Similar to this question, when needing to measure out 1/2 tablespoon, I use my hemispherical tablespoon, but I hold mine at an angle and fill to the lowest edge. At what angle should I tilt so that my tablespoon is only half full?
Convention note: 0 degrees is level where the tablespoon holds a full tablespoon. 90 degrees would be holding the tablespoon sideways where it holds no liquid. In practice I usually hold at a little less than 45 degree angle, i.e. more level than tipped.
If the radius of the hemisphere is $R$ then the volume of the tilted spoon is
$ V = \pi ( \dfrac{2}{3} R^3 - a R^2 + \dfrac{1}{3} a^3 ) $
where $a = R \sin \theta $
This comes from the formula for the volume of a spherical cap of height $h$ in a sphere of radius $R$. The formula is listed here. What it says there is that the volume is given by
$ V = \dfrac{\pi h^2}{3} ( 3 R - h) $
In our problem, $h = R - a $. Therefore,
$ V = \dfrac{\pi (R - a)^2}{3} (2 R + a ) = \dfrac{1}{3} \pi ( R^2 - 2 a R + a^2 )(2 R + a) $
And this gives the formula above.
Now, we want to solve
$ \dfrac{\pi R^3}{3} = \pi ( \dfrac{2}{3} R^3 - a R^2 + \dfrac{1}{3} a^3 ) $
Take $R = 1 $, then
$ \dfrac{1}{3} a^3 - a + \dfrac{1}{3} = 0 $
i.e.
$ a^3 - 3 a + 1 = 0 $
Solving the general cubic equation is addressed here.
Here, our cubic equation is of the form
$t^3 + p t + q = 0 $
It is explained here how a trigonometric solution is obtained for this cubic equation. If $ 4 p^3 + 27 q^2 \lt 0 $ then there will be three solutions given by
$ t_k = 2 \sqrt{ -\dfrac{p}{3}} \cos \left( \dfrac{1}{3} \cos^{-1} \left( \dfrac{3 q }{2 p} \sqrt{ \dfrac{-3}{p}} \right) - \dfrac{2 \pi k }{3 } \right) \hspace{10pt} \text{ for } k = 0, 1, 2 $
This formula is due to the French mathematician François Viète.
Substituting $ p = -3 , q = 1 $ we first verify that there is indeed three real roots, because $4p^3 + 27 q^2 \lt 0 $. Then we have
$ t_k = 2 \cos \left( \dfrac{1}{3} \cos^{-1} \left( -\dfrac{1 }{2 } \right) - \dfrac{2 \pi k }{3 } \right) \hspace{10pt} \text{ for } k = 0, 1, 2 $
This simplifies further to
$ t_k = 2 \cos \left( \dfrac{2 \pi}{9} - \dfrac{2 \pi k }{3 } \right) \hspace{10pt} \text{ for } k = 0, 1, 2 $
So the three roots are
$t_1 = 2 \cos \left( \dfrac{2 \pi}{9} \right) = 1.53208888$
$t_2 = 2 \cos \left( \dfrac{4 \pi}{9} \right) = 0.3472963 $
$ t_3 = 2 \cos \left( \dfrac{ 10 \pi }{9 }\right) = -1.87938524 $
The only valid solution is $t_2$. Hence, we have
$ a = \sin \theta = 2 \cos \left( \dfrac{4 \pi}{9} \right) $
From which
$ \theta = \sin^{-1} \left( 2 \cos \left( \dfrac{4 \pi}{9} \right) \right) = \sin^{-1}(0.3472963) = \boxed{20.322^\circ}$
