# How do you find the measures of the angles of the triangle whose vertices are A = (-1,0), B = (3,3) and C = (3, -2)?

Nov 23, 2016

#### Explanation:

It will not change anything with regard to side lengths or angles, if we add 1 to all of the x coordinates:

$A = \left(0 , 0\right) , B = \left(4 , 3\right) , \mathmr{and} C = \left(4 , - 2\right)$

The length of side "a" (from C to B), is easy, because B and C have the same x coordinate, therefore, you just use the y coordinate difference:

$a = \left(3 - - 2\right)$
$a = 5$

The length of side "b" (from A to C) is made easier by A being the origin:

$b = \sqrt{{4}^{2} + {\left(- 2\right)}^{2}} = \sqrt{20}$

The same is true for the length of side "c" (from A to B):

$c = \sqrt{{4}^{2} + {3}^{2}} = 5$

It is an isosceles triangle so we need to find $\angle B$ and we know that $\angle A = \angle C$

${b}^{2} = {a}^{2} + {c}^{2} - 2 \left(a\right) \left(c\right) \cos \left(\angle B\right)$

$20 = 25 + 25 - 50 \cos \left(\angle B\right)$

$\frac{20}{50} = 1 - \cos \left(\angle B\right)$

$\cos \left(\angle B\right) = \frac{3}{5}$

$\angle B = {\cos}^{-} 1 \left(\frac{3}{5}\right) \approx {53}^{\circ}$

$\angle A = \angle C = \frac{{180}^{\circ} - {53}^{\circ}}{2} = {63.5}^{\circ}$