Maths Problem - Impossible?

GregoryNimmo

My Maths teacher for higher level says that a question s impossble to do because it is misprinted. But i'm sure theres a way around it.

If anyone has "Text and Tests 4" Higher Level its page 72 question 9.

Responses on how to solve it would be much appreciated.

sellerbarry

Post it up for us.

GregoryNimmo

It is Trigonometry and is accompanied by a diagram. Sorry.

pathway33

A river flows due east and a tower [cd] stands on its left-hand bank. From a point a upstream and on the same bank as the tower, the angle of elevation of the top of the tower is 60 degrees. From a point b at right angles across from a, the angle of elevation is 45 degrees. If the height of the tower is 36 metres, find the width of the river, correct to the nearest metre.

Diagram given with question

UrbanSea

Post it in the mathematics forum I'd say you might have more luck.

pathway33

jubella

not impossible:

use tan to find |ad| and |bd|
because you already have the angle and the opposite side

then use pythagoras' theorem to find |ab| since abd is a right angled triangle.

partyatmygaff

It's pretty tough but definitely not impossible. I got the answer (29.08569824 m) after a page or two.

The trick is to take care of triangle acd first. Use the sin rule to get ¦ac¦ then use it again to get ¦ad¦ . Once you have all three angles and all three sides move on to triangle abc and find ¦ab¦. You have to realise that triangle abc is just a right angle triangle tilted slightly.

In summary, I only used simple sin/tan/cos ratios, the sin rule, phytaghorus' theorem and some basic geometry. The only thing that makes it such an awkward question is the fact that almost all the sides are irrational.

Meteoric

Thinking about it, and I could well be wrong, there is not enough information in the question. To solve it one would have to make assumptions such as that the two banks of the river are at the same height or that the distance from a to b is the width of the river. If you assume either the answer is as posted but the question is badly written as you should not have to assume, they should have told you that.

jubella

Meteoric wrote: »

Thinking about it, and I could well be wrong, there is not enough information in the question. To solve it one would have to make assumptions such as that the two banks of the river are at the same height or that the distance from a to b is the width of the river. If you assume either the answer is as posted but the question is badly written as you should not have to assume, they should have told you that.

That's getting too technical about it, it's a leaving cert book, you just have to take the question at face value. The river mentioned is just trying to make the question more interesting. The question really just is given a few triangles and angles of elevation, find the length of another side.

pathway33

Meteoric wrote: »

the distance from a to b is the width of the river. If you assume either the answer is as posted but the question is badly written as you should not have to assume, they should have told you that.

the width of the river is from a to b as shown in diagram attached

Meteoric

jubella wrote: »

That's getting too technical about it, it's a leaving cert book, you just have to take the question at face value. The river mentioned is just trying to make the question more interesting. The question really just is given a few triangles and angles of elevation, find the length of another side.

Oh I agree, I'm just trying to understand where the teacher is coming from is all. It's the only reason I can think of someone would say the calculation is impossible.

Chuchoter

Lol when a teacher says a question is impossible its not a challenge its an invitation to move onto the next one :P

J_E

partyatmygaff wrote: »

It's pretty tough but definitely not impossible. I got the answer (29.08569824 m) after a page or two.

The trick is to take care of triangle acd first. Use the sin rule to get ¦ac¦ then use it again to get ¦ad¦ . Once you have all three angles and all three sides move on to triangle abc and find ¦ab¦. You have to realise that triangle abc is just a right angle triangle tilted slightly.

In summary, I only used simple sin/tan/cos ratios, the sin rule, phytaghorus' theorem and some basic geometry. The only thing that makes it such an awkward question is the fact that almost all the sides are irrational.

Correct, says 29 on the back of the book as well. Sometimes it's easy to overcomplicate things, especially with the 3 dimensional triangles.

randylonghorn

crayolastereo wrote: »

Lol when a teacher says a question is impossible its not a challenge its an invitation to move onto the next one :P

Never been tempted to prove 'em wrong?

I used to love doing this (back when I was actually good at maths! >.< )

GregoryNimmo

partyatmygaff wrote: »

It's pretty tough but definitely not impossible. I got the answer (29.08569824 m) after a page or two.

The trick is to take care of triangle acd first. Use the sin rule to get ¦ac¦ then use it again to get ¦ad¦ . Once you have all three angles and all three sides move on to triangle abc and find ¦ab¦. You have to realise that triangle abc is just a right angle triangle tilted slightly.

In summary, I only used simple sin/tan/cos ratios, the sin rule, phytaghorus' theorem and some basic geometry. The only thing that makes it such an awkward question is the fact that almost all the sides are irrational.

The problem I have is, I can get all three sides and angles of triangle acd.

But when I get to the triangle abc I only have ac which is 41.54 and the angle A which is 90 degrees.

jubella

GregoryNimmo wrote: »

The problem I have is, I can get all three sides and angles of triangle acd.

But when I get to the triangle abc I only have ac which is 41.54 and the angle A which is 90 degrees.

try it the way I did it... |ad| = 36/tan60, |bd| = 36/tan45 , then use pythagoras' theorem to get |ab|... much easier IMO.

GregoryNimmo

jubella wrote: »

try it the way I did it... |ad| = 36/tan60, |bd| = 36/tan45 , then use pythagoras' theorem to get |ab|... much easier IMO.

That still doesnt work out...
The awnser is 29m (nearest metre) if you got that explain what you did and how you got your awnsers.

Its a toughie

GregoryNimmo

Nevermind everybody, I worked it out, Thanks to all of you.

I appreciate it very much and cant wait to show my teacher he was wrong :mad:.

"If at first you dont suceed, try and try again...:)

irelandrover

how can the angle of elevation from point a to the top of the be 60° and the angle of elevation from point b be only 45°? surely the angle of elevation from point b must be greater

jumpguy

irelandrover wrote: »

how can the angle of elevation from point a to the top of the be 60° and the angle of elevation from point b be only 45°? surely the angle of elevation from point b must be greater

No, it would be less as it is farther away (assumedly). Think about it, if you have a rope attached to the top of a pole and attached to ground near the pole, it will have a larger angle of elevation than one attached to ground farther from it.

irelandrover

jumpguy wrote: »

No, it would be less as it is farther away (assumedly). Think about it, if you have a rope attached to the top of a pole and attached to ground near the pole, it will have a larger angle of elevation than one attached to ground farther from it.

but that is different. that applies if they are at the same height. in this case they are both the same distance from the tower but point a is higher up.
looking at teh diagram given the angle of elevation from point a must be smaller than the angle of elevation from point b

lostatsea

I'd say this problem is of a part (b) standard. With 3d triangles always lift out the separate triangles and draw them as 2d diagrams. In most LC questions on 3d triangles most of the triangles are right-angled.

BingoMingo

1 get 29.6510 mtrs

Although they don't look it - because it is 3D

ACD is Right angled triangle - AC is 41.569219 mtrs
BCD is a right angled triangle - BC is 50.911688 mtrs

ABC is a right angled triangle - AC is still 41.569219 mtrs
- BC is still 50.911688 mtrs

Use Tan or Cos to get AB

Is this right?

partyatmygaff

You're on the right track but the answer should be 29.08569824 m. Your answer would round up to 30m which would be wrong.

jubella

GregoryNimmo wrote: »

That still doesnt work out...
The awnser is 29m (nearest metre) if you got that explain what you did and how you got your awnsers.

Its a toughie

The way I did it was...
|ad| = 36/tan60 = 20.78460969

|bd| = 36/tan45 = 36

pythagoras' theorem:

(36)2 = (20.78460969)2 + |ab|2

1296 = 432 + |ab|2

864 = |ab|2

29.39387691 = |ab|

correct to nearest metre = 29m

BingoMingo

partyatmygaff wrote: »

You're on the right track but the answer should be 29.08569824 m. Your answer would round up to 30m which would be wrong.

Yes I took a shortcut I had the length of the opposite and the hypothenuse and could calculate the sine but I forget how to convert this to an angle so I used Trial and error.

The sine is 41.569219 / 50.911688 = 0.8164966

How do I turn this into an angle?
I guessed at 54.5 degrees thought it was close enough.

(been a long time)

FoxT

I agree with jubella, but prefer to work with surds & leave the actual calculation till the end:


|ad| = 36/tan60 = 36/sqrt(3)|bd| = 36/tan45 = 36

pythagoras' theorem:

(36)2 = (36/sqrt(3))2 + |ab|2

|ab|2 = (36)2 - (36/sqrt(3))2
=> 2/3(36)2 = 24.36 (ie 24 x 36)

|ab| = 6 * sqrt(24)
= 29.398

as before.

The advantage of doing it this way (for me at least) is that the answer is much easier to check. If you have finger trouble with the calculator & you want to get say sqrt(824) it is easy in the heat of the exam to just write down the answer, no matter what the calculator tells you.
Doing it as above, we just use the calc for sqrt(24) and we know the answer is going to be close to 5, so less chance of a blunder.

TimeToShine

Ah I remember that page, If I'm right the question 11 is the (c) part from 1994, probably the hardest question I've ever seen.

However, Good job on out-playing all these guys if that was your plan by using the word "impossible". As soon as these Maths nerds see that out comes the pen and paper and BOOM your teacher thinks you're a genius

HxGH

randylonghorn wrote: »

Never been tempted to prove 'em wrong?

I used to love doing this (back when I was actually good at maths! >.< )

Best feeling when you out maths your maths teacher!