Hobbyists often compete with their model rockets to determine which rocket flies the highest. On one test launch, a rocket was fired vertically upward. The angle of elevation to the top of the flight was measured from two points that were 20 m apart, on the same side of the launch site, and collinear with it. The angles measured at the two points were $66$ degrees and $37$ degrees. How high did the rocket fly, to the nearest metre?

I drew a diagram with the base floor, and a rocket flying upward, with two points on the same ground level of the floor, and a line from one point to another showing that the distance is $20~\text{m}$.

I then measured $\tan 66^\circ = h/x$, and $\tan 37^\circ = h/(20 - x)$, and solved the system of equations. I got the wrong answer though, and even verified in wolframalpha, verifying the fact that I did indeed get a wrong answer.

I looked for solutions online and people say they differentiate between the two lengths of the points as $x - 20$, and $x$, as opposed to $x$ and $20 - x$, wouldn't it be $20 - x$, and $x$, since $(20 - x) + x = 20$, which is the sum of the two sides, as opposed to $(x - 20) + x = 2x - 20$? I'm getting the wrong answers when having $20 - x$, but the right answers when having $x - 20$. Can anyone help me out here?


  • $\begingroup$ Imagine that x is, say, 37. You would agree that the two points were $17m$ and $37m$, no? So, 37 and $37-20$. $\endgroup$ – turkeyhundt Jan 22 '15 at 3:13
  • $\begingroup$ That's interesting, thanks for the explanation. How come though when thinking algebraically if you will (in terms of variables), $20 - x$, and $x$ feels like the more natural option? 20 - x + x = 20, and the distance between them is 20, but yet when a value is plugged into x, that no longer holds and x - 20 and x are the correct options. $\endgroup$ – user164403 Jan 22 '15 at 3:18
  • $\begingroup$ To get from the point in contention to the point "$x$", we are adding $20$, not $x$. Because they are $20$ apart, not $x$ apart. So, what number can we add $20$ to and get $x$? $x-20$. We should never be adding $x$ to anything. It's the farther point from the launch. EDIT: $(x-20)+20=x$ $\endgroup$ – turkeyhundt Jan 22 '15 at 3:21

The observer who measures the angle of elevation to the rocket to be $66^\circ$ is closer than the observer who measures the angle of elevation to be $37^\circ$.


Consequently, if $x$ is the distance from the closer observer to the point where the rocket is launched, then

$$\tan(66^\circ) = \frac{h}{x}$$

and, since the second observer is $20~\text{m}$ farther from the launch site,

$$\tan(37^\circ) = \frac{h}{x + 20~\text{m}}$$


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