Find The Height Of The Mountain

[chikis]

Homework Statement

A simple measuring device is used at point X and Y on the same horizontal level to measure the angles of elevation of the peak P of a certain mountain. If X is known to be 5,200m above sea level, |XY| = 4,000m and the measurement of the angles of elevation of P at X and Y are 15¤ and 35¤ respectively, find the height of the mountain. (take tan 15¤ = 0.3 and tan 35¤ = 0.7)

Homework Equations

Tan¤ = opp/adj

The Attempt at a Solution

I drew a large triangle XPY which consist of two small triangles, XPO and POY.
The distance |XY| is 4000m. |XY| is above sea level of heigt 5,200m.
The height am required to calculate is the distance |OP|. Let |OP|= h.
We need to find the distance XO and OY respectively. Let XO = (4000-b) and OY = b. Considering triangle XPO, tan15 = h/(4000-b)
---> (4000-b)0.3 = h
1200-0.3b = h
b = 1200-h/0.3 ----->(1)
Considering triangle POY tan35¤ = h/b
h = btan35¤
---> h = 0.7b
b = h/0.7 ----->(2)
Equating equation 1 and 2, you will have:
1200-h/0.3 = h/0.7
840 - 0.7h = 0.3h
0.7h + 0.3h = 840m
h = 840m
the height of the mountian will now become 840m +5,200m
= 6040m
but my answer is not the same as the answer provided for the working. How do I approach this problem?

 

[HallsofIvy]

Homework Statement

A simple measuring device is used at point X and Y on the same horizontal level to measure the angles of elevation of the peak P of a certain mountain. If X is known to be 5,200m above sea level, |XY| = 4,000m and the measurement of the angles of elevation of P at X and Y are 15¤ and 35¤ respectively, find the height of the mountain. (take tan 15¤ = 0.3 and tan 35¤ = 0.7)

Homework Equations

Tan¤ = opp/adj

The Attempt at a Solution

I drew a large triangle XPY which consist of two small triangles, XPO and POY.
This doesn't make sense until you tell us what 'O' is. Is O the foot of the perpendicular from P to the line XY?

The distance |XY| is 4000m. |XY| is above sea level of heigt 5,200m.
The height am required to calculate is the distance |OP|. Let |OP|= h.
We need to find the distance XO and OY respectively. Let XO = (4000-b) and OY = b.

You have these backwards. Y is closer to the mountain that X is so XO is larger than OY. It should be XO= 4000+ b. And, just as a matter of clarity, you should define b in "b= OY" before using it in XO.

Considering triangle XPO, tan15 = h/(4000-b)
---> (4000-b)0.3 = h

You should have 4000+ b.

1200-0.3b = h
b = 1200-h/0.3 ----->(1)
Considering triangle POY tan35¤ = h/b
h = btan35¤
---> h = 0.7b
b = h/0.7 ----->(2)
Equating equation 1 and 2, you will have:
1200-h/0.3 = h/0.7
840 - 0.7h = 0.3h
0.7h + 0.3h = 840m
h = 840m
the height of the mountian will now become 840m +5,200m
= 6040m
but my answer is not the same as the answer provided for the working. How do I approach this problem?

 

[chikis]

This doesn't make sense until you tell us what 'O' is. Is O the foot of the perpendicular from P to the line XY?

Yes O is the foot of the pependicular from P and meets the line XY at O.

You have these backwards. Y is closer to the mountain that X is so XO is larger than OY.

But the questions says that X and Y are on the same horizontal level and that XY is 4000m, that makes the it difficult to understand between X and Y which is closer and which is farer.

 

[eumyang]

But the questions says that X and Y are on the same horizontal level and that XY is 4000m, that makes the it difficult to understand between X and Y which is closer and which is farer.

In your original post:

A simple measuring device is used at point X and Y on the same horizontal level to measure the angles of elevation of the peak P of a certain mountain. If X is known to be 5,200m above sea level, |XY| = 4,000m and the measurement of the angles of elevation of P at X and Y are 15° and 35° respectively, find the height of the mountain. (take tan 15° = 0.3 and tan 35° = 0.7)

See bolded. Because the angle of elevation at Y is larger, Y is closer to O than X.

 

[chikis]

In your original post:


See bolded. Because the angle of elevation at Y is larger, Y is closer to O than X.

So if the angle of elevation at Y is larger and Y is closer to O than X, what do I make of that?

 

[chikis]

Thanks guys, I have made good of your post.

 

[chikis]

P
      /|\
     /|   \
   /  |     \
  /   |       \
 /    |         \
/__ _| _  _  _ _\
X     O           Y

That's how my diagram looks like.

 

[Mark44]

X and Y should be on the same side of the origin, (and the same side of the mountain). The diagram that you drew would not be useful to surveyors.

Point Y is closer to the mountain than point X, since the angle of elevation at Y is larger than the angle of elevation at X.

The idea is that the surveyors set up their equipment at point Y, determine the angle to the top of the mountain, and then move farther away. At point X, they get the new angle of elevation (which is smaller).

 

[eumyang]

That's how my diagram looks like.

X and Y should be on the same side of the origin, (and the same side of the mountain). The diagram that you drew would not be useful to surveyors.

chikis, see attached diagram.
(Mods, I hope this is okay. He/she's tried to draw a diagram for us.)

The idea is that the surveyors set up their equipment at point Y, determine the angle to the top of the mountain, and then move farther away. At point Y, they get the new angle of elevation (which is smaller).

I think you mean point X.

 

[Mark44]

chikis, see attached diagram.
(Mods, I hope this is okay. He/she's tried to draw a diagram for us.)

IMO, it's fine.

I think you mean point X.

Right you are - I meant X.

Edit: I've gone back and fixed that error.

 

[chikis]

chikis, see attached diagram.
(Mods, I hope this is okay. He/she's tried to draw a diagram for us.)

I think you mean point X.

But the question never said that X and Y are on the same side of the mountain. The question says that the angle of elevation of P at X is 15¤ and that the angle of elevation of P at Y is 35¤ and XY is 4000m which fits my diagram excatly.
With my diagram, if I use tan15 = h/(4000+b) tan35 = h/b, I will get 7300m as my final answer, but if use tan15 = h/(4000-b) and tan35 = h/b, I will get 6040m as the final answer. I think am actually confuse on how to get the distance XO and OY, is there any help I can get on that.

 

[eumyang]

But the question never said that X and Y are on the same side of the mountain. The question says that the angle of elevation of P at X is 15¤ and that the angle of elevation of P at Y is 35¤ and XY is 4000m which fits my diagram excatly.

Read mark44's post (#8) again. If X and Y were to be on opposite sides of O, the problem would have mentioned that.

With my diagram, if I use tan15 = h/(4000+b) tan35 = h/b, I will get 7300m as my final answer...

With your diagram, that's impossible. You would end up with |XO| being longer than |XY|.

... but if use tan15 = h/(4000-b) and tan35 = h/b, I will get 6040m as the final answer. I think am actually confuse on how to get the distance XO and OY, is there any help I can get on that.

You said it yourself in the original post!

I drew a large triangle XPY which consist of two small triangles, XPO and POY.
The distance |XY| is 4000m. |XY| is above sea level of heigt 5,200m.
The height am required to calculate is the distance |OP|. Let |OP|= h.
We need to find the distance XO and OY respectively. Let XO = (4000-b) and OY = b.

BTW, is 7,300m the answer that the book provided?

 

[chikis]

BTW, is 7,300m the answer that the book provided?

No, the book provided 6400 as the final answer which I think is not correct.

 

[eumyang]

No, the book provided 6400 as the final answer which I think is not correct.

It isn't. I'm getting 7,300 m as well (using tan 15° ≈ 0.3 and tan 35° ≈ 0.7). Must be a typo.

 

[chikis]

You must note that you will get 7,300m if only and only if X and Y are on same side of the mountain but if they at the opposite sides of the mountain as stated clearly in the question, then you will get 6040m which ever way you compute XO and OY. If you like take XO as tan15¤ = h/4000-b and tan35¤ = h/b or if you like take tan35¤ = h/4000-b and tan15¤ = h/b, you will still get 6040m.

 

[eumyang]

You must note that you will get 7,300m if only and only if X and Y are on same side of the mountain but if they at the opposite sides of the mountain as stated clearly in the question...

Really? I'm looking at the original post...

A simple measuring device is used at point X and Y on the same horizontal level to measure the angles of elevation of the peak P of a certain mountain. If X is known to be 5,200m above sea level, |XY| = 4,000m and the measurement of the angles of elevation of P at X and Y are 15° and 35° respectively, find the height of the mountain. (take tan 15° = 0.3 and tan 35° = 0.7)

... and I don't see it stated clearly that X and Y are on opposite sides of the mountain.

Once more I ask you, did you read Mark44's post (#8, emphasis mine)?

X and Y should be on the same side of the origin, (and the same side of the mountain). The diagram that you drew would not be useful to surveyors.

Point Y is closer to the mountain than point X, since the angle of elevation at Y is larger than the angle of elevation at X.

The idea is that the surveyors set up their equipment at point Y, determine the angle to the top of the mountain, and then move farther away. At point X, they get the new angle of elevation (which is smaller).

 

[chikis]

Once more I ask you, did you read Mark44's post?

Yes I did, and let me ask as well, is Mark44 a surveyor or an engineer to know that surveyors set up their equipment at point Y, determine the angle to the top of the mountain, and then move farther away. At point Y, they get the new angle of elevation (which is smaller).
Assuming X and Y are at the opposite sides of the mountain and XY is 4000m, how do you compute XO and OY respectively?

 

[Ray Vickson]

Yes I did, and let me ask as well, is Mark44 a surveyor or an engineer to know that surveyors set up their equipment at point Y, determine the angle to the top of the mountain, and then move farther away. At point Y, they get the new angle of elevation (which is smaller).
Assuming X and Y are at the opposite sides of the mountain and XY is 4000m, how do you compute XO and OY respectively?

It does not matter whether Mark44 is an surveyor, or what. By lots of experience, he can say that if the points X and Y are taken on opposite sides of the mountain, the question would have said so (or else the person setting the question was sloppy). In fact, the question leaves open the positions of X and Y; P could be due North of X and Y could be NNE of X, but closer to P, so there are probably infinitely many "solutions" that fit the original description (depending on the angle XOY). Basically, it is an unstated assumption that we want the simplest possibility that would also be physically the simplest: measuring the distance |XY| by surveying requires X and Y to be on the same side of the mountain; otherwise, some much more involved operations would be needed to measure |XY| when X and Y are on opposite sides of the mountain and thus invisible to each other. Furthermore, the mathematically simplest situation is when X, Y and O are on the same straight line.

RGV

 

[chikis]

It does not matter whether Mark44 is an surveyor, or what. By lots of experience, he can say that if the points X and Y are taken on opposite sides of the mountain, the question would have said so (or else the person setting the question was sloppy). In fact, the question leaves open the positions of X and Y; P could be due North of X and Y could be NNE of X, but closer to P, so there are probably infinitely many "solutions" that fit the original description (depending on the angle XOY). Basically, it is an unstated assumption that we want the simplest possibility that would also be physically the simplest: measuring the distance |XY| by surveying requires X and Y to be on the same side of the mountain; otherwise, some much more involved operations would be needed to measure |XY| when X and Y are on opposite sides of the mountain and thus invisible to each other. Furthermore, the mathematically simplest situation is when X, Y and O are on the same straight line.

RGV

Ok! Since the question did not state clearly the position of X and Y. We take it that they are two possible solutions, that for when X and Y are on same sides of the mountain, we have known, but what of when X and Y are at opposite sides of the mountain?

 

[Ray Vickson]

Ok! Since the question did not state clearly the position of X and Y. We take it that they are two possible solutions, that for when X and Y are on same sides of the mountain, we have known, but what of when X and Y are at opposite sides of the mountain?

Then you have a triangle XYP with interior angles of 15 degrees and 35 degrees given at X and Y, and you are given the length of the base XY. If h is the height (above the base) and |XO| = x, we have h/x = tan(15) and h/(4000-x) = tan(35). You can solve these two equations to get h and x.

RGV

 

[Mark44]

Yes I did, and let me ask as well, is Mark44 a surveyor or an engineer to know that surveyors set up their equipment at point Y, determine the angle to the top of the mountain, and then move farther away. At point Y, they get the new angle of elevation (which is smaller).

This deserves a response. No, I am not a surveyor, but I have two degrees in mathematics, so I understand the mathematics that surveyors use. Ordinary common sense would dictate that the surveyor would set up his equipment as I described in my earlier post, NOT on opposite sides of the mountain.

Assuming X and Y are at the opposite sides of the mountain and XY is 4000m, how do you compute XO and OY respectively?

This is a silly assumption to make. If X and Y are on opposite sides of the mountain, how would the surveyor determine that the points were 4000m apart? A direct measurement using a surveyor's chain would be impossible, because of the mountain being in the way.

The only reasonable interpretation here is that X and Y are on the same side of the mountain.

 

[chikis]

Then you have a triangle XYP with interior angles of 15 degrees and 35 degrees given at X and Y, and you are given the length of the base XY. If h is the height (above the base) and |XO| = x, we have h/x = tan(15) and h/(4000-x) = tan(35). You can solve these two equations to get h and x.

RGV

Thanks for that inteligent answer. Let me equally add:
If we take it that X and Y are at opposite sides of the mountain, then the figure I drew, will be more useful for the working. Working with my own figure, XY = 4000m, that I know and very sure because it has been stated already in the question. The distance XO and OY, we don't know, I just I just take it that XO could be = (4000-b) and OY could be = b, or OY = (4000-b) and XO = b. If you work with any of the assumption, you will still get (840m + 5200m) = 6040m as the heigt of the mountain.

 

[chikis]

This deserves a response. No, I am not a surveyor, but I have two degrees in mathematics, so I understand the mathematics that surveyors use. Ordinary common sense would dictate that the surveyor would set up his equipment as I described in my earlier post, NOT on opposite sides of the mountain.

This is a silly assumption to make. If X and Y are on opposite sides of the mountain, how would the surveyor determine that the points were 4000m apart? A direct measurement using a surveyor's chain would be impossible, because of the mountain being in the way.

The only reasonable interpretation here is that X and Y are on the same side of the mountain.

Mark44, don't be offended please, I only want to be sure of what am about to accept. Thanks!