Right Triangle

[wombatcube]

I have this problem on a physics review packet and I'm not sure how to solve it

I think I have the correct values for the smaller triangle but I"m not sure how to scale them up to the larger triangle to find x.
Also I got the angles for the triangle made with the arrows in a weird way with parallel lines that I'm pretty sure isn't correct and is based on wrong assumptions.

Sorry for the weird images, this is my first time posting
I've been stuck on this problem for a very long time

 

[Dr.Peterson]

So, what are your answers? The numbers I see, 23.094 and 11.547, are not the values of anything in the figure that I can see. It appears that you are thinking the 20 is a leg of a triangle, which is not what is intended.

The picture is terrible. It is not to scale (that isn't really anything close to a 60° angle!), and the right angles don't look like right angles, even the one that is labeled as such. Most important for you, the arrow labeled as 20 is supposed to represent a force, not a distance, so the fact that it reaches the horizontal line is misleading; it is only meant to be part of the small right triangle, not of anything else. And the X is a leg of the small right triangle, not the horizontal line.

You have a small right triangle with angle 60° and hypotenuse 20. What are the legs X and Y? That's all you need.

 

[wombatcube]

Thank you for the response. Before I posted this topic I emailed my teacher about a few things. She said that the arrows in the middle made a right triangle and that x was the green part. I probably should've included that in the initial post. At this point I'm not really sure if the problem is solvable the way my teacher intended it to be.

 

[Dr.Peterson]

Thank you for the response. Before I posted this topic I emailed my teacher about a few things. She said that the arrows in the middle made a right triangle and that x was the green part. I probably should've included that in the initial post. At this point I'm not really sure if the problem is solvable the way my teacher intended it to be.

No, that makes no sense, as you are given no size information at all about the large triangle or about where the block is located on it. That distance can't be determined.

Yes, the picture looks like the X means that length, but it can't possibly. And in the application you will soon be learning, the two legs of the small triangle will be what matters: Y is the normal force, which relates to the amount of friction, and X is the parallel (downslope) force, which in combination with friction determines whether and how fast the block will slide.