a)
she drew a Δy = 6 with a slope of 11°, well, based on that angle/ratio table above, 11° corresponds to a line with a slope ratio of 1/5, meaning
[tex]\bf slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{\Delta y}{\Delta x}\implies \cfrac{6}{x}=\stackrel{11^o~ratio}{\cfrac{1}{5}}\implies 30=\Delta x[/tex]
b)
she drew a Δx = 40 on a line with slope of 72°, therefore
[tex]\bf slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{\Delta y}{\Delta x}\implies \cfrac{\Delta y}{40}=\stackrel{72^o~ratio}{\cfrac{3}{1}}\implies \Delta y = 120[/tex]
c)
Δx = 23 and Δy = 23, therfore then
[tex]\bf slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{\Delta y}{\Delta x}\implies \cfrac{23}{23}\implies \stackrel{\textit{slope ratio}}{\cfrac{1}{1}}\implies \stackrel{\textit{slope angle}}{45^o}[/tex]
