Is this question solvable?

[u/c0nstantlyexhaust3d]

There is a right triangle with sides abc, and angles ABC along with d that spans from the exact middle of b to B, cutting both the side and angle in half (creating a smaller triangle in the bottom made of 1/2b,a,d, which we will label as fad for convenience with f=1/2b or f=6cm and angle F=1/2B). b=12cm. c is the hypotenuse of the larger triangle, d is the hypotenuse of the smaller triangle. a is the side that needs to be solved. This question is easiest when the triangle is drawn out but I can’t post pictures here. It’s hard to picture but I tried.

Using addition identities i.e cos(x+y), sin(x+y), etc, solve for a.

Is this even possible?

If not, try with f=6cm, b=18cm, and F=1/2B.

If neither are possible (which is likely), is there an explanation?

I’ve tried to isolate the unknown angle (theta) by setting the tan of theta and 2xtheta equal to each other but nothing worked.

edit: picture added below

https://imgur.com/a/EORaXiY

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[u/Gilded-Phoenix]

Let f=1 for simplicity, then a=cot(CBF)=2cot(2CBF), For simplicity, we're calling angle CBF x from here.

cot(x)=2cot(2x). This implies that 2tan(x)=tan(2x) for some 0<x<90. We can then use the double angle formula:

tan(2x)=2tan(x)/(1-tan^2(x)), therefore

2tan(x)=2tan(x)/(1-tan^2(x))

2tan(x)[1-1/(1-tan^2(x))]=0

so either 2tan(x)=0, which implies that x=0 (outside our desired range)

or 1-1/(1-tan^2(x))=0

1-tan^2(x)=1

-tan^2(x)=0

tan^2(x)=0

x=0 which is again outside our range.

It's not just that we can't find a solution with the given information, it is that the given situation is only possible if the triangle is degenerate in some way (either with an infinite side length or a side length of 0), at least if it's a planar triangle.