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) ![=\sqrt{(4-9)^2+(2-2)^2}= \sqrt{25}= 5](https://tex.z-dn.net/?f=%3D%5Csqrt%7B%284-9%29%5E2%2B%282-2%29%5E2%7D%3D%20%5Csqrt%7B25%7D%3D%205)
Length of the side AC (b) ![= \sqrt{(7-9)^2 +(5-2)^2}= \sqrt{4+9}=\sqrt{13}](https://tex.z-dn.net/?f=%3D%20%5Csqrt%7B%287-9%29%5E2%20%2B%285-2%29%5E2%7D%3D%20%5Csqrt%7B4%2B9%7D%3D%5Csqrt%7B13%7D)
Length of the side AB (c) ![= \sqrt{(7-4)^2 +(5-2)^2} =\sqrt{9+9}=\sqrt{18}](https://tex.z-dn.net/?f=%3D%20%5Csqrt%7B%287-4%29%5E2%20%2B%285-2%29%5E2%7D%20%3D%5Csqrt%7B9%2B9%7D%3D%5Csqrt%7B18%7D)
We need to find ∠ABC or ∠B . So using <u>Cosine rule</u>, we will get...
![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](https://tex.z-dn.net/?f=cosB%3D%20%5Cfrac%7Ba%5E2%2Bc%5E2-b%5E2%7D%7B2ac%7D%20%5C%5C%20%5C%5C%20cos%20B%3D%20%5Cfrac%7B%285%29%5E2%2B%28%5Csqrt%7B18%7D%29%5E2-%28%5Csqrt%7B13%7D%29%5E2%7D%7B2%2A5%2A%5Csqrt%7B18%7D%20%7D%5C%5C%20%5C%5C%20cosB%3D%20%5Cfrac%7B25%2B18-13%7D%7B10%5Csqrt%7B18%7D%7D%20%3D%5Cfrac%7B30%7D%7B10%5Csqrt%7B18%7D%7D%3D%5Cfrac%7B3%7D%7B%5Csqrt%7B18%7D%7D%5C%5C%20%5C%5C%20cosB%3D%5Cfrac%7B3%7D%7B3%5Csqrt%7B2%7D%7D%20%3D%5Cfrac%7B1%7D%7B%5Csqrt%7B2%7D%7D%5C%5C%20%5C%5C%20B%3D%20cos%5E-%5E1%28%5Cfrac%7B1%7D%7B%5Csqrt%7B2%7D%7D%29%3D%2045%20degree)
So, the measure of angle ABC is 45°