/ Maths help please

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Deadeye - on 08 Apr 2019

Embarrassingly I'm struggling to solve a straightforward triangles problem.

Two triangles, one sides a b and 694; the other sides a b and 294.

The angle between sides b and 694 is the same as the angle between a and 294 = arctan(4.2)

I need a and b ... and seem to be lost in trigonometric identities.

Thanks

antdav - on 08 Apr 2019
In reply to Deadeye:

right angled triangles?

Deadeye - on 08 Apr 2019
In reply to antdav:

No

stevevans5 on 08 Apr 2019
In reply to Deadeye:

What links the two triangles? can you draw a picture to make things a bit clearer?

67hours - on 08 Apr 2019
In reply to Deadeye:

I'm not sure it's right angled triangles.

Maybe a picture would help? I made a quick sketch of how the triangles sit together...

https://imgur.com/a/6GtfzYi 

Post edited at 12:13
67hours - on 08 Apr 2019
In reply to Deadeye:

Actually, I drew the angles in the wrong place. Here's the correct version!
https://imgur.com/a/PJxgREH

Johnhi - on 08 Apr 2019
In reply to 67hours:

It's symmetrical(edit: not any more) split it in half and use the right angled triangle to find A then a couple of Pythagoras to get B

Post edited at 12:15
Deadeye - on 08 Apr 2019
In reply to Johnhi:

I haven't explained well. They are two separate triangles, which happen to have two sides and one angle the same

Deadeye - on 08 Apr 2019
In reply to 67hours:

> Actually, I drew the angles in the wrong place. Here's the correct version!https://imgur.com/a/PJxgREH


yes - that works

Deadeye - on 08 Apr 2019
In reply to Deadeye:

And I mean arctan(10/42) for the angle.. which amazingly you had autocorrected!

Luke90 on 08 Apr 2019
In reply to Deadeye:

Do the two side Bs form a straight line, as it appears they do in the diagram?

If so, you should be able to treat the whole thing as a single triangle to find B using the two known sides and known angle.

Edit: Only just noticed that the diagram wasn't yours. Guess the straight line appearance is just a coincidence.

Post edited at 12:34
67hours - on 08 Apr 2019
In reply to Deadeye:

Ha! I made a mistake and entered the angle as arccotan(4.2) by accident, which corrected your inverse error .

antdav - on 08 Apr 2019
In reply to Deadeye:

I think it's solved by dropping a couple of perpendicular lines in on 694 and A, and it ends up as a pair of simultaneous equations.

Ramblin dave - on 08 Apr 2019
In reply to Deadeye:

Out of interest, is this a problem that's arisen in a real world situation where the numbers are what they are, or is it a constructed one where the numbers have been carefully chosen to give nice round answers?

It seems to be possible by a fairly dumb approach of adding an extra line to each triangle that divides it into two separate right-angled triangles, titting around with trig identities and pythagoras to get a relationship between a and b for each triangle and then substituting the relationship from one triangle into the relationship from the other, but the numbers got ugly fairly fast when I actually started doing the maths...

antdav - on 08 Apr 2019

A=593 B=299

Deadeye - on 08 Apr 2019
In reply to Ramblin dave:

It's a real world issue (clearances in woodwork).

I got as far as simulataneous equations but with some very big numbers flying about.  Then made an error somewhere because got a blatantly wrong answer.

Deadeye - on 08 Apr 2019
In reply to antdav:

Thank you - what approach did you use?

Deadeye - on 08 Apr 2019
In reply to antdav:

Hmmm.  Ive just drawn that and it's not correct... which may mean I've mis-specified the problem somewhere.

I will try to draw the actual setup

https://imgur.com/a/DjkXPlD

Post edited at 14:18
Johnhi - on 08 Apr 2019
In reply to Deadeye:

Had some success, but disagrees with the above answer?  

Setting n=694, m=294 and x=arctan(4.2)

Cosine law gives equations

a^2=b^2 +n^2-2nbcosx

b^2=a^2+m^2-2macosx

Sub in a^2 and √a^2 into other eq.  Then rearrange to get the √ on one side, square it and get a quadratic eq which I just used excel to get a=1265.36 and b=1230.95

Deadeye - on 08 Apr 2019
In reply to Johnhi:

Hi and I mislead you it's arctan (10/42) not (42/10)

sorry...

antdav - on 08 Apr 2019
In reply to Deadeye:

I could easily have mistakes, its certainly not an elegant solution.

Tough to explain without being able to use superscript but see if you get the gist:

Take top triangle and split into 2x right angled triangles by adding a perpendicular from A.

New top right triangle has a side of 294 and angle of 13.39°. Use trig to find other sides which gives 286 (right part of A) and 68 on the new line drawn in.

Top left triangle now has sides of B and 69. Pythagorus gives remaining part of A.

Now you can define A in terms of B: A=286+sqrt(Bsquared + 68squared)

Onto bottom triangle, drop another perpendicular in. Left traingle has an angle and B so can the other lengths are Bcos(13.39) and Bsin(13.39).

Right triangle has sides of A and BSin13.39, pythagorus gives the last side as sqrt(Asquared+BSin13.39squared).

So 694 = BCos13.39+sqrt(Asquared+BSin13.39squared)

Sub in A=286+sqrt(Bsquared + 68squared) and solve for B.

694 = BCos13.39+sqrt((286+sqrt(Bsquared + 68squared))squared+BSin13.39squared)

Johnhi - on 08 Apr 2019
In reply to Deadeye:

The method should still be correct, touch wood, just need to use the correct angle when you work out the quadratic formula

stevevans5 on 08 Apr 2019
In reply to Deadeye:

I went for the cosine rule and simultaneous equation approach, and it gets numbers that work for one of the triangles but not the other... I ended up with a = 611 and b = 332 which seems to work for the a b 294 triangle but not the a b 694. 

Andy Hardy on 08 Apr 2019
In reply to Deadeye:

Daft question, but could you not draw it to scale, to find your answer (should be easy on CAD)?

Johnhi - on 08 Apr 2019
In reply to stevevans5:

213.73 & 488.60

stevevans5 on 08 Apr 2019
In reply to Johnhi:

I agree with that, I found an error in my maths!

antdav - on 08 Apr 2019
In reply to stevevans5:

me too, got my pythag backwards and didnt realise excel doesnt work in degrees.

Deadeye - on 08 Apr 2019
In reply to Johnhi:

Thank you everyone!

67hours - on 08 Apr 2019
In reply to Andy Hardy:

How do you think I drew the images above ;-)

Martin Hore - on 10 Apr 2019
In reply to Johnhi:

> 213.73 & 488.60

Got there too, about a day and a half late! Good problem to keep the geriatric mind alert.

I started by adding the two cosine formula expressions which gives a relationship between "a" and "b" (the squared terms cancel out). Did anyone else do that? I expect so.

I made a silly error moving on from there the first time which gave me impossible answers. 

Martin


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