When calculating the pieces in a triangle with only two sides and an intermediate angle is known, one must solve a quadratic equation. By solving the equation are 2, 1 or 0 solutions, as described below (A-D). These four options is illustrated in Figure 1-4 below. You must write at the bottom of the options (A-D) belongs to the illustrations (1-4)
A: there should be 2 solutions (both positive)
B: there must be 2 solutions (one positive and one negative)
C: there must be a solution
D: must be 0 solutions
Problem is that I have no idea how to start or how to find solutions. I hope anyone who wants to 'make one so I can follow them and do the rest.
If you can trust the pictures, then no calculation is necessary. The first picture, for example, depicts a "side-side-angle" setup in which no choice of the angle $C$ will yield a valid triangle. If you want to verify this numerically, then probably you should check that one of the relations that holds in triangles fails.
Here law of cosines might help: if there were such a triangle and hence a valid choice for $c$, the length of the side opposite $C$, then we'd have \begin{align*} 5^2 &= 13^2 + c^2 - 2 \cdot 13c\,\cos29^\circ \\ 0 &= c^2 - (26\,\cos29^\circ)c + 144 \end{align*} Look at the discriminant: $(26\,\cos29^\circ)^2 - 4 \cdot 144$. Is it negative?