Question

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)?

Answer 1

Please see the explanation.

Explanation:

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

`A = (0, 0), B = (4, 3), and C = (4, -2)`

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 = (3 - -2)`
`a = 5`

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

`b = sqrt(4^2 + (-2)^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`

Use the Law of Cosines:

`b^2 = a^2 + c^2 - 2(a)(c)cos(angleB)`

`20 = 25 + 25 - 50cos(angleB)`

`20/50 = 1 - cos(angleB)`

`cos(angle B) = 3/5`

`angle B = cos^-1(3/5) ~~ 53^@`

`angle A = angle C = (180^@ - 53^@)/2 = 63.5^@`