0
$\begingroup$

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?

Thanks!

$\endgroup$
3
  • $\begingroup$ 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
1
$\begingroup$

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 $37^\circ$.

two_angles_of_elevation_to_the_same_point

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,

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.