Edit: Ok the question has now been amended by my tutor as I highlighted it was not possible to solve.
A pole is tensioned at point A as shown below, if the angles are to be kept the same, then using a suitable double angle identity determine the distance $d$:
Edit: Note that triangle ACB is not a right angled triangle. Should now read:
Note that triangle ACD is not a right angled triangle.
By using $$\cos 2x=2\cos^2 x-1$$
My working is:
$$\cos x=\frac d3$$
$$\cos 2x=\frac d3$$
therefore:
$$\frac d3=2(\frac d3)^2-1$$
Now becomes solvable as a quadratic equation
Best Answer
I will interpret the problem as "given a point $C$ on a segment $BD$ such as $BC=b$ and $CD=c$, what is the distance $d$ from $A$ to $B$ such that $BC$ and $CD$ have the same angular diameter?"
Using the law of cosines in triangles $ABC$ and $ACD$ we get $$c^2=2^2+3^2-2\times 2\times 3 \cos x$$ $$b^2=d^2+2^2-4d\cos x$$ so that $$\cos x=\dfrac{13-c^2}{12}$$ and $$b^2=d^2+4-\dfrac{4d(13-c^2)}{12}$$ The last equation is quadratic in $d$ and easy to solve.