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

Question

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 $$\begin {align*} \sin A&=x\\ A&=\sin^{-1} (x)\\ \sin B&=y\\ B&=\sin^{-1} (y)\\ \sin C&=z\\ C&=\sin^{-1} (z). \end{align*} $$ So, what's next?

Answer 1

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 $\pi/2, \pi/3, 0$.

Answer 2

HINT

Use the low of sines to prove 1) as the correct answer.

$2R =\frac a {\sin A}=\frac b {\sin B}$ then $2R = \frac a {\sin A}=\frac b {\sin A + x} = \frac {b-a} x$ therefore $b-a=2Rx$ where $x$ is the ratio of AP

Answer 3

Your answer as long as A+B+C = $\pi$

Answer 4

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.