The width of the river, given by the segment [tex]\overline{fe}[/tex], is the
altitude of the Δfgh.
Correct response:
A possible diagram showing the situation is attached.
From the diagram, we have;
∠g = 90° - 48° = 42°
The width of the river = [tex]\overline{ef}[/tex]
Using the law of sines, we have;
[tex]\mathbf{\dfrac{\overline{ef}}{sin(42^{\circ})}} = \dfrac{39}{sin(48^{\circ})}[/tex]
Therefore;
[tex]The \ width \ of \ the \ river, \ \overline{ef} = sin(42^{\circ}) \times \dfrac{39}{sin(48^{\circ})} \approx \mathbf{ 35.12}[/tex]
T
he options are; 30yd, 32 yd, 35 yd, 37 yd, 39 yd, 40 yd, and 50 yd.
Learn more about the laws of sines here:
https://brainly.com/question/16555495