Trig - I'm stuck

I'm pretty confused. My last line of working was $\cos \frac{\theta}{2} = 0$, which I wrote down the wrong solution $\theta = 90^o$. There was something else that could have been 0 instead which I'm working on now, but I don't see any reason why I would accidentally come up with the right answer.

I'm pretty confused. My last line of working was $\cos \frac{\theta}{2} = 0$, which I wrote down the wrong solution $\theta = 90^o$. There was something else that could have been 0 instead which I'm working on now, but I don't see any reason why I would accidentally come up with the right answer.

Let φ=CBA=BCA
AD=1 and we take a parallel line to BC from D which intersects CA at E
DE= x
cosφ = x/2
sinφ = (x+sqrt(6))tan15/sqrt((sqrt(6)-x)^2+(x+sqrt(6))^2*tan^2(15))
sin^2φ+cos^2φ=1 which gives x thus φ thus θ

I know

Let φ=CBA=BCA
AD=1 and we take a parallel line to BC from D which intersects CA at E
DE= x
cosφ = x/2
sinφ = (x+sqrt(6))tan15/sqrt((sqrt(6)-x)^2+(x+sqrt(6))^2*tan^2(15))
sin^2φ+cos^2φ=1 which gives x thus φ thus θ

I think I did this and ended up with a quartic. Is that what you got?

So it looks this can be solved by getting help to solve a quartic, which is what I thought.

The challenge is to find a nicer method that gives the answer directly. It could be that there isn't one.

Wha?

I just checked using the sine and cosine rules and 90 was fine

90 works? Now I'm confused.

I'll have another look later when I have time.