therefore, angle Q = arccos(√3/2) = 30 degrees.

So angle at Q is 30 degrees.

Then, angle at S:

Opposite side PQ (2√3). So using Law of Cosines:

PQ² = PS² + QS² - 2*PS*QS*cos(angle S)

(2√3)^2 = 2^2 + 4^2 - 2*2*4*cos(angle S)

12 = 4 + 16 - 16 cos(angle S)

12 = 20 - 16 cos(angle S)

16 cos(angle S) = 20 - 12 = 8

cos(angle S) = 8/16 = 1/2

Therefore, angle S = arccos(1/2) = 60 degrees.

Finally, angle at P:

Remaining angle. Since triangle angles sum to 180 degrees, angle P = 180 - 30 - 60 = 90 degrees.

Alternatively, using Law of Cosines:

PS² + PQ² - 2*PS*PQ*cos(angle P) = QS²

2^2 + (2√3)^2 - 2*2*(2√3)*cos(angle P) = 4^2

4 + 12 - 8√3 cos(angle P) = 16

16 - 8√3 cos(angle P) = 16

-8√3 cos(angle P) = 0

cos(angle P) = 0

Therefore, angle P = 90 degrees.

So the three angles are 30°, 60°, and 90°. Therefore, the smallest angle is 30 degrees.

But let me just confirm once again to make sure I didn't make any calculation errors.

Compute angle Q again:

Law of Cosines:

PS² = PQ² + QS² - 2*PQ*QS*cos(angle Q)

4 = 12 + 16 - 2*(2√3)*4*cos(angle Q)

4 = 28 - 16√3 cos(angle Q)

16√3 cos(angle Q) = 24

cos(angle Q) = 24/(16√3) = (3)/(2√3) = √3/2

Yes, that's correct. arccos(√3/2) is 30 degrees.

Angle S: cos(angle S) = 1/2, which is 60 degrees. Correct.

Angle P: 90 degrees. Correct.

Therefore, the smallest angle is 30 degrees. So the answer is 30 degrees.

But just to be thorough, maybe I can check using coordinates. Let's compute vectors and use the dot product to confirm the angles.

Compute vectors at point Q:

Points Q(-2, 0), P(1, √3), S(2, 0).

Vectors from Q to P: (1 - (-2), √3 - 0) = (3, √3)

Vectors from Q to S: (2 - (-2), 0 - 0) = (4, 0)

The angle between vectors (3, √3) and (4, 0).

Dot product: (3)(4) + (√3)(0) = 12 + 0 = 12

Length of first vector: sqrt(3² + (√3)²) = sqrt(9 + 3) = sqrt(12) = 2√3

Length of second vector: sqrt(4² + 0²) = 4

So cos(theta) = 12 / (2√3 * 4) = 12 / (8√3) = (3)/(2√3) = √3/2

Therefore, theta = 30 degrees. Correct.

Similarly, check angle at S:

Vectors from S to P: (1 - 2, √3 - 0) = (-1, √3)

Vectors from S to Q: (-2 - 2, 0 - 0) = (-4, 0)

Dot product: (-1)(-4) + (√3)(0) = 4 + 0 = 4

Length of first vector: sqrt((-1)^2 + (√3)^2) = sqrt(1 + 3) = 2

Length of second vector: sqrt((-4)^2 + 0^2) = 4

cos(theta) = 4 / (2*4) = 4 / 8 = 0.5

Therefore, theta = 60 degrees. Correct.

Angle at P: 90 degrees, as computed earlier. So all angles check out.

Therefore, the smallest angle is 30 degrees.

**Final Answer**
The degree measure of the smallest angle in triangle \( PQS \) is \boxed{30}.
Therefore, the smallest angle in triangle \( PQS \) is $\boxed{30^\circ}$.
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