Answer:
Step-by-step explanation:
(A)From the figure drawn, It is given that AB is the height of the cell tower that is equal to=150 feet and the angle formed between the wire and the ground is ∠C=48°.
Then, the length of the guy wire will be:
[tex]\frac{AB}{AC}=sin48^{\circ}[/tex]
⇒[tex]\frac{150}{AC}=0.743[/tex]
⇒[tex]AC=\frac{150}{0.743}[/tex]
⇒[tex]AC=201.8 feet[/tex]
thus, the length of the guy wire is 201.8 feet.
(B) The distance of the guy wire from the base of the tower is given s:
[tex]\frac{AB}{BC}=tan48^{\circ}[/tex]
⇒[tex]\frac{150}{BC}=1.110[/tex]
⇒[tex]BC=\frac{150}{1.110}[/tex]
⇒[tex]BC=135.1ft[/tex]
Thus, The distance of the guy wire from the base of the tower is 135.1 feet.
