Answer:
Step-by-step explanation:
Vertices of ΔABC are,
A(-3, 6), B(2, 1) and C(9, 5)
Use the formula to get the distance between two points [tex](x_1,y_1)[/tex] and[tex](x_2,y_2)[/tex],
Distance = [tex]\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]
By using the formula,
AB = [tex]\sqrt{(1-6)^2+(2+3)^2}[/tex]
= [tex]\sqrt{50}[/tex] units
BC = [tex]\sqrt{(5-1)^2+(9-2)^2}[/tex]
= [tex]\sqrt{65}[/tex] units
AC = [tex]\sqrt{(6-5)^2+(-3-9)^2}[/tex]
= [tex]\sqrt{145}[/tex]
Use cosine rule to find the measure of ∠ABC.
AC² = AB² + BC²- 2(AB)(BC)cos(B)
[tex](\sqrt{145})^2=(\sqrt{50})^2+(\sqrt{65})^2-2(\sqrt{50})(\sqrt{65})\text{cosB}[/tex]
145 = 50 + 65 - 2(√3250)cosB
cos(B) = [tex]-(\frac{145-115}{2\sqrt{3250}})[/tex]
= -0.26312
B = [tex]\text{cos}^{-1}(-0.26312)[/tex]
B = 105.26°
