trigonometry
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
Here is the situation as it seems to be described.
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The Law of Sines says that $$ \sin(\phi)=\frac{r+h}{r}\sin(\theta)\tag{1} $$ The two values of $\phi$ specified in $(1)$ correspond to the two points of intersection of the line with the circle. The one pictured above is $$ \phi=\pi-\arcsin\left(\frac{r+h}{r}\sin(\theta)\right)\tag{2} $$ The arclength from the base of the tower to the ground point of the cable is $$ \begin{align} r(\pi-\theta-\phi) &=r\left(\arcsin\left(\frac{r+h}{r}\sin(\theta)\right)-\theta\right)\\ &=6371\left(\arcsin\left(\frac{6371+100}{6371}\frac12\right)-\frac\pi6\right)\\ &=57.89\text{ km}\tag{3} \end{align} $$ where the angles in $(3)$ are given in radians.
Just as a point of comparison, the answer without considering the curvature of the earth would be $$ \begin{align} h\tan(\theta) &=100\frac1{\sqrt{3}}\\ &=57.74\text{ km}\tag{4} \end{align} $$
The formula in $(3)$ becomes the formula in $(4)$ when $r\to\infty$: $$ \begin{align} \lim_{r\to\infty}r\left(\arcsin\left(\frac{r+h}{r}\sin(\theta)\right)-\theta\right) &=h\lim_{r\to\infty}\frac{\arcsin\left(\left(1+\frac hr\right)\sin(\theta)\right)-\theta}{\frac hr}\\ &=h\lim_{t\to0}\frac{\arcsin((1+t)\sin(\theta))-\theta}{t}\\ &=h\lim_{t\to0}\frac{\sin(\theta)}{\sqrt{1-(1+t)^2\sin^2(\theta)}}\\ &=h\tan(\theta) \end{align} $$
Roughly speaking, in the formula $s=r\theta$, the symbol $s$ is the size (length) of the object you are observing, and $r$ is the distance from your eye to the object.
So in the mallard question, we have $s=18''$, and "how high it is flying" is $r$. Thus $r=\dfrac{s}{\theta}$. since you will want to give the answer in feet, you can do one of two things: (i) Set $s=18$, and compute $r$. The answer will be in inches, and you convert to feet by dividing by $12$; or (ii) Convert the $18''$ to feet ($1.5$) and compute.
For the hill, the distance you are from it ("$r$") is known, and you want $s$, the size of the hill. We use the formula $s=r\theta$. We know $r$, and $\theta$, so we multiply.
Best Answer
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.