The measure of angle ABC is 45°
Explanation
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) [tex]=\sqrt{(4-9)^2+(2-2)^2}= \sqrt{25}= 5[/tex]
Length of the side AC (b) [tex]= \sqrt{(7-9)^2 +(5-2)^2}= \sqrt{4+9}=\sqrt{13}[/tex]
Length of the side AB (c) [tex]= \sqrt{(7-4)^2 +(5-2)^2} =\sqrt{9+9}=\sqrt{18}[/tex]
We need to find ∠ABC or ∠B . So using Cosine rule, we will get...
[tex]cosB= \frac{a^2+c^2-b^2}{2ac} \\ \\ cos B= \frac{(5)^2+(\sqrt{18})^2-(\sqrt{13})^2}{2*5*\sqrt{18} }\\ \\ cosB= \frac{25+18-13}{10\sqrt{18}} =\frac{30}{10\sqrt{18}}=\frac{3}{\sqrt{18}}\\ \\ cosB=\frac{3}{3\sqrt{2}} =\frac{1}{\sqrt{2}}\\ \\ B= cos^-^1(\frac{1}{\sqrt{2}})= 45 degree[/tex]
So, the measure of angle ABC is 45°
