Respuesta :
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.
Answer:
B, A, C
Step-by-step explanation:
We first find the length of each side using the distance formula. For AB,
[tex]d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\\\=\sqrt{(3-6)^2+(2-6)^2}\\\\=\sqrt{(-3)^2+(-4)^2}\\\\=\sqrt{9+16}=\sqrt{25}=5[/tex]
For BC,
[tex]d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\\\=\sqrt{(6-0)^2+(6-14)^2}\\\\=\sqrt{6^2+(-8)^2}\\\\=\sqrt{36+64}=\sqrt{100}=10[/tex]
For CA,
[tex]d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\\\=\sqrt{(0-3)^2+(14-2)^2}\\\\=\sqrt{(-3)^2+12^2}\\\\=\sqrt{9+144}=\sqrt{153}\approx 12.4[/tex]
Angle A will be across from BC. Angle B will be across from CA. Angle C will be across from AB.
The side lengths, in order from largest to smallest, are CA, BC and AB. This makes the angles, in order from largest to smallest, B, A, C.
