We are asked to determine the sinT. To do that let's remember that the function sine is defined as:
[tex]\sin x=\frac{opposite}{hypotenuse}[/tex]
In this case, we have:
[tex]\sin T=\frac{VU}{VT}[/tex]
To determine the value of VU we can use the Pythagorean theorem which in this case would be:
[tex]VT^2=VU^2+TU^2[/tex]
Now we solve for VU first by subtracting TU squared from both sides:
[tex]VT^2-TU^2=VU^2[/tex]
Now we take the square root to both sides:
[tex]\sqrt[]{VT^2-TU^2^{}}=VU[/tex]
Now we plug in the values:
[tex]\sqrt[]{(6)^2+(\sqrt[]{36^{}})^2}=VU[/tex]
Solving the squares:
[tex]\sqrt[]{36+36}=VU[/tex]
Adding the values:
[tex]\sqrt[]{2(36)}=VU[/tex]
Now we separate the square root:
[tex]\sqrt[]{2}\sqrt[]{36}=VU[/tex]
Solving the square root:
[tex]6\sqrt[]{2}=VU[/tex]
Now we plug in the values in the expression for sinT:
[tex]\sin T=\frac{6\sqrt[]{2}}{6}[/tex]
Now we simplify by canceling out the 6:
[tex]\sin T=\sqrt[]{2}[/tex]
And thus we obtained the expression for sinT.
