Hello!
We know that this is a right triangle and the angle C is 45º.
Knowing it, we have:
Considering the information above, we must use the sine of 45º to calculate the value of side BC, look:
As we know, the sine of 45º is:
Let's replace all the values in the formula:
![\begin{gathered} \sin(45\operatorname{\degree})=\frac{\mathrm{oppos\imaginaryI te}}{\mathrm{hypotenuse}} \\ \\ (√(2))/(2)=\frac{10}{\mathrm{BC}} \\ \\ \mathrm{BC}√(2)=10\cdot2 \\ \mathrm{BC}√(2)=20 \\ BC=(20)/(√(2)) \\ \\ BC=(20\cdot√(2))/(√(2)\cdot√(2))=(20√(2))/(√(4))=(20√(2))/(2)=\boxed{10√(2)\text{ ft}} \end{gathered}]()
Answer:
Alternative B.