Question

The vertices of a triangle are A(7, 5), B(4, 2), and C(9, 2). What is m`/_ABC`?

Answer 1

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 = $\sqrt{(x1 - x2)^2 + (y1 - y2)^2}$

Length of side AB, c= $\sqrt{(7-4)^2 + (5-2)^2}$

= $\sqrt{(3)^2 + (3)^2}$

= $\sqrt{9+9}$

= $\sqrt{18}$

Length of side BC, a= $\sqrt{(4-9)^2 +(2-2)^2}$

= $\sqrt{(-5)^2 + 0}$

= $\sqrt{25}$

= 5

Length of the side AC, b = $\sqrt{(7-9)^2 +(5-2)^2}$

= $\sqrt{(-2)^2 + (3)^2}$

= $\sqrt{4+ 9} = \sqrt{13}$

Cos B = $\frac{a^2 + c^2 -b^2}{2ac}$

Cos B = $\frac{(5)^2 + (\sqrt{18})^2 - (\sqrt{13})^2}{2*5*\sqrt{18}}$

Cos B = $\frac{25+18-13}{10\sqrt{18}}$

Cos B= $\frac{30}{10\sqrt{18}}$

Cos B = $\frac{3}{\sqrt{18}} = \frac{3}{3\sqrt{2}} = \frac{1}{\sqrt{2}}$

By taking Inverse Cosine function,

B= Cos⁻¹ ($\frac{1}{\sqrt{2}}$)

B= 45°

The measure of angle ABC is 45°