[solved... sort of] Trigonometry Problem

[Ruairi]

I have been stuck on the problem shown below. I've made many attempts but the answer will not show itself!

 

 

 

I'm solving for d

Edited June 29, 2017 by Ruairi

[studiot]

This is homework help so you need to post both the actual problem and at least one of your failed attempts (perhaps in outline).

 

Your jpg has not taken so I can't see the problem.

 

There is a moderator online right now (swansont) so ask him for help getting your question posted.

[Ruairi]

Ok thanks.

[studiot]

I presume that those right angles at the centre are a mistake?

 

This problem is known to surveyors and navigators as the resection problem.

that is they determine their position by taking bearing sights on three objects of known position.

 

One solution method is due to Tienstra

 

https://en.wikipedia.org/wiki/Tienstra_formula

 

Here is an online solver to play with

 

http://mesamike.org/geocache/GC1B0Q9/tienstra/

[Ruairi]

I have checked the textbook again. The question has the two right angles on either side.

 

This won't interfere with the resources you have allocated?

[studiot]

How many right angles in a straight line?

[Ruairi]

A straight line has 180 degrees so, two right angles???

[studiot]

A straight line has 180 degrees so, two right angles???

 

So what does that make the angle opposite the side you have labelled as d in the bottom triangle?

 

In other words what does that make angles alpha and beta?

Edited June 29, 2017 by studiot

[Ruairi]

That makes the angle opposite to d equal 180 degrees. But it can't because angles of a triangle add up t 180 degrees which would make alpha and beta equal to 0 which can't be right.

[studiot]

Exactly.

 

So something is wrong with the problem, or with your reading of it.

[Ruairi]

Thank you very much.

[studiot]

I suggest you ignore the right angles.

 

So you can easily establish alpha and beta, giving the angle opposite d (call it gamma).

 

This will give you d by the sine rule as asked for if you know y.

 

To get y by the sine rule you need to solve the upper right hand triangle.

 

Can you do this?

 

I made

 

alpha = 23.57818 degrees

 

beta = 29.55192 degrees

 

gamma = 126.86990 degrees

[imatfaal]

 

I think you have to assume that this a problem in 3d. You are looking into a right-angled corner

y is along the floor and h is up the corner of two walls etc. This would allow three right angles in the corner

 

1. Establish length of y

Tan(theta) = opposite/adjacent = h/y = 7.8/y

0.6 = 7.8/y

y= 7.8/0.6 = 13

 

2. Use sine rule

y/sin(alpha) = d / sin(90) = d/1

d= y/sin(alpha) = 13/ sin(alpha)

d= 13/(0.4) = 32.5

 

 

 

 

actually upon further inspection (ie sin(alpha+beta) =.08) I realise I made an improper assumption. Correction follows

 

 

 

 

Whilst this is a 3d problem one must not assume that this is a right angled corner - in fact the point that sin(alpha+beta) is not 1 means that angle opposite d cannot be right angle. To visualise image two walls with an angle not equal to 90 degrees between them - h is the vertical join of theses walls, y is a horizontal, the plane in which d is scribed is the floor

 

1. Establish length of y

Tan(theta) = opposite/adjacent = h/y = 7.8/y

0.6 = 7.8/y

y= 7.8/0.6 = 13

 

2. Use sine rule (gamma is angle opposite d)

y/sin(alpha) = d / sin(gamma)

d= sin(gamma)*y/sin(alpha)

 

2a. determine sin(gamma)

Note Sine( Pi - Angle) = Sine (Angle) as sine curve is symmetric around pi/2

But gamma + alpha + beta = Pi

gamma = Pi - (alpha + beta)

so

Sin(gamma) = Sin (Pi - (alpha+beta))

per above equality

Sin(gamma) = Sin (alpha +beta)

 

3. continue sine rule

d= sin(gamma)*y/sin(alph)

d= sin(alpha+beta)*y/sin(alpha)

d= .8*(13/.4)

d= 26

 

 

 

 

@studiot - I concur with your answers for the angles. Yet I still fail to see how you can solve the upper right hand triangle - you only know one angle and a side; that can create any number of triangles, all with different length y. Could you specify for me how you do this without assuming that the angle between y and h is a right angle?

[studiot]

 

I think you have to assume that this a problem in 3d. You are looking into a right-angled corner

y is along the floor and h is up the corner of two walls etc. This would allow three right angles in the corner

 

1. Establish length of y

Tan(theta) = opposite/adjacent = h/y = 7.8/y

0.6 = 7.8/y

y= 7.8/0.6 = 13

 

2. Use sine rule

y/sin(alpha) = d / sin(90) = d/1

d= y/sin(alpha) = 13/ sin(alpha)

d= 13/(0.4) = 32.5

 

 

 

 

actually upon further inspection (ie sin(alpha+beta) =.08) I realise I made an improper assumption. Correction follows

 

 

 

 

Whilst this is a 3d problem one must not assume that this is a right angled corner - in fact the point that sin(alpha+beta) is not 1 means that angle opposite d cannot be right angle. To visualise image two walls with an angle not equal to 90 degrees between them - h is the vertical join of theses walls, y is a horizontal, the plane in which d is scribed is the floor

 

1. Establish length of y

Tan(theta) = opposite/adjacent = h/y = 7.8/y

0.6 = 7.8/y

y= 7.8/0.6 = 13

 

2. Use sine rule (gamma is angle opposite d)

y/sin(alpha) = d / sin(gamma)

d= sin(gamma)*y/sin(alpha)

 

2a. determine sin(gamma)

Note Sine( Pi - Angle) = Sine (Angle) as sine curve is symmetric around pi/2

But gamma + alpha + beta = Pi

gamma = Pi - (alpha + beta)

so

Sin(gamma) = Sin (Pi - (alpha+beta))

per above equality

Sin(gamma) = Sin (alpha +beta)

 

3. continue sine rule

d= sin(gamma)*y/sin(alph)

d= sin(alpha+beta)*y/sin(alpha)

d= .8*(13/.4)

d= 26

 

 

 

 

@studiot - I concur with your answers for the angles. Yet I still fail to see how you can solve the upper right hand triangle - you only know one angle and a side; that can create any number of triangles, all with different length y. Could you specify for me how you do this without assuming that the angle between y and h is a right angle?

 

The only way I can make this work is to assume the pretty picture with the shading and no identifying letters is not a plane figure but a 3D picture of an internal corner.

 

Labelling points

A by angle alpha,

B by angle beta,

C at the top and

O at the centre/ origin / internal corner.

 

OC forms a vertical axis cut by plane CAB at C.

 

Triangle CAO is a right angled triangle (at O) as is CBO, but triangle OAB is not.

 

Thus triangles COA and COB are at right angles to the plane of AOB.

 

Thus in triangle CBO y = h/tan(theta)

 

This give three known angles and one side in triangle AOB, which can now be solved by the sine rule for d.

Edited June 29, 2017 by studiot

[Ruairi]

Studiot, I have attempted your modification of the question my methodology is shown below:

 

In triangle CBO:

 

θ = tan^-1(0.6)

θ = 30.96375653...

 

 

tan(θ) = h/y

y = h/tan(θ)

y = 7.8/tan(30.96375653...)

y = 13

 

 

In triangle AOB:

 

alpha = 23.57818 degrees

beta = 29.55192 degrees

gamma = 126.86990 degrees

 

a/sin(a) = b/sin(b) = c/sin© [sine rule]

 

sin(alpha)/y = sin(gamma)/d

 

d/sin(126.86990) = 13/sin(23.57818)

 

d = (13/sin(23.57818))*sin(126.86990)

 

d = 25.99999762

 

d = 26 (nearest whole number)

Imatfaal, Your proposal matches the answer in the back of the book. Though I'm not familiar with using pi in this topic.

[imatfaal]

 

Studiot, I have attempted your modification of the question my methodology is shown below:

 

In triangle CBO:

 

θ = tan^-1(0.6)

θ = 30.96375653...

 

 

tan(θ) = h/y

y = h/tan(θ)

y = 7.8/tan(30.96375653...)

y = 13

 

 

In triangle AOB:

 

alpha = 23.57818 degrees

beta = 29.55192 degrees

gamma = 126.86990 degrees

 

a/sin(a) = b/sin(b) = c/sin© [sine rule]

 

sin(alpha)/y = sin(gamma)/d

 

d/sin(126.86990) = 13/sin(23.57818)

 

d = (13/sin(23.57818))*sin(126.86990)

 

d = 25.99999762

 

d = 26 (nearest whole number)

Imatfaal, Your proposal matches the answer in the back of the book. Though I'm not familiar with using pi in this topic.

 

 

@Ruairi I tend to work in radians rather than degrees. The total internal angle of a triangle is pi radians. If you have not covered radians yet then just substitute 180 degrees for Pi and 90 degrees for Pi/2 and it will all make sense.

 

You will also note that you do not need to use a calculator - nor work out any trigonometrical values; all the values needed are given and you can do this with a pencil and paper. It is massively important to learn when you can do this - you are likely to be asked exactly this sort of question in exams in which you are not allowed a calculator!

 

@studiot OK so that's exactly how I proceeded above. Your post started with the injunction to ignore right angles so I just could not see what you were doing. thanks

[studiot]

 

 

@studiot OK so that's exactly how I proceeded above. Your post started with the injunction to ignore right angles so I just could not see what you were doing. thanks

 

Yes I saw this morning when I opened your spoiler that you had reached the 3D conclusion before I did.

 

I don't know if this was obvious in the textbook (for instance a chapter on 3D trig).

 

Also this question demonstrates the need for proper labelling of diagrams.

 

If the pretty picture was a scan from the textbook I would take proper labelling over beauty any time.

 

I think the authors were unduly lazy.

[imatfaal]

 

Yes I saw this morning when I opened your spoiler that you had reached the 3D conclusion before I did.

 

I don't know if this was obvious in the textbook (for instance a chapter on 3D trig).

 

Also this question demonstrates the need for proper labelling of diagrams.

 

If the pretty picture was a scan from the textbook I would take proper labelling over beauty any time.

 

I think the authors were unduly lazy.

 

Agree completely - I remember watching examiners attempt an ad hoc rationalization of similarly poorly written question; it was all intended to help draw out/highlight proper understanding of the subject etc. Then someone piped up - Was the fact that there was no section c, there were three spelling mistakes, and the diagram was wrongly labelled all part of the same teaching strategy; at that point the staff admitted that everyone thought someone else was checking the questions for basic quality control.

 

Did you see the news piece around 6 months ago about errors in exam revision texts? Some O'level and A'level science and maths primers had enormous error ratios in the self-test answers.

 

http://www.bbc.co.uk/news/uk-wales-38721478

[studiot]

 

Agree completely - I remember watching examiners attempt an ad hoc rationalization of similarly poorly written question; it was all intended to help draw out/highlight proper understanding of the subject etc. Then someone piped up - Was the fact that there was no section c, there were three spelling mistakes, and the diagram was wrongly labelled all part of the same teaching strategy; at that point the staff admitted that everyone thought someone else was checking the questions for basic quality control.

 

Did you see the news piece around 6 months ago about errors in exam revision texts? Some O'level and A'level science and maths primers had enormous error ratios in the self-test answers.

 

http://www.bbc.co.uk/news/uk-wales-38721478

 

 

Thank you

 

Several of my former clients were retired university lecturers who undertook A level marking for some extra income.

 

Most have since quit in disgust in disputes over unsatisfactory marking schemes and model answers, not to mention the online antics with computer based inputting.

 

(Did you see that recent question here as to why everything needs to be online?)

[imatfaal]

 

..

(Did you see that recent question here as to why everything needs to be online?)

 

No I missed that - but I can imagine. I am no old fogey - but I do think that a fashion in teaching allied to insidious cost cutting has led to a real decline in the methodology of examination. I do not hold with the "everything is getting easier" mantra that is so easy for those of us with pockets full of exams to trot out every August - I think children, young adults, and students in further/higher education work harder (and with less financial) support than I ever did.

 

I think we are at peak-online; we are realising that whilst online access can be world-changing somethings are just better off-line.

Once it is no longer a challenge to get stuff online it stops being an end in and of itself, and going to an online scenario is chosen because of a positive cost-benefit analysis rather than an online-good ideology.

[studiot]

 

No I missed that - but I can imagine. I am no old fogey - but I do think that a fashion in teaching allied to insidious cost cutting has led to a real decline in the methodology of examination. I do not hold with the "everything is getting easier" mantra that is so easy for those of us with pockets full of exams to trot out every August - I think children, young adults, and students in further/higher education work harder (and with less financial) support than I ever did.

 

I think we are at peak-online; we are realising that whilst online access can be world-changing somethings are just better off-line.

Once it is no longer a challenge to get stuff online it stops being an end in and of itself, and going to an online scenario is chosen because of a positive cost-benefit analysis rather than an online-good ideology.

 

 

I hope you are right, but fear otherwise for the underlined bit.

 

 

http://www.scienceforums.net/topic/107448-viruses-ransoms-trojans-spys-and-other-malware-into-corporations/