[syhprum (OP)]

This puzzle appears on Facebook adverts as part of a scam with the comment that 97% of the viewers cannot work out the angles a and c for the moment this includes me.
I am assuming the sides AB and AC are equal
I feel that simultaneous equations are needed or is there a more simple way ?

[RD]

... I feel that simultaneous equations are needed or is there a more simple way ?

I think it is down to simultaneous equations.

I am assuming the sides AB and AC are equal

IMO the fact the outer triangle is isosceles isn't helpful.

The values below seem to fit numerically ...
 [ Invalid Attachment ]
a = 120^o
c = 100^o
x =   60^o
y =   20^o

but they don't match the geometry of the diagram where angle y>x

[syhprum (OP)]

I omitted to label the angles that you have labelled x and y which I should have done, the assumption that it is Isosceles makes the calculations of the angles at B and C possible.
I am tempted to cheat and gun up Mathematica  on simultaneous equations

PS Y+X must equal 80°

[RD]

Just measured the angles on the diagram and they are quite accurate : within +/- 2^o,
So the solution must have y>x , (x approx 10^o), (so my answer, which has x>y, must be wrong  [:I] ) 

[Bored chemist]

It's a trick.
The three angles at A, B and C have to add to 180 degrees
B is 50, C is 50 so A must be 80
i.e. it's 180-(50+50)
But x+y = 80 and y =80 so x = zero.
there is no second triangle a, c, C.

[syhprum (OP)]

Y=70° X=10°   c= 150°  a=70 all done by guess work not proper maths it would not get marks in an exam bring on a maths expert and let us have a formal solution.

[CliffordK]

Facebook, may in fact be right.
97% of the viewers may be CORRECT!!!
and 3% of the viewers may be WRONG!!!

I came up with 4 equations, 4 unknowns.
1) a+c = 360 - 140 = 220  (360° around the central part)
2) x+c = 180 - 20 = 160 (180° around right most triangle)
3) y+a = 180 - 40 = 140 (180° around left most triangle)
4) x+y = 80 (80°is given for angle A, also can calculate with A+B+C = 180)

Can anybody see a fifth independent equation?

So, solving my equations:
1) a+c=220
2) x+c=160
3) y+a=140
4) x+y=80

Subtract (4) from (2), and one gets:
1) a+c=220
2-4) c-y=80
3) y+a=140

Add (2-4) + (3), and one gets:
1) a+c=220
2-4+3) a+c=220

SO,
My 4 equations and 4 unknowns were not independent.  And trying to solve them, I end up with 1 equation and 2 unknowns.

All angles have to be positive.
So, if you take the last equation (4), x+y = 80°

You can choose any x & y < 80°, and solve for a & c.

Unless there is another equation that I'm missing.

For example:
x = 1
y = 79
c = 159
a = 61

or
y = 1
x = 79
c = 81
a = 139

And, anything in between.
Thus, there is no unique answer.

[CliffordK]

BC,
I think you misinterpreted the original diagram.
Note the semi-circle going from lines AB to line AC, thus indicating the entire angle A  = 80°,
and x+y=80°
x must be non-zero, otherwise, one could not have the 20° component at C.

[CliffordK]

Whew, I'm rusty on my trig.

I should be able to figure out the lengths of sides for the lower triangle.
Then, from that determine the lengths of sides of the other two triangles.

[syhprum (OP)]

I think that calculation of the length of lines or distance of the central point from the long sides should pin down an unique solution.
as you point out x cannot be zero.
I should have kept my school Tech drawing kit!

[David Cooper]

I started off thinking it was meant to be a tetrahedron, but no - it has to be flat. If you name the point where c and a meet as point D, you can then draw in a vertical line down there to a point E on BC. Armed with that, you can start to assign lengths for all sides of the triangle BCD based on simple soh cah toa stuff, and then you can apply more to the triangle ABC in the same manner. That will show you if BA and BD are the same length. Whatever the case though, it will certainly force a single answer out.

Edit:-

Starting point: BD=1

BE = cos 10
DE = sin 10
EC = DE * tan 60
BC = EC + BE
BF = 0.5 * BC (F is the point half way along BC)
BA = BF / cos 40

So BA appears to be about 0.85 the length of  BD, if I've crunched the numbers correctly. Have I got it right thus far?

[CliffordK]

Ok,

I pounded a bit of brute force through it.

 [ Invalid Attachment ]

The outer triangle is an isosceles triangle, so one can set the outer edges to length 1.
From this one can calculate the length of the base.

Then I made the inner triangles into right triangles, and worked from the little triangle bottom right (30-60-90 triangle), and worked back around counter clockwise, up to where I could solve for the triangle with Y in it.

I ended up with y=70° (to the numerical precision of my calculations).

Then, from my equations above, one can solve the rest.
y=70°
x=10°
a=70°
c=150°

These angles seem a little bit bizarre, as if there is some geometric voodoo that I have forgotten.

All the numbers seem plausible based on the diagram, but I could always have made a mistake somewhere.

[syhprum (OP)]

Really hearty congratulations on a formal solution these are the same values that I produced by successive approximation guesswork.
According to the scam you are now eligible to pay out £4.50 a week to receive more puzzles

[David Cooper]

Ah, found my silly error - the last bit should have been BA = BF / cos 50 and not BA = BF / cos 40. Yes, BA=BD.

[CliffordK]

Ok,
Sorry about editing my diagram a couple of times, I had managed to let a layer shift, then added some inconsequential parenthesis to one of the equations.

Looking back at my diagram above.
I calculated that the length of d/sin(10) = `1  (length from B to middle).

This made my left most triangle into an isosceles triangle.

Thus, the angles a = y, and one easily calculates the angle to be 70°

Anyway, that simplifies the calculations a bit, but I've still got some thinking on why this all worked out like it did.

No idea why I need to pay £4.50 for new puzzles, when all I need is Syhprum to scramble my brain, and remind me that I'm no longer taking 11^th grade math!!!   [xx(]

[syhprum (OP)]

No SOH CAH TOA for us we had to memorise Peter Has (perpendicular over  hypotenuse)  Been Here (base over hypotenuse) Playing Billiards  (perpendicular over base) which seems to have worked  Knowing it after 70+ years

[CliffordK]

For me, the SOH, CAH, TOA mnemonic has been sufficient to remember my sin, cos, and tan relations long-term. 
But, it is infrequent enough that I do trig problems like this that I have to work a bit to remember how to use all the equations.  And, it never hurts to write my mnemonic down as I'm pulling out the equations.

I am still struggling a bit on whether or not one could have solved this "puzzle" without using the sin/cos/tan trig functions, or perhaps using them only superficially.

As far as math stuff.  I do wonder about future generations.  There is so much that one can look up online.  I missed the slide-rules, and did very little table lookup and interpolation, but landed early in the calculator generation.

We didn't have fancy graphing calculators, or calculators that could symbolically solve/simplify equations.  So, one learned things like a quadratic makes a U, a 3^rd order makes a sideways S, a 4^th order makes a W, etc.

Does having a calculator do it all for you make it easier, or harder?  I suppose half of the battle would be learning how to use the calculator.

[syhprum (OP)]

I grew up in the era when we had a complex money system that made domestic financial calculations difficult and the base unit the pound was very large about half my fathers weekly wage when I was a boy.
I did not like slide rules but found log tables very useful for interest calculations etc.
I do not think this calculation could be done without the use of trigonometry as the key to the whole thing is the bottom triangle about which we have complete information.
What I like about this modern era are spell checkers which have made me much more confident about writing previously being rather dyslectic. 

[damocles]

In my opinion there is a fairly simple solution. It involves use of the sin rule -- Y/sin y = Z/sin z

To make it explicit:

Triangle ABC is isosceles, so AB = AC

In the triangle ABX --- AX/sin 40° = AB/sin a
In the triangle ACX --- AX/sin 20° = AC/sin c

therefore --- sin a/sin c = sin 40°/sin 20° = 2 cos 20° (double angle formula)

moreover a + c = 220° (angles in a full circle)

This provides 2 equations in 2 unknowns.

[damocles]

The last few lines of my last post left me with a bit of work to do:

therefore --- sin a/sin c = sin 40°/sin 20° = 2 cos 20° (double angle formula)

moreover a + c = 220° (angles in a full circle)

This provides 2 equations in 2 unknowns.

Substituting (220° – c) for a leads to

(sin 220° cos c – cos 220°sin c) /sin c = 2 cos 20°

cot c sin 220° – cos 220° = 2 cos 20°

cot c = (cos 220° + 2 cos 20°)/sin 220° = (cos 40° –  2 cos 20°)/sin 40°

From tables I find that cos 40° = 0.76604444   cos 20° = 0.93969262   sin 40° = 0.64278761

Therefore cot c = (0.76604444 – 2 * 0.93969262) / 0.64278761 = –1.7320508

And therefore c = 150° and a = 70°

*****
The fact that the solutions prove to be such convenient round numbers convinces me that there should be an even simpler way of tackling this with geometric constructions rather than trig equations. More anon maybe?