The measure of angle ABC is 45°
<em><u>Explanation</u></em>
Vertices of the triangle are: A(7, 5), B(4, 2), and C(9, 2)
According to the diagram below....
Length of the side BC (a) 
Length of the side AC (b) 
Length of the side AB (c) 
We need to find ∠ABC or ∠B . So using <u>Cosine rule</u>, we will get...

So, the measure of angle ABC is 45°