How to construct an isosceles triangle given the base angle and height to one side?

Question

I've got this math exercise. It reads:

Construct an isosceles triangle given the base angle and height to one side. How many solutions exist? Why?

I made a sketch, but still can't figure out how to do the construction. Any help is appreciated. $v$ and $\angle CAB$ are known.

Answer 1

The hint:

Firstly, construct $\Delta ALB.$

Now, let $M$ be a midpoint of $AB$, $M\in l$ and $l\perp AB$.

Let $l\cap BL=\{C\}$.

Thus, $\Delta ABC$ is the needed triangle.

Answer 2

note that all angles are known in the triangle $\Delta ABL$

Answer 3

Suppose the triagle is $\Delta ABC$ Let $\angle ABC=k$ be the base angle and $AC=BC$. Mapping the triangle in the coordinate system let $A=(0,0), B=(b,0), C=\left(\frac{b}{2},a\right)$ for some $b,a\in\mathbb{R}$. Then the equation of $BC$ is obtained by using the slope $\tan(\pi-x)$ and the point $B$ is $y=\tan(\pi-k)x-\tan(\pi-k)b$, while if we have used the same slope but the point $C$ we would have obtained the equation of $B$ as $y=\tan(\pi-k)x-\tan\left(\pi-\frac{k}{2}\right)\frac{b}{2}+a$. Hence equating coeffcient we get $$\tan(\pi-k)b=\tan\left(\pi-\frac{k}{2}\right)\frac{b}{2}-a$$ $$\text{or,}~~a=b\left(\frac{1}{2}\tan\left(\pi-\frac{k}{2}\right)-\tan(\pi-k)\right)$$ Now we are given that the height from $A$ on $BC$ is $h$, or $$\frac{\tan(\pi-k)b}{\sqrt{1+\tan^2(\pi-k)}}=h$$ Thus $a$ and $b$ are determined uniquely by $h$ and $k$.