Answer: [tex]BC=4\sqrt{2}[/tex]
Step-by-step explanation:
For this exercise you need to analize the diagram given in the exercise.
You can see that the triangle BCG is a Right triangle, because it has an angle that measures 90 degrees.
Then, you need to use the following Trigonometry Identity:
[tex]sin\alpha=\frac{opposite}{hypotenuse}[/tex]
In this case you can identify that:
[tex]\alpha=45\°\\\\opposite=CG=4\\\\hypotenuse=BC[/tex]
Now, you must substitute those values into [tex]sin\alpha=\frac{opposite}{hypotenuse}[/tex], as below:
[tex]sin(45\°)=\frac{4}{BC}[/tex]
And finally, you must solve for BC in order to find its value.
Then, you get that this is:
[tex](BC)(sin(45\°))=4\\\\BC=\frac{4}{sin(45\°)}\\\\BC=4\sqrt{2}[/tex]
