Answer:
[tex]Height\ of\ monument=555.5\ ft[/tex]
Explanation:
[tex]In\ the\ digram\ AC\ is\ the\ monument\ and\ DE\ is\ the\ student.\\\\AC=AB+BC\\\\Let\ AB=x\\\\BC=Height\ of\ the\ eye\ of\ the\ student=DE=5\ ft\\\\\Rightarrow AC=x+5.................................eq(1)\\\\Student\ is\ standing\ at\ a\ distance\ of\ 25\ ft.\\\\CD=25\ ft\\\\BE=CD=25\ ft\\\\Angle\ of\ elevation=87.4\textdegree\\\\\angle AEB=87.4\textdegree\\\\Now\ in\ \triangle ABE\\\\\tan \angle AEB=\frac{opposite}{adjacent}=\frac{AB}{BE}\\\\\tan 87.4=\frac{x}{25}\\\\x=25\times \tan 87.4\\[/tex]
[tex]x=25\times 22.02171001\\\\x=550.54275\ ft\approx 550.5\ ft\\\\Height\ of\ monument=AC=x+5\\\\Height\ of\ monument=550.5+5\\\\Height\ of\ monument=555.5\ ft[/tex]
