HELP ASAP! GIVING BRAINLIEST!!∆ABC has A(-3, 6), B(2, 1), and C(9, 5) as its vertices. The length of side AB is A) (50)^1/2B) (65)^1/2C) (105)^1/2D) (145)^1/2units. The length of side BC isA) (50)^1/2B) (65)^1/2C) (105)^1/2D) (145)^1/2units. The length of side AC isA) (50)^1/2B) (65)^1/2C) (105)^1/2D) (145)^1/2units.∠ABC ≈ ° A) 55.21B) 85.16C) 105.26D) 114.11

Answer:1. Option A2. Option B3. Option D4. Option CStep-by-step explanation:The given vertices of triangle ABC are A(-3, 6), B(2,1), C(9, 5).We have to fine the distance AB, BC and AC.To calculate the distance between two vertices we will use the formula [tex]d=\sqrt{(x-x')^{2}+(y-y')^{2}}[/tex]For the length of side ABAB=[tex]\sqrt{(2+3)^{2}+(1-6)^{2}}=\sqrt{5^{2}+5^{2}}=\sqrt{50}[/tex]Option A. is the correct optionFor the length of side BC[tex]BC=\sqrt{(9-2)^{2}+(5-1)^{2}}=\sqrt{7^{2}+4^{2}}=\sqrt{65}[/tex]Option B is the answer.For the length of side AC[tex]AC=\sqrt{(9+3)^{2}+(5-6)^{2}}=\sqrt{12^{2}+(-1)^{2}}=\sqrt{145}[/tex]Option D is the answer.For ∠ABC we will use the formula [tex]tan\theta =\frac{m_{1}-m_{2}}{1+m_{1}m_{2}}[/tex]Since angle ABC is formed by two sides AB and BCSo we will find the slopes of these two lines and find the angleNow slope of AB, [tex]m_{1}=\frac{y-y'}{x-x'}=\frac{1-6}{2+3}=\frac{-5}{5}=-1[/tex]Slope of BC, [tex]m_{2}=\frac{5-1}{9-2}=\frac{4}{7}[/tex][tex]tan\theta =\frac{m_{1}-m_{2}}{1+m_{1}m_{2}}[/tex][tex]tan\theta =\frac{(-1)-(\frac{4}{7})}{1+(-1)(\frac{4}{7})}=\frac{\frac{-11}{7}}{1-\frac{4}{7}}=\frac{\frac{-11}{7}}{\frac{3}{7}}=\frac{-11}{7}\times \frac{7}{3}=-\frac{11}{3}=-3.67[/tex][tex]\theta =tan^{-1}(-3.67)=74.75[/tex]Since angle between them [tex]tan\theta[/tex] is negative that means angle theta will be obtuse angle.So the angle between AB and BC = (180 - 74.75) = 105.26°Therefore Option C. 105.26° is the answer.