Answer:
Part 1) [tex]y=3\ units[/tex]
Part 2) The length of the hypotenuse is 6 units
Step-by-step explanation:
Part 1) we know that
In the right triangle of the figure
[tex]sin(60^o)=\frac{y}{2\sqrt{3}}[/tex] ----> by SOH (opposite side divided by the hypotenuse)
solve for y
[tex]y=sin(60^o)2\sqrt{3}[/tex]
Remember that
[tex]sin(60^o)=\frac{\sqrt{3}}{2}[/tex]
substitute
[tex]y=(\frac{\sqrt{3}}{2})2\sqrt{3}[/tex]
[tex]y=3\ units[/tex]
Part 2)
Let
h ----> the length of the hypotenuse
we know that
In the right triangle of the figure
[tex]sin(45^o)=\frac{3\sqrt{2}}{h}[/tex] ----> by SOH (opposite side divided by the hypotenuse)
[tex]h=\frac{3\sqrt{2}}{sin(45^o)}[/tex]
Remember that
[tex]sin(45^o)=\frac{\sqrt{2}}{2}[/tex]
substitute
[tex]h=3\sqrt{2}:\frac{\sqrt{2}}{2}[/tex]
[tex]h=6\ units[/tex]
