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)?

Question

17546 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-11-23T01:25:43

Please see the 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 $angle A = angle C$

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

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

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

$cos(angle B) = 3/5$

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

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