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)?

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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:

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

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:

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

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

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

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

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

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

$b^{2} = a^{2} + c^{2} - 2 (a) (c) cos (∠ B)$ $b^2 = a^2 + c^2 - 2(a)(c)cos(angleB)$

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

$\frac{20}{50} = 1 - cos (∠ B)$ $20/50 = 1 - cos(angleB)$

$cos (∠ B) = \frac{3}{5}$ $cos(angle B) = 3/5$

$∠ B = {cos}^{- 1} (\frac{3}{5}) \approx 53^{\circ}$ $angle B = cos^-1(3/5) ~~ 53^@$

$∠ A = ∠ C = \frac{180^{\circ} - 53^{\circ}}{2} = {63.5}^{\circ}$ $angle A = angle C = (180^@ - 53^@)/2 = 63.5^@$