Question

A pilot in a helicopter sights an ambulance heading toward an accident scene. He measures the angles of depression to the ambulance and the accident to be 21o and 15o, respectively. If the helicopter is 4000 ft from the ambulance, how far does the ambulance have to travel to get to the accident?

Answer 1

Answer:

the ambulance have to travel 4507.847 feet to get to the accident

Explanation:

Let D represent position of ambulance

Let C represent position of accident scene

The situation is represented in the below attached figure

In the figure we have

AB = 4000 feet

$<XAD=<ADB=15^{o}$

$<XAC=<ACB=21^{o}$

Thus in triangle ABC we have

$tan(21)=\frac{AB}{BC}\\\\\therefore BC=\frac{AB}{tan(21)}=\frac{4000}{tan(21)}=10420.356feet$

In triangle ABD we have

$tan(15)=\frac{AB}{BD}\\\\\therefore BD=\frac{AB}{tan(15)}=\frac{4000}{tan(15)}=14928.20feet$

Thus the distance between the ambulance and accident scene will be 14925.20-10420.356 = 4507.847 feet