# Can I solve this problem?

Dealing with oblique projections, I encountered a problem that I can formulate like this:

I have a parallelogram with two sides of 1 unit and two unknown. I also know the angles. There is a right triangle sharing the parallelograms unknown side, like in the image. Is it possible to figure out the lengths of the triangle, especially x?

• That is not a possible triangle, the hypotenuse (with length $2$) can't be shorter than one of its legs (with length $4$?) – user170231 Mar 25 at 18:45
• @user170231 those are letters, not numbers. – Paul Mar 25 at 18:49
• @user170321 I'm sorry for my writing, the "2" is a z and the "4" is y – Andrei Agache Mar 25 at 18:49

If I am reading your handwriting correctly, you have 1 angle and 1 side length (though it is a parallelogram and that gives you the opposite side length and all 4 angles). In this case, just imagine lengthing the side labeled $$z$$ by a lot (think 100x as long), then $$x$$ would be made much longer, without changing the length of the side labeled 1 or any angles. Thus $$x$$ cannot be found with only 1 angle and 1 side length from this parallelogram.
Though you could solve for $$x$$ in terms of $$z$$.
• I am unfamiliar with geometry. I know with Pythagoras theorem you can solve for $x$ in terms of $y$ and $z$. I know with trig functions, I can solve for $x$ in terms of $z$. What is another way to solve for $x$ in terms of $z$? – name Mar 25 at 19:29
Yes, you could find $$x$$ and $$y$$ in terms of $$z,$$ but you don't know what $$z$$ is.