Laying on the beach while sunseting question

1. The problem statement, all variables and given/known data
Suppose that, while lying on a beach near the equator of a far-off planet watching the sun set over a calm ocean, you start a stopwatch just as the top of the sun disappears. You then stand, elevating your eyes by a height H = 1.58 m, and stop the watch when the top of the sun again disappears. If the elapsed time is t = 10.9 s, what is the radius r of the planet to two significant figures? Notice that duration of a solar day at the far-off planet is the same that is on Earth.


2. Relevant equations
86400 s=1 day

3. The attempt at a solution
I attempted to solve this by setting the angle, but I think I got the wrong angle. (.0454°)
cos(0.0454°)=((r)/(r+h)) h=1.58
When I solved for the radius I got 1.5e^3, but it is not the right answer. Please help me figure where I went wrong! Thank you!

 

Hey,
Well I'm not quite sure as to how to get the angle. I tried two different ways so far: First, I tried: 10.9s/86400s=1.26e-4°. The second was me trying something crazy and doesn't make any sense: 10.9s*(1min/60s)*(1hr/60min)*(360/24hr)=.0454°

 

Okay, so that means that 1.26E^-4*360°=.04536° for the angle? Then I would still be getting my final answer as 1.5E^3

 

Okay, so when I solve for the angle I get .0454°
Then take the cos of the angle in which I get ≈ .9897
Then I set that equal to R/(R+h) h=1.58 → .9897=R/(R+1.58)
Solve for R by multiplying .9897(R+1.58)=R
Then get like terms of one side 1.59=R-R(.9897)
A R can be taken out 1.59=R(1-.9897)
Then I divide 1.59/(1-.9897)=R
R=154m
Is this the correct setup?

 

Wow, I would've never done that. Even my calculus professor didn't tell me that. Thank you so much guys for your help!

 

I have no preference about length of finding a solution as long as it's the correct one. Thank you so much once again!