Question

1. On a quiet day without many witnesses, a 4.5 ft victim is shot by a sniper that
was in a tower 780 ft away. If the angle of elevation is 48° from the victim, at
what vertical distance did the sniper shoot from?

Answer 1

Answer:

The sniper shoot from 870.775 feet vertical distance.

Step-by-step explanation:

Refer the attached figure

Height of victim = AB = 4.5 ft.

Distance between tower and victim BC = 780 feet

The angle of elevation =$\angle DAE = 48^{\circ}$

Let DE be x

AB = CE = 4.5 feet

BC =AE = 780 feet

Height of tower = DE+CE=x+4.5

In ΔDAE

$\frac{Perpendicular}{Base}=Tan\theta\\\frac{DE}{AE}=Tan 48^{\circ}\\\frac{x}{780}=1.11061\\x=1.11061 \times 780\\x=866.275$

Height of tower = DE+CE=x+4.5=866.275+4.5=870.775 feet

Hence  the sniper shoot from 870.775 feet vertical distance.