Triangle abc has vertices a(3,2) b(6,6) and c(0,14). list the angles in order from largest to smallest. brainly

Accepted Solution

b>a>c. We first calculate the side lengths of the triangle, using coordinates and the Pythagorean Theorem. Distance ab=\sqrt((6-3)^2,(6-2)^2)=5. ac=\sqrt((3-0)^2,(14-2)^2)=\sqrt(153), bc=\sqrt((6-0)^2,(14-6)^2)=10. Since bc^2+ab^2>ac^2, we know that b is an obtuse angle, and by similar calculation, a and c are acute angles, so b is the largest. Then to compare a and c, we use the law of sines, sin(a)/bc=sin(c)/ab. Since bc(10) is longer than ab(5), sin(a) is larger than sin(c), then a is larger than c. So b>a>c.