# Distances

• December 10th 2010, 01:08 AM
Evaldas
Distances
I don't understand this:
http://img837.imageshack.us/img837/4438/85154834.jpg
The plan shows the roads between towns A, B and C. Find the distance:
a) from A to BC;
b) from B to AC;
c) from C to AB.
• December 10th 2010, 01:26 AM
ark600
from A to BC probably means the shortest distance from point A to some point on BC.
• December 10th 2010, 01:28 AM
Prove It
And the shortest distance to a line occurs at a right angle to that line.
• December 10th 2010, 01:31 AM
Evaldas
They give such answers: a) 15 km, b) 7 and 1/17 km (that's one number), c) 8 km.
• December 10th 2010, 01:33 AM
Prove It
The first thing you will have to do is to use the cosine rule to evaluate each angle.
• December 10th 2010, 01:38 AM
Evaldas
It's an exercise from an 8th grade book, where trigonometry functions (sin, cos, tg) are two years away, that's why I posted it not into trigonometry section. You can't use triangle similarity at this stage as well. This exercise is given at the end of the "Pythagoras theorem" chapter, where usually mixed exercises are, from earlier (not necessarily about the theorem). And I personally can't see how you can apply Pythagoras theorem here.
• December 10th 2010, 02:20 AM
HallsofIvy
You apply the Pythatorean theorem to show that the angle at B is a right angle!
• December 10th 2010, 02:23 AM
Evaldas
I noticed that, but what does it give me? But I deduced that the fact that there's no small square indicating that B is a right angle, they want us to solve it differently...
• December 10th 2010, 07:49 AM
Soroban
Hello, Evaldas!

I'll do part (b).

Quote:

$\text{The plan shows the roads between towns }A, B, C.$
$\text{Find the distance: (b) from B to AC.}$
Code:

                    B                     o                   * |*                 *  | *         15  *    |  *  8             *      |h  *           *        |    *         *          |    *       *            |      *   A o * * * * * * * o * * * o C     :    17-x      D  x  :

We have $\Delta ABC$ with: . $AB = 15,\;BC = 8,\;AC = 17$

Draw $BD \perp AC.\;\text{ Let }h = BD.$

Let $x = DC,\;17-x = AD.$

In right triangle $BDC\!:\;h^2 + x^2 \:=\:8^2 \quad\Rightarrow\quad h^2 \:=\:64 - x^2$ .[1]

In right triangle $BDA\!:\;h^2 + (17-x)^2 \:=\:15^2$ .[2]

Substitute [1] into [2]: . $(64-x^2) + (17-x)^2 \:=\:225$

. . which simplifies to: . $34x \:=\:128 \quad\Rightarrow\quad x \:=\:\frac{64}{17}$

Substitute into [1]: . $h^2 \:=\:64 - \left(\frac{64}{17}\right)^2 \:=\:\frac{14,\!400}{289}$

. . Therefore: . $h \:=\:\sqrt{\frac{14,\!400}{289}} \:=\:\frac{120}{17} \;=\;7\frac{1}{7}$

• December 10th 2010, 07:56 AM
Evaldas
Thank you Soroban. I get it now.
• December 10th 2010, 03:29 PM
bjhopper
distances
Soroban's method does not require ABC to be a right triangle. If you notice that 17^2 = 15^2+ 8^2 ABC is aright triangle
Area = 4x15 = 60 km^2
60 =a1 x17/2 a1=120/17
60 =a2 x8/2 a2=120/8
60 = a3 x15/2 a3 = 120/15 where a's are altitudes to coressponding sides

bjh