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I was a regular triangle before I broke it down to 2 right triangles. The regular triangle is not isosceles or scalene. I'm trying to find x using trig functions.

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    You can't really do that unless you know one of the sides.. then you can use the Law of Sines. But since you have two triangles I guess you can do that with the equivalent side alpha... find x and 1-x using the LoS and then solve. – 2017-01-23
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    Something is definitely wrong with your diagram. If the two labeled sides were in fact both equal to $h$, then the two triangles would be congruent to each other, and the two angles marked as $40°$ and $25°$ would be congruent. – 2017-01-23
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    He states that the triangle is not isoceles; it's just a placeholder or something. – 2017-01-23
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    He may have stated it's not isosceles, but he also states it's a "regular triangle", and that it's also "not scalene". It's pretty hard to make sense out of that. – 2017-01-23
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    I am assuming he means the triangle need not be isoceles nor need be scalene. What is written is written incorrectly. Every triangle is either isoceles or scalene. None are neither. – 2017-01-23

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Don't break down :) , you are not a regular triangle. You have to consider the full trig triangle situation by solving the for entire triangle HPQ as given by an SAA method in an improved sketch I redrew for you as above.

Start with

$$ h1 \cos 40 ^0 + h2 \sin 25^0 = 1, \quad d ( \cot 40^0 + \cot 25^0) =1 $$

using Sine / Cosine Rules when required.

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As I mentioned in the comments, the diagram is impossible if in fact the two sides labeled $h$ are equal. This is because if two right triangles have congruent hypotenuses and also one pair of congruent legs, then (by the Pythagorean Theorem) the other pair of legs would also have to be congruent, so the entire triangles are congruent, and therefore the angles marked $40°$ and $25°$ would have to be equal.

Moreover, your description of the triangle as "not scalene" doesn't make sense. If all three angles in a triangle are unequal, then the three sides are unequal as well, so the triangle is scalene.

So in what follows I am going to assume the diagram is mis-labeled, and that the sides labeled $h$ are not supposed to be the same length.

Now: using the left-hand triangle, you know that $\tan 40° = \frac{d}{x}$, so $$d = x\tan 40°$$ Using the right-hand triangle, you know that $\tan 25° = \frac{d}{1-x}$, so $$d = (1-x)\tan 25°$$ Setting these equal to each other, we get $$x\tan 40° = (1-x)\tan 25°$$ Can you take it from here and solve for $x$?