# Thread: Super Slanted wooden board beam problem

1. ## 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.

2. ## Triangles

Hello KevinVM20
Originally Posted by KevinVM20

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.

3. 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?

4. Hello KevinVM20
Originally Posted by KevinVM20
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.)