Answer:
A(7,5)
Step-by-step explanation:
This is correct because the area of a triangle is b × h = a.
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,
[tex]d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}[/tex]
Length of side AB, A(7,5) and B(4,2)
[tex]c= \sqrt{(7-4)^2 + (5-2)^2} \\c= \sqrt{(3)^2 + (3)^2} \\c= \sqrt{9+9} \\c= \sqrt{18}[/tex]
Length of side BC, B(4,2) and C(9,2)
[tex]a= \sqrt{(4-9)^2 +(2-2)^2} \\a= \sqrt{(-5)^2 + 0} \\a= \sqrt{25} \\a= 5[/tex]
Length of the side AC, A(7,5) and C(9,2)
[tex]b = \sqrt{(7-9)^2 +(5-2)^2}\\ b= \sqrt{(-2)^2 + (3)^2} \\b= \sqrt{4+ 9}\\b=\sqrt{13}[/tex]
By the Law of Cosines,
[tex]\cos B=\frac{a^2 + c^2 -b^2}{2ac}[/tex]
Substitute the values,
[tex]\cos B=\frac{(5)^2 + (\sqrt{18})^2 - (\sqrt{13})^2}{2\times 5\times \sqrt{18}}[/tex]
[tex]\cos B =\frac{25+18-13}{10\sqrt{18}}[/tex]
[tex]\cos B=\frac{30}{10\sqrt{18}}[/tex]
[tex]\cos B =\frac{3}{\sqrt{18}}[/tex]
[tex]\cos B= \frac{3}{3\sqrt{2}}[/tex]
[tex]\cos B= \frac{1}{\sqrt{2}}[/tex]
Taking Inverse Cosine function,
[tex]B= \cos^{-1}( \frac{1}{\sqrt{2}})[/tex]
[tex]B=45^\circ[/tex]
Therefore, The measure of ∠ABC is 45°.
