Question

LAWS OF SINE

Answer 1

By the law of sines,

$\dfrac{\sin49^\circ}{AB}=\dfrac{\sin53^\circ}{18\,\mathrm{ft}}\implies AB=\dfrac{(18\,\mathrm{ft})\sin49^\circ}{\sin53^\circ}\approx17.0\,\mathrm{ft}$