what is the sine of an isosceles triangle with sides 10^7 and h=1

Question

The angle of an isosceles triangle with sides = 10^7 and h = 1 is according to Wolfram $$ 1.000~000~000~000~000~4 \times 10^{-7} \quad \text{radians} $$ which corresponds to $$ 5.727 \times 10^{-6} \quad \text{degrees} $$ but that values gives the sine equal to $$ 9.9955 \times 10^{-8} $$ I imagined that the wolfram value is grossly approximated and spent a long time trying to find how many degrees give that sine but to no avail. The closest I could get was 5.729577951279..... no matter how many 9 I add I can't get that value

can anyone explain this riddle?

EDIT

I'd like to know with max precision the angle of the triangle and the value of its sine. The sine has fourteen 9'sif we use the radians and only 3 if we use the degrees, isn't that a huge difference for a calculator like wolphram's?

Answer 1

Which riddle? The two numbers are much the same: $1 * 10^{-7}=10 * 10^{-8}$

The two numbers differ for less than $0.1\%$, which is due to approximation.

Answer 2

Consider the expansion $$\sin^{-1}(x)=x+\frac{1}{6}x^3+O\left(x^5\right)$$ or better

$$\sin^{-1}(x)=\frac{x-\frac{17 }{60}x^3}{1-\frac{9 }{20}x^2}$$ Both of them would give (in degrees) the value after multiplying by $\frac {180}\pi$