Answer:
2720 ft.
Step-by-step explanation:
Let John measure angle of depression to his house (at D) of 35° and to the parade (at C) 42° from a height of building AB where B is the bottom and A is the top of the building.
Now, given that BD = 3500 ft,
∠ OAD = ∠ ADB = 35° {Alternate angles}
and ∠ OAC = ∠ ACB = 42° {Alternate angles}
Now, from Δ ABD,
[tex]\tan 35 = \frac{AB}{BD} = \frac{x}{3500}[/tex] {Where AB = x feet.}
⇒[tex]x = 3500 \tan 35 = 2450.72[/tex] ft.
Now, taking Δ ABC,
[tex]\tan 42 = \frac{AB}{BC} = \frac{x}{BC} = \frac{2450.72}{BC}[/tex]
⇒ [tex]BC = \frac{2450.72}{\tan 42} = 2721.8[/tex] ft.
Therefore, the parade is 2720 ft far from John. (Answer)
{To the nearest ten feet}
