Answer:
102 ft.
Step-by-step explanation:
See the diagram attached.
From right triangle Δ ADC,
[tex]\sin 30 = \frac{DC}{AC} = \frac{80}{AC}[/tex]
⇒ AC = 160 ft. {Since [tex]\sin 30 = \frac{1}{2}[/tex] }
Now, given that BC = 42 ft.
Hence, AB = AC - BC = 160 - 42 = 118 ft.
Now, from right triangle Δ AEB we can write
[tex]\cos 30 = \frac{AE}{AB} = \frac{AE}{118}[/tex]
⇒ [tex]AE = 118 \cos 30 = 102.19[/tex] ft ≈ 102 ft.
Therefore, the closest distance from point A on one bank to point E on the opposite bank is 102 ft. (Answer)