UKH

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

right angled triangles?

No

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

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 

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

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

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

yes - that works

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

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.

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

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.

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...

A=593 B=299

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.

Thank you - what approach did you use?

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

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

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

sorry...

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)

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

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. 

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

213.73 & 488.60

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

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

Thank you everyone!

How do you think I drew the images above

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