isosceles triangle height and base given only angles and area If we know only that:
$A=22$
$v_1=30^\circ$ and $v_2=75^\circ$ and $v_3=75^\circ$
How do I get height and base without knowing at least one of the sides?
Edit: v are angles and A is the area.


We have that $A$ = $22$
Let $M$ be the midpoint of $AC$. Then $A = 2\cdot \frac{1}{2}\cdot AM \cdot MB = AM\cdot MB$
You also have the relationship $AM = AB\cos(75^{\circ})$ and $AC = 2AM$.
Let $AB = a \Rightarrow BC = a$
Let $AM = b$
Then we have the area of the triangle = $22 = \frac{1}{2}$ X base X height= $b\sqrt{a^{2}-b^{2}}$
We also have that $a\cos 75^{\circ} = b$
Subbing this in, we get $a^{2}\cos75^{\circ}\sin75^{\circ}=22$
You can use this to find $a$, then the rest all follows.
Note:
I get the base of the triangle = $2(\sqrt{33}-\sqrt{11})$
the height = $\sqrt{33} + \sqrt{11}$
