The accompanying figure shows two ships at points $P$ and $Q$ , which are in the same vertical plane as an airplane at point $R$ . When the height of the airplane is 4,000 ft, the angle of depression to $P$ is 38° and that to $e$ is 15°. Find the distance between the two ships.
A) 54,900 ft
B) 2,050 ft
C) 20,050 ft
D) 80,430 ft
E) 4,200 ft

Question

The accompanying figure shows two ships at points $P$ and $Q$ , which are in the same vertical plane as an airplane at point $R$ . When the height of the airplane is 4,000 ft, the angle of depression to $P$ is 38° and that to $e$ is 15°. Find the distance between the two ships. 
A) 54,900 ft 
B) 2,050 ft 
C) 20,050 ft 
D) 80,430 ft 
E) 4,200 ft

Quizplus · Accepted Answer

Let's call the distance between the airplane and ship P "x" and the distance between the airplane and ship Q "y". We can create two equations based on the angles of depression:
$$\tan(38^\circ)=\frac{4000}{x}$$
$$\tan(15^\circ)=\frac{4000}{y}$$
We can solve for x and y in terms of each other:
$$x=\frac{4000}{\tan(38^\circ)}\approx 5066.83\text{ ft}$$
$$y=\frac{4000}{\tan(15^\circ)}\approx 13763.75\text{ ft}$$
Finally, we can use the Pythagorean Theorem to find the distance between the two ships:
$$\sqrt{(y-x)^2+4000^2}\approx \boxed{\textbf{(C) }20,050\text{ ft}}.$$

The Accompanying Figure Shows Two Ships at Points PPP And QQQ , Which Are in the Same Vertical Plane as an Q

The Accompanying Figure Shows Two Ships at Points $P$ And $Q$ , Which Are in the Same Vertical Plane as an Q