The vertices of a triangle are A(7, 5), B(4, 2), and C(9, 2)
First we need to find the length of each side of the triangle using Distance formula, d = [tex] \sqrt{(x1 - x2)^2 + (y1 - y2)^2}
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Length of side AB, c= [tex] \sqrt{(7-4)^2 + (5-2)^2} [/tex]
= [tex] \sqrt{(3)^2 + (3)^2} [/tex]
= [tex] \sqrt{9+9} [/tex]
= [tex] \sqrt{18} [/tex]
Length of side BC, a= [tex] \sqrt{(4-9)^2 +(2-2)^2} [/tex]
= [tex] \sqrt{(-5)^2 + 0} [/tex]
= [tex] \sqrt{25} [/tex]
= 5
Length of the side AC, b = [tex] \sqrt{(7-9)^2 +(5-2)^2} [/tex]
= [tex] \sqrt{(-2)^2 + (3)^2} [/tex]
= [tex] \sqrt{4+ 9} = \sqrt{13} [/tex]
Cos B = [tex] \frac{a^2 + c^2 -b^2}{2ac} [/tex]
Cos B = [tex] \frac{(5)^2 + (\sqrt{18})^2 - (\sqrt{13})^2}{2*5*\sqrt{18}} [/tex]
Cos B = [tex] \frac{25+18-13}{10\sqrt{18}} [/tex]
Cos B= [tex] \frac{30}{10\sqrt{18}} [/tex]
Cos B = [tex] \frac{3}{\sqrt{18}} = \frac{3}{3\sqrt{2}} = \frac{1}{\sqrt{2}} [/tex]
By taking Inverse Cosine function,
B= Cos⁻¹ ([tex] \frac{1}{\sqrt{2}} [/tex])
B= 45°
The measure of angle ABC is 45°
Answer:
The measure of the angle is 45°
Step-by-step explanation:
Correct on Plato
