To find the measure of [tex]\(\angle ABC\)[/tex] in the triangle with vertices [tex]\(A(7,5)\)[/tex], [tex]\(B(4,2)\)[/tex], and [tex]\(C(9,2)\)[/tex], we need to follow these steps:
1. Determine the lengths of the sides of the triangle using the distance formula:
[tex]\[
AB = \sqrt{(B_x - A_x)^2 + (B_y - A_y)^2}
\][/tex]
[tex]\[
BC = \sqrt{(C_x - B_x)^2 + (C_y - B_y)^2}
\][/tex]
[tex]\[
AC = \sqrt{(C_x - A_x)^2 + (C_y - A_y)^2}
\][/tex]
Using the vertices given:
[tex]\[
AB = \sqrt{(4 - 7)^2 + (2 - 5)^2} = \sqrt{(-3)^2 + (-3)^2} = \sqrt{9 + 9} = \sqrt{18} \approx 4.242640687119285
\][/tex]
[tex]\[
BC = \sqrt{(9 - 4)^2 + (2 - 2)^2} = \sqrt{5^2 + 0^2} = \sqrt{25} = 5
\][/tex]
[tex]\[
AC = \sqrt{(9 - 7)^2 + (2 - 5)^2} = \sqrt{2^2 + (-3)^2} = \sqrt{4 + 9} = \sqrt{13} \approx 3.605551275463989
\][/tex]
2. Apply the Law of Cosines to find [tex]\(\cos \angle ABC\)[/tex]:
[tex]\[
\cos(\angle ABC) = \frac{AB^2 + BC^2 - AC^2}{2 \times AB \times BC}
\][/tex]
Substituting the lengths we have calculated:
[tex]\[
\cos(\angle ABC) = \frac{(4.242640687119285)^2 + (5)^2 - (3.605551275463989)^2}{2 \times 4.242640687119285 \times 5}
\][/tex]
[tex]\[
\cos(\angle ABC) = \frac{18 + 25 - 13}{2 \times 4.242640687119285 \times 5} = \frac{30}{42.42640687119285} \approx 0.7071067811865476
\][/tex]
3. Find the angle [tex]\( \angle ABC \)[/tex] using the arccosine function:
[tex]\[
\angle ABC = \arccos(0.7071067811865476)
\][/tex]
[tex]\[
\angle ABC \approx 45.0^{\circ}
\][/tex]
So, the measure of [tex]\(\angle ABC\)[/tex] is closest to [tex]\(45^{\circ}\)[/tex].
Therefore, the correct answer is [tex]\(B\)[/tex].
