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Math Help - Super Slanted wooden board beam problem

  1. #1
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    Super Slanted wooden board beam 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 8 feet from the point where the ramp meets the ground. The slope of the ramp is 2/5 (this means that for every 2 feet it goes up, it goes 5 to the side). 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.

    Find the distances and plot the points, so you can use the distance formula on the points on either side of the metal beam to find its length!

    Ok, so I know that there is a triangle in the middle of the board. Then I must use pythagorean theorem to find the missing side. Then I use the distance formula to find the distance between the points. Is this correct thus far?

    Thank you very much for any potential help.
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  2. #2
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    Triangles

    Hello KevinVM20
    Quote Originally Posted by KevinVM20 View Post


    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 8 feet from the point where the ramp meets the ground. The slope of the ramp is 2/5 (this means that for every 2 feet it goes up, it goes 5 to the side). 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.

    Find the distances and plot the points, so you can use the distance formula on the points on either side of the metal beam to find its length!

    Ok, so I know that there is a triangle in the middle of the board. Then I must use pythagorean theorem to find the missing side. Then I use the distance formula to find the distance between the points. Is this correct thus far?

    Thank you very much for any potential help.
    As I understand it the question requires us to find the distance ED in the diagram I have attached, given that AB = 8', BD = 1' and the \angle ABD = \angle AED =90^o, and the slope of AE is \frac{2}{5}; i.e. 5 across to 2 up.

    You'll see that I've produced BD to meet AE produced at C, and I've marked \angle ACB as \theta.

    Then, in \triangle ABC, \tan\theta = \frac{AB}{BC} = \frac{5}{2}=2.5, given the slope of AE is \frac{2}{5}

    \Rightarrow \theta = 68.20^o

    and BC = \frac{AB}{2.5}= 3.2'

    \Rightarrow CD = 3.2 - 1 = 2.2'

    So from \triangle CDE, ED = CD\sin\theta = 2.2\sin68.2^o = 2.04' to 2 d.p.

    Grandad

    Attached Thumbnails Attached Thumbnails Super Slanted wooden board beam problem-untitled.jpg  
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  3. #3
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    Thank you, Grandad. If I am correct there are three proofs we are working with here. One is the distance formula. What is two and three? Thanks again

    Can a tangent and sin be a proof?
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  4. #4
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    Hello KevinVM20
    Quote Originally Posted by KevinVM20 View Post
    Thank you, Grandad. If I am correct there are three proofs we are working with here. One is the distance formula. What is two and three? Thanks again

    Can a tangent and sin be a proof?
    I used tangent and sine because I thought it was easier to follow, but if you like you can use ratios (fractions) with similar triangles and Pythagoras Theorem. Like this:

    The slope of AC is \frac25 \Rightarrow \frac{CB}{AB}=\frac{2}{5}

    \Rightarrow CB = \tfrac25\times 8 = 3.2

    \Rightarrow DC = 3.2-1 = 2.2

    In \triangle\text{'s } ABC, DEC:

    \angle ABC = \angle DEC = 90^o (given)

    \angle C is common

    \Rightarrow \triangle\text{'s } ABC, DEC are similar (equal angles)

    \Rightarrow \frac{ED}{AB}= \frac{DC}{AC} (ratios of corresponding sides are equal)

    \Rightarrow ED = \frac{AB\times DC}{AC}

    Using Pythagoras: AC^2 = 8^2 + 3.2^2 = 74.24

    \Rightarrow AC = 8.616

    \Rightarrow ED = \frac{8\times 2.2}{8.616}= 2.04 (2 d.p.)

    Grandad
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