To determine the type of triangle formed by the cities A, B, and C, we need to analyze the lengths of the sides of the triangle. The distances between the cities are as follows:
- Distance between city A and city B, [tex]\( AB = 22 \)[/tex] miles.
- Distance between city B and city C, [tex]\( BC = 54 \)[/tex] miles.
- Distance between city A and city C, [tex]\( AC = 51 \)[/tex] miles.
We will use the properties of the triangle inequality and the scalar product law of cosine to determine the type of triangle: whether it is acute, right, or obtuse.
1. Calculate the squares of the sides:
[tex]\[
AB^2 = 22^2 = 484
\][/tex]
[tex]\[
BC^2 = 54^2 = 2916
\][/tex]
[tex]\[
AC^2 = 51^2 = 2601
\][/tex]
2. Compare the sum of the squares of two sides with the square of the third side:
- First, check [tex]\(AB^2 + AC^2\)[/tex] and compare it to [tex]\(BC^2\)[/tex]:
[tex]\[
AB^2 + AC^2 = 484 + 2601 = 3085
\][/tex]
[tex]\[
AB^2 + AC^2 > BC^2 \quad \text{as} \quad 3085 > 2916
\][/tex]
This indicates that the triangle satisfies the condition for being an acute triangle for these specific sides. For the triangle to be acute, the sum of the squares of any two sides must be greater than the square of the third side.
- Next, check [tex]\(AB^2 + BC^2\)[/tex] and compare it to [tex]\(AC^2\)[/tex]:
[tex]\[
AB^2 + BC^2 = 484 + 2916 = 3400
\][/tex]
[tex]\[
AB^2 + BC^2 > AC^2 \quad \text{as} \quad 3400 > 2601
\][/tex]
- Finally, check [tex]\(BC^2 + AC^2\)[/tex] and compare it to [tex]\(AB^2\)[/tex]:
[tex]\[
BC^2 + AC^2 = 2916 + 2601 = 5517
\][/tex]
[tex]\[
BC^2 + AC^2 > AB^2 \quad \text{as} \quad 5517 > 484
\][/tex]
Since the inequality [tex]\(AB^2 + AC^2 > BC^2\)[/tex] holds true and shows that all sides result in sums greater than the square of the remaining side, the triangle is confirmed to be an acute triangle.
Hence, the correct answer is:
- An acute triangle, because [tex]\( 22^2 + 51^2 > 54^2 \)[/tex].
