Question

Not sure f either of my answers would satisfy the question in the second picture

Answer 1

$\bf \begin{array}{llll} tan(\alpha)=-\cfrac{80}{15}\qquad II\\\\ x=-15\\ y=8\\ r=17 \end{array} \qquad \qquad \begin{array}{llll} cos(\beta)=\cfrac{5}{6}\qquad I\\\\ x=5\\ y=\sqrt{11}\\ r=6 \end{array}\\\\ -----------------------------\\\\ sin({{ \alpha}} + {{ \beta}})=sin({{ \alpha}})cos({{ \beta}}) + cos({{ \alpha}})sin({{ \beta}}) \\\\\\ sin({{ \alpha}} + {{ \beta}})=\cfrac{8}{17}\cdot \cfrac{5}{6}+\cfrac{-15}{17}\cdot \cfrac{\sqrt{11}}{6}\implies \cfrac{40-15\sqrt{11}}{102}$



$\bf cos({{ \alpha}} + {{ \beta}})= cos({{ \alpha}})cos({{ \beta}})- sin({{ \alpha}})sin({{ \beta}}) \\\\\\ \cfrac{-15}{17}\cdot \cfrac{5}{6}-\cfrac{8}{17}\cdot \cfrac{\sqrt{11}}{6}\implies \cfrac{-75-8\sqrt{11}}{102}$





$\bf tan({{ \alpha}} + {{ \beta}}) = \cfrac{tan({{ \alpha}})+ tan({{ \beta}})}{1- tan({{ \alpha}})tan({{ \beta}})} \\\\\\ tan({{ \alpha}} + {{ \beta}}) = \cfrac{-\frac{8}{15}+\frac{\sqrt{11}}{5}}{1-\left( -\frac{8}{15}\cdot \frac{\sqrt{11}}{5} \right)}\implies \cfrac{\frac{-8+3\sqrt{11}}{15}}{1+\frac{8\sqrt{11}}{75}} \\\\\\ tan({{ \alpha}} + {{ \beta}}) =\cfrac{\frac{-8+3\sqrt{11}}{15}}{\frac{75+8\sqrt{11}}{75}}\implies \cfrac{-8+3\sqrt{11}}{15}\cdot \cfrac{75}{75+8\sqrt{11}}$

$\bf tan({{ \alpha}} + {{ \beta}}) =\cfrac{-8+3\sqrt{11}}{1}\cdot \cfrac{3}{75+8\sqrt{11}}\implies \cfrac{15\sqrt{11}-40}{75+8\sqrt{11}}$