Question

The vertices of a triangle are A(7,5) B(4,2) and C(9,2) what is mABC

Answer 1

Answer:

A(7,5)

Step-by-step explanation:

This is correct because the area of a triangle is b × h = a.

Answer 2

Answer:

The measure of ∠ABC  is 45°.

Step-by-step explanation:

Given : The vertices of a triangle are A(7,5) B(4,2) and C(9,2).

To find : What is ∠ABC ?

Solution :

First we side the length of the sides,

Using Distance formula,

$d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}$

Length of side AB, A(7,5) and B(4,2)

$c= \sqrt{(7-4)^2 + (5-2)^2} \\c= \sqrt{(3)^2 + (3)^2} \\c= \sqrt{9+9} \\c= \sqrt{18}$

Length of side BC, B(4,2) and C(9,2)

$a= \sqrt{(4-9)^2 +(2-2)^2} \\a= \sqrt{(-5)^2 + 0} \\a= \sqrt{25} \\a= 5$

Length of the side AC, A(7,5) and C(9,2)

$b = \sqrt{(7-9)^2 +(5-2)^2}\\ b= \sqrt{(-2)^2 + (3)^2} \\b= \sqrt{4+ 9}\\b=\sqrt{13}$

By the Law of Cosines,

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

Substitute the values,

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

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

$\cos B=\frac{30}{10\sqrt{18}}$  

$\cos B =\frac{3}{\sqrt{18}}$  

$\cos B= \frac{3}{3\sqrt{2}}$  

$\cos B= \frac{1}{\sqrt{2}}$  

Taking Inverse Cosine function,

$B= \cos^{-1}( \frac{1}{\sqrt{2}})$  

$B=45^\circ$  

Therefore, The measure of ∠ABC  is 45°.