Help!!!! A Geometry problem!

School_girl

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Jun 6, 2005
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A wooden board is placed so that it leans against a loading dock to provide a ramp.

The board is supported by a metal beam perpendicular to the ramp and placed on a 1 ft. tall support.

The bottom of the support is 7 feet from the point where the ramp meets the ground. The slope of the ramp is 2/5.

Find the length of the beam to the nearest hundredth of a foot. Note that the 1 ft. support is vertical, but the metal beam is not.

Thanks,

ME
 
Hello, School_girl!

This is a truly evil problem . . .

A wooden board is placed so that it leans against a loading dock to provide a ramp.
The board is supported by a metal beam perpendicular to the ramp and placed on a 1 ft. tall support.
The bottom of the support is 7 feet from the point where the ramp meets the ground.
The slope of the ramp is 2/5.

Find the length of the beam to the nearest hundredth of a foot.
Note that the 1 ft. support is vertical, but the metal beam is not.
Code:
                     *B
                   * |            The board is AB.
                 *   |            
               *     |            The support is DE:
            F*       |               DE = 1, AE = 7
           *  \b     |
         *      \    |           The beam is DF = b
       *         *D  |           
     *     *     |   |          Let angle DAF = theta  
   *  *          |1  |      
 *---------------+---+-
 A       7       E   C
. . . . . . . . . . . . . . . . . . . . . . . . ______ . . . . ._
In right triangle DEA: . AD .= .√7<sup>2</sup> + 1<sup>2</sup> .= .5√2

. . . . . . . . . . . . . . . . . . . . . . . . . . . b . . . . . . . . . . . . . ._
In right triangle DFA: . sin θ . = . ---<u>--</u>- . ---> . b .= .(5√2)(sin θ)
. . . . . . . . . . . . . . . . . . . . . . . . . . 5√2

If we can find sin θ, we're done!


Note that: . tan(<u>/</u>BAC) = 2/5, . and . tan(<u>/</u>DAE) = 1/7

We have: . θ .= .<u>/</u>BAC - <u>/</u>DAE

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . tan(<u>/</u>BAC) - tan(<u>/</u>DAE)
Then: . tan θ . = . tan(<u>/</u>BAC - <u>/</u>DAE) . = . -----------------------------
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 + tan(<u>/</u>BAC)tan(<u>/</u>DAE)

. . . . . . . . . . . . . . . . (2/5) - (1/7) . . . . . 9
Hence: . tan θ . = . ------------------ . = . ---
. . . . . . . . . . . . . . . 1 + (2/5)(1/7) . . . . 37


And we have: . θ .= .13.6713<sup>o</sup> . . ---> . . sin θ .= .0.23635 . . . There!


[Oh well .... I probably over-explained it, as usual.]
 
Geometry Help!!

Hi, I am kind of confused with my previouse question.


Can you show me how to do it with similar triangles as well??

And I would like to see what is the answer clearly..

Thanks!
 
I can't draw it si I'll try to describe my attack.
Draw a line from the support to the ramp. It meets it at .4 times the distance from the end = 1/.4 = 2.5. That leaves 4.5 of the 7'. Drop a line from the ramp-brace intersection. That creates two similar triangles. Call the long part of the 4.5' a and the short part b, the height h.
From the similar triangles
h/a=b/h
From the slope
h = .4a
.4a/a = b/(.4a)
a=b/.16
a+b=4.5
(b/.16)+b = 4.5
1.16b=4.5*.16
b=4.5*.16/1.16 = .62069
h=.4*b/.16 = 1.5517
sqrt(b^2+h^2) = 1.6712

(Soroban didn't mention that the last step is multiply sin(theta) by sqrt(50) for the final answer and then we agree.)
 
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