Answer:
[tex]\large \boxed{\text{c. 405.3 ft}}[/tex]
Step-by-step explanation:
[tex]\begin{array}{rcr}\angle A + \angle B + \angle C & = & 180^{\circ}\\82^{\circ} + 78^{\circ} +\angle C & = & 180^{\circ}\\160^{\circ} + \angle C & = & 180^{\circ}\\\angle C & = & 20^{\circ}\\\end{array}[/tex]
[tex]\begin{array}{rcl}\dfrac{\sin A}{a} & = &\dfrac{\sin C}{c}\\\\\dfrac{\sin82^{\circ}}{a} & = &\dfrac{\sin20^{\circ}}{140}\\\\\dfrac{0.9903}{a} & = &\dfrac{0.3420}{140}\\\\a & = & \dfrac{0.9903 \times140}{0.3420}\\\\& = & \mathbf{405.3 ft}\\\end{array}\\\text{The distance from Ship B to the signal fire is $\large \boxed{\textbf{405.3 ft}}$}[/tex]