If in a triangle $$ABC$$, the altitudes from the vertices $$A, B, C$$ on opposite sides are in H.P., then $$\sin A, \sin B, \sin C$$ are in

Let the sides opposite to the angles $$A, B, C$$ be $$a, b, c$$ respectively and let the area of the triangle be $$\Delta$$.

Altitude from vertex $$A$$ on side $$BC$$ (whose length is $$a$$) is
$$h_a = \frac{2\Delta}{a}$$.
Similarly,
$$h_b = \frac{2\Delta}{b},\quad h_c = \frac{2\Delta}{c}$$.

The statement “altitudes $$h_a, h_b, h_c$$ are in H.P.” means that their reciprocals are in A.P.:
$$\frac{1}{h_a},\; \frac{1}{h_b},\; \frac{1}{h_c}\ \text{are in A.P.}$$

But
$$\frac{1}{h_a} = \frac{a}{2\Delta},\; \frac{1}{h_b} = \frac{b}{2\Delta},\; \frac{1}{h_c} = \frac{c}{2\Delta}.$$
Since the common factor $$\frac{1}{2\Delta}$$ is the same for all three terms, the condition is equivalent to saying that
$$a,\; b,\; c\ \text{are in A.P.}$$

Now apply the Sine Rule:
$$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R,$$
where $$R$$ is the circum-radius. Hence
$$a = 2R\sin A,\quad b = 2R\sin B,\quad c = 2R\sin C.$$

The common factor $$2R$$ is again the same for all three terms, so the progression of $$a,b,c$$ transfers directly to the sines:
if $$a, b, c$$ are in A.P., then $$\sin A,\; \sin B,\; \sin C$$ are also in A.P.

Therefore $$\sin A, \sin B, \sin C$$ are in arithmetic progression.

Option B which is: A.P.