# plane geometry

• Nov 12th 2008, 12:57 AM
TWoods
plane geometry
State the rule which connects BT, EA, and BA.

this is a topic of plane geometry and we were given a directed investigation 'How far can you see the horizon?' this is it....the circle represents a cross-section of the Earth. imagine you are in a balloon at B. BT shows the furthest point, T, on the Earth's surface that you could see. imagine that you have just dropped a very heavy, sharp object from your balloon. the object, hits the ground at E, directly below you, boring its way through Earth coming to a stop just as it reaches the antipodal point A.

here's a diagram: i have an attachment and its called doc. please someone answer this, its due monday and its worth 25% of my grade.
• Nov 13th 2008, 05:36 PM
Soroban
Hello, TWoods!

Quote:

How far can you see to the horizon?

The circle represents a cross-section of the Earth.
Imagine you are in a balloon at $B.$
$BT$ shows the furthest point, $T$, on the Earth's surface that you could see.

Code:

                B                 o                 |\                 | \               h|  \                 |  \                 |    \               * * *  \           *    |    *\ T         *      |      o       *      R|    *  *                 |  *R       *        | *      *       *        o        *       *        O        *       *                *         *              *           *          *               * * *

The balloon is at $B,\;h$ miles above the Earth.
The center of the Earth is $O$; the radius is: $OT = R$

The angle at $T$ is 90°.

Using Pythagorus on $\Delta BTO\!:\;\;BT^2 + OT^2 \:=\:BO^2 \quad\Rightarrow\quad BT^2 \:=\:BO^2 - OT^2$

. . $BT^2 \;=\;(R+h)^2 - R^2 \;=\;2Rh + h^2$

Therefore: . $d = BT$, the distance from the balloon to the horizon

. . is given by: . $\boxed{d \;=\;\sqrt{2Rh + h^2}}$ miles

where: . $h$ = height of balloon, $R$ = radius of the Earth (both in miles).