Answer:
1)
[tex] \sin(60) = \frac{9 \sqrt{3} }{y} [/tex]
[tex]y = \frac{9 \sqrt{3} }{ \sin(60) } [/tex]
[tex]y = - 51.14[/tex]
[tex] \tan(60) = \frac{9 \sqrt{3} }{x} [/tex]
[tex]x = \frac{9 \sqrt{3} }{ \tan(60) } [/tex]
[tex]x = 48.71[/tex]
[tex]x \div y = 48.71 \div ( - 51.14)[/tex]
[tex]x \div y = - 0.95[/tex]
2) let y = the line connecting both triangles
[tex] \sin(30) = \frac{y}{4 \sqrt{3} } [/tex]
[tex]y = \sin(30) \times 4 \sqrt{3} [/tex]
[tex]y = - 6.85[/tex]
[tex] \sin(45) = \frac{ - 6.85}{x} [/tex]
[tex]x = \frac{ - 6.85}{ \sin(45) } [/tex]
[tex]x = - 8.05[/tex]
