Answer:
Part 1) [tex]x=8\sqrt{2}\ units[/tex]
Part 2) [tex]y=4\sqrt{6}\ units[/tex]
Step-by-step explanation:
see the attached figure with letters to better understand the problem
In the right triangle ABD
Find the length side BD
[tex]sin(45\°)=BD/AB[/tex]
[tex]BD=(AB)sin(45\°)[/tex]
we have
[tex]AB=8\ units[/tex]
[tex]sin(45\°)=\frac{\sqrt{2}}{2}[/tex]
substitute the given values
[tex]BD=(8)\frac{\sqrt{2}}{2}[/tex]
[tex]BD=4\sqrt{2}\ units[/tex]
In the right triangle DBC
Find the length side BC
[tex]sin(30\°)=BD/BC[/tex]
[tex]BC=BD/sin(30\°)[/tex]
we have
[tex]sin(30\°)=\frac{1}{2}[/tex]
[tex]BD=4\sqrt{2}\ units[/tex]
substitute the given values
[tex]BC=4\sqrt{2}/\frac{1}{2}[/tex]
[tex]BC=8\sqrt{2}\ units[/tex]
therefore
The value of x is
[tex]x=8\sqrt{2}\ units[/tex]
In the right triangle DBC
Find the length side DC
[tex]cos(30\°)=DC/BC[/tex]
[tex]DC=(BC)cos(30\°)[/tex]
we have
[tex]BC=8\sqrt{2}\ units[/tex]
[tex]cos(30\°)=\frac{\sqrt{3}}{2}[/tex]
substitute the given values
[tex]DC=(8\sqrt{2})\frac{\sqrt{3}}{2}[/tex]
[tex]DC=4\sqrt{6}\ units[/tex]
therefore
the value of y is
[tex]y=4\sqrt{6}\ units[/tex]
