Calculate third point of Isosceles triangle given two points and angle In this image: (Just for annotations.. The actual triangle can be pointing to any direction)

I know the coordinates of "red" base points and the "blue" vertex angle $\beta$, and I want to calculate the coordinates of the "blue" vertex point.
I did come up with a solution but it looks unnecessarily complex so I'm hoping there is an easier solution..
What I did was, using the Law of sines, I got $\frac{a}{sin \alpha} = \frac{b}{sin \beta}$ and since $sin \frac{\beta}{2} = \frac{b}{2a}$, I have $sin \beta = 2sin \frac{\beta}{2} cos\frac{\beta}{2} = \frac{bL}{a^2}$ (Eq. 1)
From Pythagoras' Theorem I have $L^2 + \frac{b^2}{4} = a^2$, I rewrite Eq. 1 as $sin \beta = \frac{bL}{(\frac{b^2}{4} + L^2)}$ which then gives me a quadratic equation in terms of L.. Then finding the positive root of this equation, I can calculate the normal vector along L and find the position of peak point.. But I'm hoping there is an easier way for it :)
Thanks for any help!
 A: Addendum added to respond to the comment of wololoo.

$\displaystyle \alpha = \frac{180 - \beta}{2}.$
$L = a \times \sin(\alpha).$
Therefore, the problem reduces to finding $~\displaystyle \frac{b}{2}$, which is immediate, by the Law of Cosines.
$b^2 = 2a^2 - 2a^2\cos(\beta).$
Edit
Alternatively, $~\displaystyle \frac{b}{2} = a\cos(\alpha).$

Addendum
Responding to the comment of wololoo.

...then what do you mean by the phrase location of the upper point?


The illustration is just one i found on Google, the actual triangle does not have to be "upright", the points' y coordinates can be different. I want a general solution to find the coordinates of the "peak" point.

Under the assumption that the base of the triangle makes an angle of $(0^\circ)$ with the horizontal, then (as stated) the Cartesian coordinates of the upper vertex are $~\displaystyle \left[a\cos(\alpha), a\sin(\alpha)\right].~$ Here, it is understood that $~\displaystyle \frac{b}{2} = a\cos(\alpha), ~L = a\sin(\alpha).$
In effect you are questioning my assumption that the base of the triangle makes an angle of $(0^\circ)$ with the horizontal, and asking what the effects would be if the base of the triangle was rotated $(\theta)^\circ$ (counter-clockwise).
The effect would be that each point on the triangle would be similarly rotated.  To visualize this, see the diagram to the right in this section of wikipedia trig identies : angle sum identities.
To actually determine the resulting Cartesian coordinates, the easiest approach is to go back and forth between polar coordinates and Cartesian coordinates.
Assuming that the base of the triangle makes an angle of $(0^\circ)$ with the horizontal, the polar coordinates of the vertex are simply $(a,\alpha)$, where $a$ is the magnitude of the distance, and $\alpha$ is the angle that the line segment makes with the horizontal.
The polar coordinates of $(a,\alpha)$ are equivalent to the Cartesian coordinates of $[ ~a\cos(\theta), a\sin(\theta) ~].$
Under the assumption that the base of the triangle (instead) makes an angle of $(\theta)$ with the horizontal, you have that the adjusted polar coordinates of the vertex are simply $(a,\alpha + \theta).$  These adjusted polar coordinates translate to adjusted Cartesian coordinates of
$$[ ~a\cos(\alpha + \theta), a\sin(\alpha + \theta) ~]. \tag1 $$
It is unclear to me whether (1) above is the type of expression that you are looking for.  You can remove all reference to $\alpha$ in (1) above, replacing it with $~\displaystyle \frac{b}{2}$ and $L$.  However, I question the logic of that, because the magnitudes of $~\displaystyle \frac{b}{2}$ and $L$ were themselves computed via the cosine and sine functions applied against the angle $\alpha$.
Anyway, if that is your wish, you can use that

*

*$\cos(\alpha + \theta) = \cos(\alpha)\cos(\theta) - \sin(\alpha)\sin(\theta).$

*$\sin(\alpha + \theta) = \sin(\alpha)\cos(\theta) + \cos(\alpha)\sin(\theta).$
Then, using the above identies, you can replace $a\cos(\theta)$ with $~\displaystyle \frac{b}{2}~$ and replace $a\sin(\alpha)$ with $L$.  This allows the expression in (1) above to be equivalently expressed as
$$\left[ ~\frac{b}{2}\cos(\theta) - L\sin(\theta), ~L\cos(\theta) + \frac{b}{2}\sin(\theta) ~\right].$$
A: The bisector $L$ divides the isosceles triangle into two right triangles.
From one of those right triangles,
$$ \frac b2 = L \tan\frac\beta2, $$
so
$$ L = \frac b2 \cot\frac\beta2. $$
