△ABC has vertices A(−1, 6), B(2, 10), and C(7, −2) . Find the measure of each angle of the triangle. Round decimal answers to the nearest tenth. [HELP THIS IS ON MY QUIZ AND IVE BEEN ON IT FOR 45 MINUTES]

<A = 90°, <B = 71.6° <C = 18.4°

Step-by-step explanation:

Given the vertices of a △ABC to be A(−1, 6), B(2, 10), and C(7, −2)

Before we can get each angle of the triangle, we need to get its sides first.

Using the formula fie finding the distance between two points

D = √(x2-x1)²+(y2-y1)²

For side AB:

Given A(−1, 6), B(2, 10)

AB = √{2-(-1)}²+(10-6)²

AB =√3²+4²

AB = √25

AB = 5

For side AC:

Given A(-1,6) and C(7, −2)

AC = √{7-(-1)}²+(-2-6)²

AC = √8²+8²

AC = √128

AC = √64×2

AC = 8√2

For side BC:

Given B(2,10) and C(7, −2)

BC = √(7-2)²+(-2-10)²

BC = √5²+12²

BC = √25+144

BC = √169

BC = 13

First we need to get theta using cosine rule

|AC|² = |AB|+|BC| - 2|AB||AC| cos theta

(8√2)² = 5²+13²-2(5)(13)cos theta

128 = 169-130costheta

128-169 = -130costheta

-41= -130costheta

Costheta = -41/-130

Cos theta = 0.315

theta = arccos 0.315

theta = 71.64°

<B = 71.6°

Using sine rule to get <A,

BC/sin A = AC/sinB

13/sinA = 8√2/sin71.6°

8√2sinA = 13sin71.6°

SinA = 12.34/8√2

SinA = 1.09°

A = arcsin1

<A = 90.0°

<C = 180°-(71.6+90)

< C = 180-161.6

<C = 18.4°