A 50,60,70 triangle is a triangle in which two of the side lengths, are different numbers taken from the set (50, 60, 70) , with the remaining number being one of the angles in the triangle, in degrees. This angle doesn’t have to be in between the two side lengths. For instance if two of the sides are 50 and 60 and the 70 degree angle is in between the 50 and un known the perimeter is roughly 164.4.
Of all the possible 50, 60, 70 triangles, one has a smaller perimeter than the rest. What is this perimeters.

We are tasked with finding the smallest perimeter of a "50, 60, 70" triangle. In such a triangle, two of the values from the set {50, 60, 70} represent side lengths, and one represents the angle in degrees.

### Step 1: Overview of possible configurations
We need to test different combinations of side lengths and angles and find the one that gives the smallest perimeter.

We will apply the Law of Cosines to solve for the third unknown side of the triangle. The formula for the Law of Cosines is:

\[
c^2 = a^2 + b^2 - 2ab \cdot \cos(C)
\]

Where:
- \(a\) and \(b\) are the known sides,
- \(C\) is the known angle,
- \(c\) is the unknown side.

The perimeter is \(P = a + b + c\).

### Step 2: Consider the cases
We will explore all three possible configurations for the sides and angle:

1. **Sides 50 and 60, angle 70°:**
    \[
    c^2 = 50^2 + 60^2 - 2(50)(60) \cdot \cos(70^\circ)
    \]
    \[
    c^2 = 2500 + 3600 - 6000 \cdot \cos(70^\circ)
    \]
    \[
    c^2 = 6100 - 6000 \cdot 0.3420 \approx 6100 - 2052
    \]
    \[
    c^2 \approx 4048
    \]
    \[
    c \approx 63.6
    \]
    The perimeter is:
    \[
    P \approx 50 + 60 + 63.6 = 173.6
    \]

2. **Sides 50 and 70, angle 60°:**
    \[
    c^2 = 50^2 + 70^2 - 2(50)(70) \cdot \cos(60^\circ)
    \]
    \[
    c^2 = 2500 + 4900 - 7000 \cdot \cos(60^\circ)
    \]
    \[
    c^2 = 7400 - 7000 \cdot 0.5 = 7400 - 3500
    \]
    \[
    c^2 = 3900
    \]
    \[
    c \approx 62.45
    \]
    The perimeter is:
    \[
    P \approx 50 + 70 + 62.45 = 182.45
    \]

3. **Sides 60 and 70, angle 50°:**
    \[
    c^2 = 60^2 + 70^2 - 2(60)(70) \cdot \cos(50^\circ)
    \]
    \[
    c^2 = 3600 + 4900 - 8400 \cdot \cos(50^\circ)
    \]
    \[
    c^2 = 8500 - 8400 \cdot 0.6428 \approx 8500 - 5399
    \]
    \[
    c^2 \approx 3101
    \]
    \[
    c \approx 55.68
    \]
    The perimeter is:
    \[
    P \approx 60 + 70 + 55.68 = 185.68
    \]

### Step 3: Conclusion
The smallest perimeter occurs for the triangle with sides 50 and 60, and the angle of 70°. The perimeter in this case is approximately:

\[
\boxed{173.6}
\]

Would you like to discuss this in more detail, or have further questions? Here are some related questions to consider:

1. How is the Law of Cosines derived from the Pythagorean theorem?
2. What other uses does the Law of Cosines have in trigonometry?
3. How does changing the angle between two sides affect the perimeter of a triangle?
4. What is the difference between the Law of Cosines and the Law of Sines?
5. What happens if all angles in a triangle are smaller than 90 degrees?

**Tip:** The Law of Cosines is particularly useful when you know two sides and the included angle in non-right triangles.

A 50,60,70 triangle is a triangle in which two of the side lengths are different numbers taken from the set (50, 60, 70), with the remaining number being one of the angles in the triangle, in degrees. This angle doesn’t have to be in between the two side lengths. For instance if two of the sides are 50 and 60 and the 70 degree angle is in between the 50 and unknown the perimeter is roughly 164.4. Of all the possible 50, 60, 70 triangles, one has a smaller perimeter than the rest. What is this perimeter?

Discover how to find the smallest perimeter of a 50-60-70 triangle using the Law of Cosines. Learn the step-by-step solution to calculate the perimeter by solving different side and angle configurations. Perfect for students in Grades 9-12 studying trigonometry and geometry.

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