1. ## Trigonometry Word Problem

A sailor at the seashore watches a ship with a smokestack 30 meters above water level as the ship steams out to sea. The sailor's eye level is 4 meters above water level. About how far is the ship from shore when the stack disappears from the sailor's view? (The radius of Earth is about 6400 kilometers.)

2. Originally Posted by strummer
A sailor at the seashore watches a ship with a smokestack 30 meters above water level as the ship steams out to sea. The sailor's eye level is 4 meters above water level. About how far is the ship from shore when the stack disappears from the sailor's view? (The radius of Earth is about 6400 kilometers.)
hope you have a sketch drawn ...

let point S be the sailor's eye position

point K = top of smokestack.

let "line of sight" segment SK be tangent to the earth at point T

center of the earth = point O

two adjacent right triangles are formed, $\Delta OTS$ and $\Delta OTK$

for $\Delta OTS$ ...

OT = 6400 km
OS = 6400.004 km

line of sight distance, $ST = \sqrt{OS^2 - OT^2} \approx 7.16 \, km$

arc length $ST = 6400 \cdot \arccos\left(\frac{6400}{6400.004}\right) \approx 7.16 \, km$

note that the line of sight distance is very close to the arclength along the earth's surface.

do the same for $\Delta OTK$ and sum the distances (or the arclengths) ST and TK.

3. Hello, strummer!

A sailor at the seashore watches a ship with a smokestack 30 meters above water level
as the ship steams out to sea. . The sailor's eye level is 4 meters above water level.
About how far is the ship from shore when the stack disappears from the sailor's view?
Code:
        A           C             B
o - - - - * o * - - - - o
4\   *     |     *   /30
*       |       *
*  \     |R    /  *
R \   |   / R
*       \ | /       *
*         o         *
*         O         *

*                 *
*               *
*           *
* * *

The sailor is at $A$, the top of the smokestack is $B.$
The line of sight is tangent to the Earth at $C.$

The center of Earth is $O$, its radius is $R$ (6,400,000 m).

In right triangle $OCA\!:\;\;AC^2 + R^2 \:=\:(R+4)^2 \quad\Rightarrow\quad AC \:=\:\sqrt{8R + 16}$

In right triangle $OCB\!:\;\;CB^2 + R^2\:=\: (R+30)^2\quad\Rightarrow\quad CB \:=\:\sqrt{60R + 900}$

Hence: . $AB \:=\:AC + CB \:=\:\sqrt{8R + 16} + \sqrt{60R + 900}$

. . . . . . . . . $= \;\sqrt{8(6,\!400,\!000) + 16} + \sqrt{60(6,\!400,\!000) + 900}$

. . . . . . . . . $= \;26,\!751.35955$ m.

Therefore, the distance is about 26.75 kilometers.

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Edit . . .

Ah, I see that skeeter did it correctly (ship-to-shore distance).

Using the Law of Cosines, $\theta = \angle AOB$ is given by:

. . $\cos\theta \;=\;\frac{6400.004^2 + 6400.03^2 - 26.75^2}{2(6400.004)(6400.03) } \:=\:0.999991264$

Hence: . $\theta \:=\:0.00417989$ radians.

Therefore: . $\overline{ACB} \;=\;r\theta \;=\;(6400)(0.00417989) \;=\;26.751296 \;\approx\;26.75$ km