The Goblins Fan Forum

It should be able to be calculated, as there's only one triangle that could work. However, my math knowledge seems to be failing me. 150 is a close estimate, but it may be a few degrees short of it based on actually building such a triangle. Looked to me to be around 147 degrees.

If people will have troubles solving this, I will provide a hint.

Perhaps, the key may be to define the lengths of the sides and apply the laws of sines, cosines, and tangents to get lengths of the legs of the triangles, and from there the measurement of the angles involved. Further, there is no reason not to create points E, F, and G by elongating line segments until they cross through the outer lines of the outer triangle.

Davecom3 wrote:Perhaps, the key may be to define the lengths of the sides and apply the laws of sines, cosines, and tangents to get lengths of the legs of the triangles, and from there the measurement of the angles involved. Further, there is no reason not to create points E, F, and G by elongating line segments until they cross through the outer lines of the outer triangle.

Yes, something like that came into my mind too, but I just did not have the time.

Since right and left line are equal, we might be able to calculate the other lines to provide the line length as the missing 5th formula I need.

I tried with E-F-G but had not enough information to get anything out of them (without line length).

Well if you extend the line CD it creates another isosceles triangle with sides equal to CB and AB. If the point of intersection along AB is labelled E, then CB = CE = AB.

Furthermore, the triangle AED is also isosceles, with AE = ED. Meaning also that BE = CD.

But it's been ages since I've done any real trigonometry so turning this into something more useful is going to take some time, when I have it. Right now the best I have is that the angle we're looking for is equal to the angle ABD + 80 degrees.

Well, the sine of angle BCD is equal to the sine of 20 degrees divided by the square root of ((the sine of 80 degrees multiplied by the sine of 30 degrees divided by the product of the sine of 50 degrees and the sine of 100 degrees) squared + (1 - (the sine of 80 degrees multiplied by the sine of 30 degrees divided by the product of the sine of 50 degrees and the sine of 100 degrees)) squared - (2 * (the sine of 80 degrees multiplied by the sine of 30 degrees divided by the product of the sine of 50 degrees and the sine of 100 degrees) * (1 - (the sine of 80 degrees multiplied by the sine of 30 degrees divided by the product of the sine of 50 degrees and the sine of 100 degrees)) * the cosine of 80 degrees)). What that means the angle equals I leave to somebody who has a functioning graphic calculator.

I think i did something wrong there. Going to recheck my work. Sine somehow wasn't positive when I managed to find batteries for my graphing calculator, so obviously I must have done something wrong.

Given that the law of sines starts us off.by, if we define the lengths of the side we know as 1, stating the length of the side opposite the 80 degree angle, which I shall define as x:
sin(50)/1=sin(80)/x
sin(50)*x=sin(80)
x=sin(80)/sin(50)

Now, given this we endeavor to prove the point where CD would bisect AB, which we will define as point E, which happens at a 100 degree angle for angle AEC.

We again use the law of sines to measure the length between A and this point, which I shall define as y:
x/sin(100)=y/sin(30)
x*sin(30)=y*sin(100)
x*sin(30)/sin(100)=y
(sin(80)/sin(50)*sin(30)/sin(100))=y

Now, BE measures 1-y, and given we know that 100 degrees+40degrees=140 degrees and that's 40 degrees short of completing a triangle, and given properties of isoscoles triangles we know DE is length y. We can next proceed by using the 80 degrees of BED to determine the length of BD, which I shall define as length z, using the law of cosines.

z=(y²+(1-y)²-2*(y)*(1-y)*cos(80))^1/2
z=(((sin(80)/sin(50)*sin(30)/sin(100))²+(1-(sin(80)/sin(50)*sin(30)/sin(100)))²-2*((sin(80)/sin(50)*sin(30)/sin(100)))*(1-(sin(80)/sin(50)*sin(30)/sin(100)))*cos(80))^1/2)

From this measure z, we can calculate, using the measure the angle of BDC by plugging values into the law of sines.

z/sin(20)=1/sin(BDC)
sin(20)=z*sin(BDC)
sin(20)/z=sin(BDC)
sin^-1(sin(20)/z)=BDC
sin^-1(sin(20)/((((sin(80)/sin(50)*sin(30)/sin(100))²+(1-(sin(80)/sin(50)*sin(30)/sin(100)))²-2*((sin(80)/sin(50)*sin(30)/sin(100)))*(1-(sin(80)/sin(50)*sin(30)/sin(100)))*cos(80))^1/2)))=BDC
This reveals that the angle we're looking for is 28.334490435743 according to my calculator, but this is obviously incorrect. That said, if we subtract this number from 180, we get 151.665509564257 degrees, which is rather close to the correct angle.

Did you switch the labels of some of the points? It seems like you start referring to point B as A halfway through there. ÔêáBDC is what we're looking for in the original problem. ÔêáADC is 140. ÔêáAED is ÔêáAEC, 100. ÔêáBED is 80.

Yes, I admit I did start calling it by the wrong name somewhere there, but nowhere that affects the calculation of the angle. Kept thinking of A as the top point for most of that.

And, yes, that was the answer I'm going to give, unless you require we be more precise.

FYI sin(80)/sin(100)=1 (for obvious reasons). And you calculated your expression wrong.

Can I use super threadromancy skills to revive this topic with a new riddle?

I won't block you but your sight,
You see me most in morning light.

Quite easy, I'd say ;-)

Go ahead I would say. I studied math, and some seem to have done the same or are at least on the tops of school math, but still I see no solution or further attemps for nearly a year.

For a while I meant to get back to the triangle problem and really give it some attention, but never quite had the time/inclination and then I just forgot about it entirely.

As for Guus' new riddle:
Eyelids? They block my sight and my eyes are generally at least half-closed for most of the morning.

Haha, I like your guess, but no

Sunglasses?

Fog.

Fog, yes!
Was it a good riddle? I made it up myself, like most other riddles I use for D&D sessions

Good enough for me.

I don't have any riddles, so whoever wants to post one is welcome to do so.
Or you can try and solve one that killed the thread. Here is a hint: you need to plot an additional regular polygon.

But they're all number riddles And I suck at those
A new normal riddle!

I'll be with you for hours on end,
Which could be bad, most at the end.
People still come back to me,
For need or joy the case may be.
What am I?

Bed?

Nope, but it fits the riddle ridiculously well

alky, *urp*, alcohol