Find the angles of a triangle whose vertices are 0 0 5 2 1 4

Find the angles of a triangle whose vertices are (0, 0), (5, -2), (1, -4).

14) We have given triangles vertics (0,0),(5,-2),(1,-4)

Let A=(0,0),B=(5,-2) and C=(1,-4)

the distance between AB=c=sqrt(25+4)=sqrt(29)

the distance between BC=a=sqrt((1-5)^2+(-4+2)^2)=sqrt(16+4)=sqrt(20)

the distance between CA =b=sqrt(1+16)=sqrt(17)

by using Law of cosines

Let angles of triangle are <A,<B and <C

substitute a=sqrt(20),b=sqrt(17) and c=sqrt(29)

cosA=[b^2+c^2-a^2]/2bc

=[17+29-20]/(2*sqrt(17)*sqrt(29)

A=arccos(26/(2*sqrt(17)*sqrt(29)))

A=54.162347 degrees

cosB=[c^2+a^2-b^2]/2ca

cosB=[29+20-17]/(2*sqrt(29)*sqrt(20))

cosB=32//(2*sqrt(29)*sqrt(20))

B=arccos(32//(2*sqrt(29)*sqrt(20)))=48.3664607 degrees

<C=180-(<A+<B)=180-(54.162347+48.3664607)

C=77.4711923 degrees

the given triangle\'s angles are <A=54.162347⁰,<B=48.3664607⁰ and <C=77.4711923⁰