Question

The vertices of AABC are A(-1,-2), B(-1,2) and

C(6,0). Which conclusion can be made about the
angles of AABC?
1) m/A=m/B 2) m/4= m/C
3) m/ACB = 90 4) m/ABC=60

Answer 1

The angles of the AABC triangle cannot be determined based on the given information.

Step-by-step explanation:

The given triangle has vertices A(-1,-2), B(-1,2), and C(6,0). Using the distance formula, we can find the lengths of each side:

AB = sqrt((-1-(-1))^2 + (2-(-2))^2) = sqrt(0 + 16) = 4

AC = sqrt((-1-6)^2 + (-2-0)^2) = sqrt((-7)^2 + (-2)^2) = sqrt(53)

BC = sqrt((-1-6)^2 + (2-0)^2) = sqrt((-7)^2 + 4^2) = sqrt(53)

The angles of a triangle can be determined using the Law of Cosines: cos(A) = (b^2 + c^2 - a^2) / (2bc). Let's find the angles:

angle A = cos^-1((4^2 + sqrt(53)^2 - sqrt(53)^2) / (2 * 4 * sqrt(53)))

angle B = cos^-1((sqrt(53)^2 + sqrt(53)^2 - 4^2) / (2 * sqrt(53) * sqrt(53)))

angle C = cos^-1((sqrt(53)^2 + 4^2 - sqrt(53)^2) / (2 * sqrt(53) * 4))

The calculated angles of the triangle are not equal, so none of the given conclusions can be made about the angles of AABC.

Learn more about Triangle angles