The value of $$\sin 10° \sin 30° \sin 50° \sin 70°$$ is:

We have to find the numerical value of the product $$\sin 10^\circ\,\sin 30^\circ\,\sin 50^\circ\,\sin 70^\circ.$$

First, notice that the three angles $$10^\circ,\;50^\circ,\;70^\circ$$ can be written in the symmetric form $$x,\;60^\circ-x,\;60^\circ+x$$ if we choose $$x = 10^\circ.$$ This observation suggests using a standard identity for the product $$\sin x\;\sin(60^\circ-x)\;\sin(60^\circ+x).$$

To derive that identity, we begin with the product-to-sum formula

$$\sin A\;\sin B = \tfrac12\bigl[\cos(A-B)-\cos(A+B)\bigr].$$

Put $$A = 60^\circ+x,\;B = 60^\circ-x.$$ Then

$$\sin(60^\circ+x)\,\sin(60^\circ-x) = \tfrac12\Bigl[\cos\bigl((60^\circ+x)-(60^\circ-x)\bigr) -\cos\bigl((60^\circ+x)+(60^\circ-x)\bigr)\Bigr].$$

Simplifying the arguments of the cosines, we get

$$\cos\bigl((60^\circ+x)-(60^\circ-x)\bigr)=\cos(2x),$$ $$\cos\bigl((60^\circ+x)+(60^\circ-x)\bigr)=\cos(120^\circ).$$

Because $$\cos120^\circ=-\tfrac12,$$ the product becomes

$$\sin(60^\circ+x)\,\sin(60^\circ-x) = \tfrac12\bigl[\cos 2x - (-\tfrac12)\bigr] = \tfrac12\cos 2x + \tfrac14.$$

Multiplying this result by $$\sin x,$$ we obtain

$$\sin x\,\sin(60^\circ-x)\,\sin(60^\circ+x) = \sin x\bigl(\tfrac12\cos 2x + \tfrac14\bigr).$$

Next we rewrite $$\cos 2x$$ via the double-angle identity $$\cos 2x = 1 - 2\sin^2 x.$$ Substituting,

$$\tfrac12\cos 2x + \tfrac14 = \tfrac12(1 - 2\sin^2 x) + \tfrac14 = \tfrac12 - \sin^2 x + \tfrac14 = \tfrac34 - \sin^2 x.$$

Hence

$$\sin x\,\sin(60^\circ-x)\,\sin(60^\circ+x) = \sin x\bigl(\tfrac34 - \sin^2 x\bigr) = \tfrac34\sin x - \sin^3 x.$$

Now we recall the triple-angle formula

$$\sin 3x = 3\sin x - 4\sin^3 x.$$

Dividing both sides of this formula by 4 gives

$$\tfrac14\sin 3x = \tfrac34\sin x - \sin^3 x.$$

Comparing with the expression just obtained, we see that

$$\boxed{\;\sin x\,\sin(60^\circ-x)\,\sin(60^\circ+x) = \tfrac14\sin 3x\;}.$$

Now set $$x = 10^\circ.$$ Then $$60^\circ-x = 50^\circ$$ and $$60^\circ+x = 70^\circ.$$ Therefore

$$\sin10^\circ\,\sin50^\circ\,\sin70^\circ = \tfrac14\sin 3(10^\circ) = \tfrac14\sin30^\circ.$$

Because $$\sin30^\circ = \tfrac12,$$ the triple product equals

$$\sin10^\circ\,\sin50^\circ\,\sin70^\circ = \tfrac14 \times \tfrac12 = \tfrac18.$$

The given expression contains one more factor $$\sin30^\circ=\tfrac12.$$ Multiplying this with the above result, we find

$$\sin10^\circ\,\sin30^\circ\,\sin50^\circ\,\sin70^\circ = \bigl(\tfrac18\bigr)\times\bigl(\tfrac12\bigr) = \tfrac1{16}.$$

Hence, the correct answer is Option B.