Respuesta :
Answer:
The length of side AB is [tex]\sqrt{50}[/tex] units.
The length of side BC is [tex]\sqrt{65}[/tex] units.
The length of side AC is [tex]\sqrt{145}[/tex] units.
[tex]\angle ABC\approx 105^\circ[/tex].
Step-by-step explanation:
Given : ∆ABC has A(-3, 6), B(2, 1), and C(9, 5) as its vertices.
To find : The length of side AB is units. The length of side BC is units. The length of side AC is units. ∠ABC ≈ ° ?
Solution :
The distance formula between two point is given by,
[tex]d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]
The distance between A(-3, 6) and B(2, 1) is
[tex]c=\sqrt{(2--3)^2+(1-6)^2} \\c=\sqrt{(2+3)^2+(-5)^2} \\c=\sqrt{25+25}\\c=\sqrt{50}[/tex]
The distance between B(2, 1), and C(9, 5) is
[tex]a=\sqrt{(9-2)^2+(5-1)^2} \\a=\sqrt{(-7)^2+(4)^2} \\a=\sqrt{49+16}\\a=\sqrt{65}[/tex]
The distance between C(9, 5) and A(-3, 6) is
[tex]b=\sqrt{(9--3)^2+(5-6)^2} \\b=\sqrt{(12)^2+(-1)^2} \\b=\sqrt{144+1}\\b=\sqrt{145}[/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{(\sqrt{65})^2 + (\sqrt{50})^2 - (\sqrt{145})^2}{2\times \sqrt{65}\times \sqrt{50}}[/tex]
[tex]\cos B=\frac{65+50-145}{2\times \sqrt{65}\times \sqrt{50}}[/tex]
[tex]\cos B=\frac{-30}{2\times \sqrt{65}\times \sqrt{50}}[/tex]
[tex]\cos B=-0.26[/tex]
[tex]B=\cos^{-1}(-0.26)[/tex]
[tex]B=105.07^\circ[/tex]
Therefore, [tex]\angle ABC\approx 105^\circ[/tex].
