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=√(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{√(11)}{6}\implies \cfrac{40-15√(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{√(11)}{6}\implies \cfrac{-75-8√(11)}{102}`






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


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