Math Is Fun Forum

Hi Mukesh,

Got it!  The improved diagram below shows there are two cases, which lead to the same answer.  With my software (Sketchpad) I've only got the accuracy set to the nearest unit so one quadrilateral gives 17 and the other 16, but with greater accuracy they would be the same and equal to your required answer.

For the left hand diagram, you can calculate angle A.

Get two cosine rule expressions for BD and equate them.  [BD^2 = 3^2 + 6^2 - 2 * 3 * 6 * cos A etc]

Replace angle C with 120 - A and use compound angle formula to get an equation with cos A and sinA, find cos(A+ alpha) where alpha is known and thus find A. 

Then C = 120 - A.

Then use the half side x side x sine angle formula to get the area of ABD and also CBD and add them together.

You'll then have to do it all again for the second diagram.

Bob

Hi John,

I think I may have had a breakthrough!

See diagram below.  Is this what you want?

The angle C is to the X axis whereas before it was to the Y axis.  So I tried recomputing using 90 - C instead of C.  I then noticed that the results for E agree well if you replace E by 90 - E.  Here's what I mean in the table.

A    B    C               D                       E                  90-E                measured
12    20    80    -19.5965742     84.37256573    5.627434266            5.4
18    25    57    -23.91656506    63.78097927    26.21902073            24.9
32    14    70    -11.93889455    73.79421699    16.20578301            15.2

Do you think this might be it?

Bob

Hi John,

It's been a while .... I've been thinking and checking.

I downloaded Google Sketchup.  Try it; it's free and I think you'll find it useful.

I constructed a cube and did the rotations on it.  It went according to plan ... no anomalies found.

Then I added matrix multiplication to my Excel of your trial angles to check the points move as required.  They do.

You can get a copy of my cube and Excel from www.bundy.demon.co.uk/Puzzles/PuzzleIndex.htm

You'll also find my email on this site if you want to communicate directly.

Bob

Hi

That is brief!  I think a few added details plus diagram below will make it clear .

Draw perpendicular bisector of AE and place Q on this line so that AQE = 20
This triangle will be isosceles so the bottom angles are both 80.

The base AE = BC so AQE and ABC are congruent triangles.
Hence AB = AC = AQ  = EQ
So centre A, radius AC also goes through Q (and B)

QAE = 80 => QAC = 60 making AQC an equilateral triangle.

AQC - AQE = 60 - 20 = 40.

Then in the isosceles triangle QEC the angle QEC = 70.
Add  QEA = 80 and you're there.

I thought I had a solution but then it evaporated when I realised I'd assumed something that looked right but minus any proof!

But now I think I've cracked it.  It's a long proof so apologies.  If you've got a quicker way I'd be glad to see it.

Hi John,

We both have the same formula.  TAN C = SIN C / COS C.  Yours is just a simplification of mine.

Still the variation in values though?

So I got EXCEL to work.  (image below)

My formulas for E are pretty complex because you have to convert to RADIANS then back to DEGREES.

E (Bobs) is =DEGREES(ATAN(COS(RADIANS(A3))*SIN(RADIANS(C3))/(COS(RADIANS(D3))*COS(RADIANS(C3))+SIN(RADIANS(D3))*SIN(RADIANS(A3))*SIN(RADIANS(C3)))))

and E (Johns) is =DEGREES(ATAN(COS(RADIANS(A3))*TAN(RADIANS(C3))/(COS(RADIANS(D3))+SIN(RADIANS(A3))*TAN(RADIANS(C3))*SIN(RADIANS(D3)))))

As you can see they give the same values.

Hope that helps

Bob

OK, here we go.

General background:

In transformation geometry the transform is usually shown before the thing it operates on.

So that forces us to use position vectors (3 rows by 1 column) rather than coordinates.

A 3 by 3 matrix will transform a 3 by 1 vector to produce another 3 by 1 vector.

As M x 0 = 0 only transformations that leave the origin invariant are possible. (rotations around O, reflections if the line goes through O, stretches and enlargements centred on O and shears where the invariant line goes through O)

Multplication reminder:

If M =  a   b   c                      and         V = p
           d   e   f                                           q 
           g   h   i                                           r

then transformed V = V' = ap + bq + cr
                                       dp + eq + fr
                                       gp + hq + ir

Sorry, cannot put large brackets; hope the alignment reaches you as it leaves me.

I looked up the rotations on Wiki at http://en.wikipedia.org/wiki/Rotational_matrix#Three_dimensions

Below the three rotations, it defines the direction of rotation.

As you want a relationship between angles, it doesn't matter how long OR and the rest are so I chose all to be 1 unit.
(OP = OQ = OR =1)

That gives OP =     sin C                               OQ =       0                     OR =   cos A
                            cos C                                           cos B                              0
                               0                                              sin B                             sin A

In my diagram, I'm imagining OX coming towards me, OY to the right and OZ straight up.

To rotate R onto OX                use  M =     cos A        0        sin A
                                                                0            1          0
                                                            -sin A          0        cos A

This gives That gives OP' =     cosAsin C                 OQ' =     sinAsinB                         OR' =   1
                                              cos C                                     cos B                                      0
                                          -sinAsinC                                 cosAsin B                                   0

Next to rotate OQ' onto the XY plane.  Call the angle D.

M = 1            0             0
       0        cosD        -sinD
       0        sinD          cosD

and we only need to work out the Z component of OQ'' and set it equal to zero

sinDcosB + cosDcosAsinB  = 0                         =>      tan D  =  -cosAtanB

After checking the direction of rotation the minus is Ok because the + direction is from Y towards Z, and we want OQ' to drop down onto the plane.

Now to figure out the final rotation around OZ of angle size E.

Apply D to P' to give P'' =            cosAsinC
                                      cosDcosC + sinDsinAsinC
                                      sinDcosC  -  cosDsinAsinC

We only need to calculate ther x component and set it to zero

M  =  cos E         -sin E        0
         sin E          cos E        0
            0               0           1

=>  cosEcosAsinC  - sinEcosDcosc - sinEsinDsinAsinC  = 0

=>  tan E =  cosAsinC / (cosDcosC + sinDsinAsinC)

And that's where I've got to.  As my formula doesn't agree with your values, either I've slipped up somewhere or you're not talking about the same angle as me.  Please have a look and see if (i) it makes sense, (ii) you can spot what's wrong.  By all means come back to me for any clarifications.  Whatever, the method will work; it's just a case of ironing out the bugs.

Bob

Hi bobbym

To find out what the angle was I first used the sine rule.  Then I spent ages trying to establish connections between the sines of angles so that I could get the values without a calculator.  Eventually did it.

Then I thought:  the sine rule is usually proved using angle properties of circles, so there ought to be a circle (circles) that lead to a solution.  Trouble is there's loads of possibilities but eventually I hit on your point as the centre of a very helpful circle and that led to a solution very similar to yours.

But then I thought:  all well defined problems like this can be solved using the sine rule (I think) so maybe they're all solvable if you can find the right circle(s).  That's as far as I've got so far.

Bob

ps.  If you've got a book with more like this, perhaps you would post one (just one please, I've got things to do!)

Hi John

I, too, am retired.  Was a maths teacher for 37 years plus some ICT in recent years.

Values helped a lot.

The negative sign is correct.  To turn as required needs -19.5966 to go towards the plane.  Might seem unimportant but the final formula is complicated and may give the wrong angle if I use the wrong sign at this stage.

What I have is tan E = (cosAsinC) / (cos DcosC + sinDsinAsinC)

             where tanD = -cosAtanB

I'm still not 100% certain on this as it doesn't give 5.4 with your values.

Would you like to see the matrices?  I can include a reminder of matrix multiplication for you.

Explaining it properly might help to validate it or highlight any error.

Bob

My method will work although I wouldn't like to have to explain it as it involves matrices and vectors.

But I have a sign mistake somewhere as I got atan(-tanBcosA) for the angle around the x axis.

I cannot decide why this has happened so, if you can be patient another 24 hours, I try to figure this out, before  giving my final result.

My final angle expression is pretty nasty; I doubt I could get it by high school trig alone.  But it is a function of sine/cos/tan on the three angles as you thought.

Bob

ps.  Do you have a known answer for some values of A, B and C, so I can test it?

Hi Machinist60

Not sure I've fully understood your problem.  Here's how I picture it.

Is this correct?

Please say what you mean by angle A.  eg A = POY or whatever.