1. The problem involves analyzing the function $$f(x) = \frac{a}{x+r} + t$$ with given information about its symmetry axis $$h(x) = -x + 3$$, points $$C=(0,3)$$ on the axis, and point $$B=(2,1)$$ on the vertical asymptote line $$x=-r$$. 2. The vertical asymptote is at $$x = -r$$, and since $$B = (2,1)$$ lies on this asymptote, we have $$-r = 2$$, so $$r = -2$$. 3. The horizontal asymptote of $$f(x)$$ is $$y = t$$. Since $$h(x) = -x + 3$$ is the symmetry line and the graph is symmetric about this line, the horizontal asymptote corresponds to the horizontal component of the symmetry line. 4. To find $$t$$, consider the point $$C = (0, 3)$$ lies on the symmetry line and on the graph's axis. From symmetry, the horizontal asymptote $$y = t$$ must satisfy the condition of symmetry along $$h(x)$$. 5. Using the symmetry property about line $$y = -x + 3$$, the center of symmetry lies on this line, meaning the horizontal asymptote $$y = t$$ is related to the constant term in the symmetry axis. 6. From the standard rational function form $$f(x) = \frac{a}{x+r} + t$$ with vertical asymptote at $$x=-r=2$$ and horizontal asymptote $$y=t$$, and given $$B=(2,1)$$ lies on the vertical asymptote, the function tends to infinity at $$x=2$$. 7. For the second question, given $$\sin \theta = 1$$ for $$0^\circ \leq \theta \leq 90^\circ$$, the solution is $$\theta = 90^\circ$$. \textbf{Final answers:} - Vertical asymptote: $$x=2$$ (hence $$r=-2$$) - Horizontal asymptote: $$y = t$$ (value depends on further info but cannot be determined exactly with given data) - Angle $$\theta = 90^\circ$$ (Note: The problem does not provide enough data to determine $$a$$ or $$t$$ explicitly.)