Height of hill when angle of elevation for each vertex is same 
The angle of elevation of the top of a hill from each of the vertices $A, B$ and $C$ of a horizontal triangle is ​​$\alpha$. Prove that the height of the hill is $\frac{a}{2} \tan\alpha \csc(A)$.

Could someone help me to approach this question. I am not getting how to start.
 A: In mathematical terms: You have a triangle $ABC$ in three space and a point $D$ (the top of the hill) not lying on the plane determined by $ABC$. Then $ABCD$ is a tetrahedron, i.e. a triangle pyramid. Let $H$ be the orthogonal projection of $D$ onto plane of $ABC$. Then $AH$ is orthogonal to plane of $ABC$. The three triangles $ADH, \, BDH, \, CDH$ share a common edge $DH$ and the following equalities hold $$\angle \, AHD = \angle \, BHD = \angle\, CHD = 90^{\circ}$$ $$\angle \, DAH = \angle \, DBH = \angle\, DCH = \alpha$$ Therefore the three triangles   $ADH, \, BDH, \, CDH$ are congruent. Hence $$AH = BH = CH$$ Consequently, the circle $k$ with center $H$ and radius $AH$ passes through all three points $A, \, B$ and $C$ which means that $k$ is the circumcircle of triangle $ABC$. By the law of sines in the triangle $ABC$ we have that $$2\,AH = \frac{BC}{\sin(\angle \, BAC)} = \frac{a}{\sin(a)} $$ assuming that $a = BC$ and $\angle \, BAC = A$. Hence $$AH = \frac{a}{2 \sin(A)}$$ By the tangent relation in right-angled triangle $ADH$ $$\frac{DH}{AH} = \tan(\alpha)$$ so 
$$DH = AH \, \tan(\alpha) = \frac{a \, \tan(\alpha)}{2 \sin(A)} = \frac{a}{2} \, \tan(\alpha) \csc(A)$$ 
For the record, I hate the functions $\csc(\alpha)$ and the other one (I cannot remember it). The most useless, annoying, awkward notation ever!  
A: radius of circumcentre                                                                       R= a/2sinA =b/2sinB= c/2sinC IN WHICH AH = BH =CH= R
