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This is the diagram.

Question: How can one express the length of $z$ in terms of $a$ and $b$ without including $x$ in the expression? If it is impossible to express $z$ in terms of $a$ and $b$ alone, please answer with an explanation of why.

More Information: $ax$ and $bx$ are angles, where $x$ is some constant and $a$ and $b$ are each being multiplied by it. In other words, the two angles are in ratio $a$ to $b$. You can assume that $a$ is greater than $b$ and both $ax$ and $bx$ are less than 90 degrees.

Motivation: I'm asking this question because I feel that the constraints on the diagram are sufficient to bind $z$ to a single value for a given $a$ and $b$. That is, I suspect knowing the lengths I have specified and knowing that the ratio of the angles is $a$ to $b$, one should be able to determine $z$ with certainty.

Thus, I suspect it is likely $z$ can be expressed in terms of only $a$ and $b$. Unfortunately, all my attempts to derive an expression for $z$ have ended up with me including $x$ in my expression or just churning out tautologies. Clearly help is needed. Thanks in advance!

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    What are $ax$ and $bx$? Segments? Angles? – 2017-01-07
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    They are angles. x is some constant and a and b are being multiplied by it. Also, you can assume a > b. – 2017-01-07
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    To anyone who can reopen questions, if you think I've added enough context, please consider reopening this. – 2017-01-12
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    The answer by @Alex shows that what you are asking for is impossible. Why do you feel it is unsatisfactory? – 2017-01-12
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    Oh, geez. You're absolutely right that this one is impossible. To fully constrain the thing, I should have specified the length of the base as well. Thanks for helping me see that. I'll accept @Alex's answer. – 2017-01-12

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Let $\alpha=ax$, $\beta=bx$. Then $$ {z\over2} = 1\cdot\sin{\alpha+\beta\over2}, $$ therefore $$ z = 2\cdot\sin{\alpha+\beta\over2} = 2\cdot\sin{x(a+b)\over2}. \tag1 $$ This expression is obtained from the fact that $z$ is the base of an isosceles triangle whose two other sides are equal to $1$. Equation $(1)$ shows that $z$ indeed depends on $x$; and so $z$ cannot be expressed in terms of $a$ and $b$ alone.

Note: On the other hand, by the Pythagorean Theorem, $$ z^2 = (\sin\alpha+\sin\beta)^2 + (\cos\beta-\cos\alpha)^2 \tag2 $$ Using $(1)$ and $(2)$ we might want to eliminate some of the unknowns. However, a closer examination shows that equations $(1)$ and $(2)$ are not independent; in fact these equations simply express the trigonometric identity $$ (\sin\alpha+\sin\beta)^2 + (\cos\beta-\cos\alpha)^2 = 4\cdot\left(\sin{\alpha+\beta\over2}\right)^2. $$