What is the thickness of the drop if its radius is r?

[Klaus von Faust]

1. A small drop of fat floats on the surface of a liquid whose surface tension is s. Surface fat tension at the air-fat interface is s1, at the fat-liquid interface is s2. Determine the thickness of the drop if its radius is r.2. $F=\sigma l$
$\delta P=\sigma (\frac 1 R_1 + \frac 1 R_2)$3. It is well known that a droplet of grease will have a "seed" shape if floating freely on water. I imaginary cut the seed in two, each half has it's contact angle, and I determined it using the fact that surface forces are in equilibrium.
$\sigma = \sigma_1 \cos\alpha_1 + \sigma_2 \cos\alpha_2$
$\sigma_1 \sin\alpha_1 =\sigma_2 \sin\alpha_2$
If I solve this system I get
$\cos\alpha_1=\frac {\sigma^2 +\sigma_1^2 -\sigma_2^2} {2 \sigma \sigma_1}$
I cannot figure out what to do next

 

[Nate810]

. Can you help me?4. Using the equations above, we can calculate the thickness of the drop by rearranging the equations and substituting in the values for surface tension, s1 and s2.The equation for the thickness of the drop, δ, is: δ = 2r(s1cosα1 + s2cosα2) / (s1sinα1 - s2sinα2)where r is the radius of the drop and α1 and α2 are the contact angles. Plugging in the values for s1, s2, r, α1 and α2, we can calculate the thickness of the drop.