Please help fast I got to turn it in

A and E can be thought of as triangles as well with angles 0°, 0° and 180°, but it will be just a line with 3 points. So it depends whether this is a question with multiple answers or not.

Step-by-step explanation:

Thinking Process:

We need to find a formula for a triangle that uses all the sides

Cosine rule uses all sides:

$c^2 = a^2 + b^2 -2ab(cosA)$

we can rearrage this formula and make the angle $A$ the subject:

$A = cos^{-1}(\frac{a^2+b^2-c^2}{2ab})$

now you can just put the values from each option in the question into this formula and see if it gives an answer or not.

Try A. (1,2,3)

put a = 1, b = 2, c = 3 (keep in mind that you can take 'a' 'b' and 'c' as 3,2 and 1 or any other arrangement within the option A(1,2,3). All we are doing here is checking whether the answer exists or not)

$A = cos^{-1}(\frac{1^2+2^2-3^2}{2(1)(2)})$

$A = 180$ , now graphically it seems that this is just a line with 3 points, or technically you can think of as the sides of the triangle are overlapping each other, and the angle between the sides is 0 0 and 180. It depends on whether the question is asking you to be technical or not. Since the question already stated there are 3 line segments, so we can think of it as a triangle.

Try B.(3,4,5)

$A = cos^{-1}(\frac{3^2+4^2-5^2}{2(3)(4)})$

$A = 90$ , this is a triangle (specifically, this is a right angled triangle)

Try C.(1,10,100)

$A = cos^{-1}(\frac{1^2+10^2-100^2}{2(1)(10)})$

from the calculator you'll see a 'math error', this shows that there is no answer, because $cos^{-1}$ can only accept values within -1 and 1 inclusive. Hence no such triangle exists with lengths (1,10,100).