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The problem presented is a geometry-based puzzle involving a **50-60-70 triangle**, where two of the side lengths are taken from the set \(\{50, 60, 70\}\) (in millimeters), and the third value is the angle (in degrees) of the triangle.

### Key details:
- A triangle is formed with two sides from the set \(\{50, 60, 70\}\) and one angle from the same set.
- The problem asks to find the triangle with the smallest **perimeter** and calculate that perimeter to **4 significant figures**.

#### Steps to solve:

1. **Understanding the structure of a 50-60-70 triangle**:
   The triangle has sides of 50 mm, 60 mm, and an angle of 70°.

2. **Using the Law of Cosines**:
   To calculate the third side, the Law of Cosines helps in non-right triangles:
   \[
   c^2 = a^2 + b^2 - 2ab \cdot \cos(\theta)
   \]
   where \(a = 50\) mm, \(b = 60\) mm, and \(\theta = 70^\circ\).

   Let's calculate the third side \(c\) using the above formula.

3. **Calculating the perimeter**:
   The perimeter \(P\) of the triangle will be the sum of all three sides.

Would you like me to compute this for you step-by-step now?

A 50-60-70 triangle is a triangle in which two of the side lengths, measured in millimeters, are different numbers taken from the set {50, 60, 70}, with the remaining number being one of the angles in the triangle, measured in degrees. Of all 50-60-70 triangles, one has a smaller perimeter than the rest. If this perimeter is P mm, what is P (to 4 sig. figs.)?

This problem involves finding the smallest perimeter of a 50-60-70 triangle, where two sides and one angle are taken from the set {50, 60, 70}. Using the Law of Cosines, we calculate the third side and the perimeter to 4 significant figures. Suitable for students in Grades 10-12.

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