Show that a formula for the altitude h from the vertex to the opposite side of a triangle is (below)?

Question

h=(a*sinB*sinC)/sinA
all A,B,C are angles.Sorry,i cant type them so i just use capital words.

Answer 1

Let AD be the altitude of ABC.
Hence, AD/AB = sinB
so AD=AB*sinB=c*sinB = a*sinB*sinC/sinA (Law Of Sine)

Answer 2

I think you mean that h runs from point A to the opposite side of the triangle "a".

In that case:

By the Law of Sines
a/sinA = b/sinB = c/sinC

From the definition of sine
sinC = h/b
h = b*sinC

But from the Law of Sines
a/sinA = b/sinB
b = a*sinB/sinA

Plugging back into h we have
h = b*sinC
h = (a*sinB/sinA)*sinC
h = (a*sinB*sinC)/sinA