Question

Can someone please help me? ​

Answer 1

Answer:

1)

$\sin(60) = \frac{9 \sqrt{3} }{y}$

$y = \frac{9 \sqrt{3} }{ \sin(60) }$

$y = - 51.14$

$\tan(60) = \frac{9 \sqrt{3} }{x}$

$x = \frac{9 \sqrt{3} }{ \tan(60) }$

$x = 48.71$

$x \div y = 48.71 \div ( - 51.14)$

$x \div y = - 0.95$

2) let y = the line connecting both triangles

$\sin(30) = \frac{y}{4 \sqrt{3} }$

$y = \sin(30) \times 4 \sqrt{3}$

$y = - 6.85$

$\sin(45) = \frac{ - 6.85}{x}$

$x = \frac{ - 6.85}{ \sin(45) }$

$x = - 8.05$