Answer: B. between 7 and 10 ft
Explanation: To find the length of \( AB \) in \(\triangle ABC\), we can use the Law of Cosines. This theorem relates the lengths of the sides of a triangle to the cosine of one of its angles. Given the sides \( AC = 10 \) ft, \( BC = 7 \) ft, and angle \( C = 68^\circ \), we can find \( AB \).
Steps:
- Law of Cosines Formula:
\[
c^2 = a^2 + b^2 - 2ab \cdot \cos(C)
\]
where \( c \) is the side opposite angle \( C \).
- Substitute the Known Values:
\[
AB^2 = 10^2 + 7^2 - 2 \cdot 10 \cdot 7 \cdot \cos(68^\circ)
\]
- Calculate:
\[
AB^2 = 100 + 49 - 140 \cdot \cos(68^\circ)
\]
- Find \(\cos(68^\circ)\):
\[
\cos(68^\circ) \approx 0.3746
\]
- Continue Calculation:
\[
AB^2 = 149 - 140 \cdot 0.3746
\]
\[
AB^2 = 149 - 52.444
\]
\[
AB^2 = 96.556
\]
- Solve for \( AB \):
\[
AB = \sqrt{96.556} \approx 9.82 \text{ ft}
\]
Since \( AB \approx 9.82 \) ft, it falls between 7 and 10 ft. Thus, the correct answer is B. between 7 and 10 ft.