# Geometry Triangles

• Aug 31st 2008, 09:35 AM
KevinVM20
Geometry Triangles
I was working on a problem. I know that I am supposed to solve for the unknown. I just don't know what exactly the unknown is.

Here is the problem-

I can look out my window and see the top of a tower. On the map, I see that it is 2 miles away. I read somewhere that the tower is 500 feet tall. As I look at the tower, I see that the top leaves of a tree sometimes get in the way of the top of the tower. The tree is 50 yards from where I sit. How tall is the tree?

How do I put this problem into an equation and what is the equation?

Thanks for all potential help.
• Aug 31st 2008, 10:42 AM
Moo
Hello,
Quote:

Originally Posted by KevinVM20
I was working on a problem. I know that I am supposed to solve for the unknown. I just don't know what exactly the unknown is.

Here is the problem-

I can look out my window and see the top of a tower. On the map, I see that it is 2 miles away. I read somewhere that the tower is 500 feet tall. As I look at the tower, I see that the top leaves of a tree sometimes get in the way of the top of the tower. The tree is 50 yards from where I sit. How tall is the tree?

How do I put this problem into an equation and what is the equation?

Thanks for all potential help.

Do you know how to use the Intercept theorem ?
• Aug 31st 2008, 11:08 AM
KevinVM20
no
• Aug 31st 2008, 11:37 AM
Soroban
Hello, KevinVM20!

. . the unknown is obviously the height of the tree.

Quote:

I can look out my window and see the top of a tower.
On the map, I see that it is 2 miles away.
I read somewhere that the tower is 500 feet tall.
As I look at the tower, I see that the top leaves of a tree
sometimes get in the way of the top of the tower. **
The tree is 50 yards from where I sit.
How tall is the tree?

** I assume this means that the top of tree lines up
. . with the line-of-sight to the top of the tower.

We further assume that your eye and the base of the tower are at the same height.

Let $\displaystyle h$ = height of the tree (in feet).
Change all units to feet.
Code:

                              *                           *  |                       *      |                   *          | 500               *  |h          |           *      |          |       * - - - - - * - - - - - *       : -  150  - :       : - - - 10,560  - - - - :

From the two similar right triangles, we have: .$\displaystyle \frac{h}{150} \:=\:\frac{500}{10,560}$

Therefore: .$\displaystyle h \;=\;\frac{75,000}{10,560} \;=\;\frac{625}{88} \;\approx\;7.1\text{ feet}$

• Aug 31st 2008, 12:20 PM
KevinVM20
Thank you very much Soroban! I now have a better understanding.

Quote:

Originally Posted by Soroban
Hello, KevinVM20!

. . the unknown is obviously the height of the tree.

** I assume this means that the top of tree lines up
. . with the line-of-sight to the top of the tower.

We further assume that your eye and the base of the tower are at the same height.

Let $\displaystyle h$ = height of the tree (in feet).
Change all units to feet.
Code:

                              *                           *  |                       *      |                   *          | 500               *  |h          |           *      |          |       * - - - - - * - - - - - *       : -  150  - :       : - - - 10,560  - - - - :

From the two similar right triangles, we have: .$\displaystyle \frac{h}{150} \:=\:\frac{500}{10,560}$

Therefore: .$\displaystyle h \;=\;\frac{75,000}{10,560} \;=\;\frac{625}{88} \;\approx\;7.1\text{ feet}$