Geometry Triangles

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August 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.
August 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 ?
August 31st 2008, 11:08 AM
KevinVM20

no
August 31st 2008, 11:37 AM
Soroban

Hello, KevinVM20!

To answer your first question:
. . 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 $h$ = 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: . $\frac{h}{150} \:=\:\frac{500}{10,560}$

Therefore: . $h \;=\;\frac{75,000}{10,560} \;=\;\frac{625}{88} \;\approx\;7.1\text{ feet}$
August 31st 2008, 12:20 PM
KevinVM20

Thank you very much Soroban! I now have a better understanding.

Quote:

Originally Posted by Soroban

Hello, KevinVM20!

To answer your first question:
. . 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 $h$ = 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: . $\frac{h}{150} \:=\:\frac{500}{10,560}$

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