At what distance does a tree 24 ft tall subtend an angle of 10'? At what distance does a tree 24 ft tall subtend an angle of 10'?
this is what I got, 
given formula: d = rθ
θ = 1/60 degree * 3.14/180 degree = 2.9 x 10^-4 or 0.00029
d = rθ = (24)(.00029) = 0.00696 mile?
here is what i got, but my answer is not correct,
the correct answer for the problem is: 1.6 mile
but I have no idea how to get 1.6 mi
 A: In the formula that you are quoting, the height of the tree is the $d$, and the $r$ is the required distance (the "radius").  That is the opposite of the interpretation taken in the posted solution. 
Since $d=r\theta$, from algebra we get that $r=\dfrac{d}{\theta}$.  
First we calculate $\theta$. The angle is $10'$, that is, $10$ minutes. So $\theta$ is $\frac{10}{60}$ of a degree. In radians, that is $\dfrac{10}{60}\cdot\dfrac{\pi}{180}$. If at this stage you want to use a calculator (myself, I would wait), we get that $\theta\approx 0.0029089$. Note this is $10$ times as large as the number you computed, because the angle is $10'$, not $1'$. 
Now divide $d$, that is, $24$, by $\theta$. We get that $r\approx 8250.5922$.
But recall that we are working in feet, since the height of the tree was measured in feet. We must convert to miles. To convert, we divide by $5280$, since there are $5280$ feet in a mile.  
We get approximately $1.5626$. Round appropriately. 
Remark: The formula $d=r\theta$ relates the length $d$ of a circular arc, in a circle with radius $r$, to the angle $\theta$ subtended by the arc an eye at the centre of the circle. 
If we are on level ground, at distance $r$ from the base of a tree growing straight up, then the actual relationship is $d=r\tan\theta$.  
For "small" angles $\theta$, like ours, $\tan\theta$ is very close to $\theta$, if $\theta$ is measured in radians. In our case, to the display accuracy of my cheap calculator ($7$ decimal places) they are equal. For larger angles, the difference can be significant. 
