Trig Word Problem Involving System of Equations
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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 $20~textm$.
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!
trigonometry
edited Jan 22 '15 at 11:27
N. F. Taussig
38.8k93153
asked Jan 22 '15 at 3:11
user164403
4691917
add a comment |Â
up vote
0
down vote
favorite
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 $20~textm$.
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!
trigonometry
edited Jan 22 '15 at 11:27
N. F. Taussig
38.8k93153
asked Jan 22 '15 at 3:11
user164403
4691917
Imagine that x is, say, 37. You would agree that the two points were $17m$ and $37m$, no? So, 37 and $37-20$.
– turkeyhundt
Jan 22 '15 at 3:13
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.
– user164403
Jan 22 '15 at 3:18
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: $(x-20)+20=x$
– turkeyhundt
Jan 22 '15 at 3:21
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
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 $20~textm$.
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!
trigonometry
edited Jan 22 '15 at 11:27
N. F. Taussig
38.8k93153
asked Jan 22 '15 at 3:11
user164403
4691917
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 $20~textm$.
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!
trigonometry
edited Jan 22 '15 at 11:27
N. F. Taussig
38.8k93153
asked Jan 22 '15 at 3:11
user164403
4691917
edited Jan 22 '15 at 11:27
N. F. Taussig
38.8k93153
edited Jan 22 '15 at 11:27
N. F. Taussig
38.8k93153
edited Jan 22 '15 at 11:27
N. F. Taussig
38.8k93153
38.8k93153
asked Jan 22 '15 at 3:11
user164403
4691917
asked Jan 22 '15 at 3:11
user164403
4691917
asked Jan 22 '15 at 3:11
user164403
4691917
4691917
Imagine that x is, say, 37. You would agree that the two points were $17m$ and $37m$, no? So, 37 and $37-20$.
– turkeyhundt
Jan 22 '15 at 3:13
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.
– user164403
Jan 22 '15 at 3:18
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: $(x-20)+20=x$
– turkeyhundt
Jan 22 '15 at 3:21
add a comment |Â
Imagine that x is, say, 37. You would agree that the two points were $17m$ and $37m$, no? So, 37 and $37-20$.
– turkeyhundt
Jan 22 '15 at 3:13
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.
– user164403
Jan 22 '15 at 3:18
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: $(x-20)+20=x$
– turkeyhundt
Jan 22 '15 at 3:21
Imagine that x is, say, 37. You would agree that the two points were $17m$ and $37m$, no? So, 37 and $37-20$.
– turkeyhundt
Jan 22 '15 at 3:13
Imagine that x is, say, 37. You would agree that the two points were $17m$ and $37m$, no? So, 37 and $37-20$.
– turkeyhundt
Jan 22 '15 at 3:13
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.
– user164403
Jan 22 '15 at 3:18
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.
– user164403
Jan 22 '15 at 3:18
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: $(x-20)+20=x$
– turkeyhundt
Jan 22 '15 at 3:21
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: $(x-20)+20=x$
– turkeyhundt
Jan 22 '15 at 3:21
add a comment |Â
1 Answer
1
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up vote
0
down vote
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
$$tan(66^circ) = frachx$$
and, since the second observer is $20~textm$ farther from the launch site,
$$tan(37^circ) = frachx + 20~textm$$
answered Jan 22 '15 at 11:25
N. F. Taussig
38.8k93153
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
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
$$tan(66^circ) = frachx$$
and, since the second observer is $20~textm$ farther from the launch site,
$$tan(37^circ) = frachx + 20~textm$$
answered Jan 22 '15 at 11:25
N. F. Taussig
38.8k93153
add a comment |Â
up vote
0
down vote
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
$$tan(66^circ) = frachx$$
and, since the second observer is $20~textm$ farther from the launch site,
$$tan(37^circ) = frachx + 20~textm$$
answered Jan 22 '15 at 11:25
N. F. Taussig
38.8k93153
add a comment |Â
up vote
0
down vote
up vote
0
down vote
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
$$tan(66^circ) = frachx$$
and, since the second observer is $20~textm$ farther from the launch site,
$$tan(37^circ) = frachx + 20~textm$$
answered Jan 22 '15 at 11:25
N. F. Taussig
38.8k93153
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
$$tan(66^circ) = frachx$$
and, since the second observer is $20~textm$ farther from the launch site,
$$tan(37^circ) = frachx + 20~textm$$
answered Jan 22 '15 at 11:25
N. F. Taussig
38.8k93153
answered Jan 22 '15 at 11:25
N. F. Taussig
38.8k93153
answered Jan 22 '15 at 11:25
N. F. Taussig
38.8k93153
answered Jan 22 '15 at 11:25
N. F. Taussig
38.8k93153
38.8k93153
add a comment |Â
add a comment |Â
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Imagine that x is, say, 37. You would agree that the two points were $17m$ and $37m$, no? So, 37 and $37-20$.
– turkeyhundt
Jan 22 '15 at 3:13
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.
– user164403
Jan 22 '15 at 3:18
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: $(x-20)+20=x$
– turkeyhundt
Jan 22 '15 at 3:21