If $\sin A$, $\sin B$, $\sin C$ are in AP, then $a$,$b$,$c$ are in If $\sin A$, $\sin B$, $\sin C$ are in AP, then $a$,$b$,$c$ are in..

$1$ AP, $2$.GP, $3$. HP, $4$.none
My attempt:
Let
$$\begin {align*}
\sin A&amp;=x\\
A&amp;=\sin^{-1} (x)\\
\sin B&amp;=y\\
B&amp;=\sin^{-1} (y)\\
\sin C&amp;=z\\
C&amp;=\sin^{-1} (z).
\end{align*}
$$
So, what's next?
If sin = 0, 1/2, 1,
then angles = 0, $\pi/6, \pi/2$.

Therefore not 1
($\pi/2-\pi/6 \ne \pi/6-0$),
not 2
(can't have a zero term),
not 3
(can't have zero).

Therefore 4. 

Note:
Same applies to cos,
with the angles being
$\pi/2, \pi/3, 0$.
We know $a/\sin A=b/\sin B=c/\sin C=2R$.
       So, by instruction $\sin A$, $\sin B$, $\sin C$ are in AP.
     So, $2\sin B=\sin A+\sin C$.
         Therefore $2b=a+c$.
       So $a$, $b$, $c$ are in also AP.