math skills needed.

[NAPOTS]

why not just go 1/1000 scale and do 69"x25.4"
thats about 6'x2' still pretty big

[NAPOTS]

Yeah, now that I look at it again, there's really no need to calculate that imaginary triangle on the right or figure out the other angles like I did above.

You can just use the SAS (Side-Angle-Side) method instead, using the known angle (24°), and the known sides (1550m).

Ah well. It was fun anyway.

Its been too long since I have been out of geometry class to remember the names of any theorems used. I bisected the shape down the long axis cutting the 24deg angle in half. I then intersected this new line from the unknown angle at 90 deg. this gave me a right triangle with an angle of 12 deg in one corner and a hypotenuse of 1550m. I can then solve for the length of the other two legs. the other triangle no has 2 legs with known length. I can calculate the length of the third. now I have all of the information I need.

[Full Otto]

I wish I was still doing Autocad I'd have that knocked out in no time
Got lazy with all the formulas and scaling after learning AC

[Helen Keller]

my brain basically dumped everything but basic math years ago , i can calculate trajectory and shit like that but im all retarded when it comes to things like "imaginary triangles"

[Altarboy]

I seriously cannot wrap my mind around such math. I guess that is why I hold a sterring wheel for a living.

[l921428x]

When can we expect pictures Helen?

[Helen Keller]

When can we expect pictures Helen?

actually, going to the hardware store to get some flat heat machine screws to start the base for it.

[N/A]

It would seem to me that the two angles marked as ∅, should actually be marked as 90 degrees. That would make it a "right angle". Thus you. could find the length by using a² + b² = c².

a = 400m
b = 1550m

400² + 1550² = c²

c² = 2562500

c = 1600.781059


ETA-- rough calculation gives a width of 677m

[NAPOTS]

It would seem to me that the two angles marked as ∅, should actually be marked as 90 degrees. That would make it a "right angle". Thus you. could find the length by using a² + b² = c².

a = 400m
b = 1550m

400² + 1550² = c²

c² = 2562500

c = 1600.781059


ETA-- rough calculation gives a width of 677m

Not quite 90 which I assume the author of the problem did just to be a dick

[N/A]

Not quite 90 which I assume the author of the problem did just to be a dick

Without knowing that angle, the problem verges on being unsolvable. Anyway, an angle of ∅ degrees is just a straight line. A ∅ degree angle doesn't exist.

[LAGC]

It would seem to me that the two angles marked as ∅, should actually be marked as 90 degrees. That would make it a "right angle". Thus you. could find the length by using a² + b² = c².

That was the mistake I made in the first place, assuming those two Ɵ angles were right-angles, but they aren't. Turns out they are slightly obtuse (greater than 90°).

Known right angles are usually designated by a little square box in the corner, if they are trying to give you that information, which I should have picked up on the first time.

The two barred O (Ɵ) symbols in that diagram were merely pointing out that both of those angles were congruent (the same as each other), which lets you know that both of those long sides of that Star Destroyer drawing are the same length (1550m) as well.

Anyway, an angle of ∅ degrees is just a straight line. A ∅ degree angle doesn't exist.

The slashed O (∅) -- "null" or "empty set" -- is not the same thing as a barred (Ɵ).

After I thought about it for awhile, I actually do recall seeing the barred O (Ɵ) symbol used to represent an unknown angle before in a triangle diagram. It's usually something in a given problem you have to solve for.

Without knowing that angle, the problem verges on being unsolvable.

Not true at all, my friend. That's what the entire study of trigonometry is all about. Solving triangles with incomplete information like that.

The Pythagorean theorem (a² + b² = c², for use with right-triangles only) isn't the only weapon in the playbook.

That's what we've been discussing in this thread. There are other methods (theorems) like the ASS and SAS that you can use to figure out angles and lengths of triangles using the Law of Sines or Cosines as well.

Check this site out for example -- TRIANGLE CALCULATOR: https://www.triangle-calculator.com/?what=sas

Just shows you how possible it is to solve all kinds of different triangles, if you know just partial information like one angle and the length of two sides.

[N/A]

I admit I haven't done any trig since high school. That's fifty years ago.
I do like to sit around at times and figure out how to work out different ways to solve math problems. It doesn't seem there is enough info if one doesn't know those two angles, or at least the area involved as the other variable. As I said tho, I fifty years removed from trig.

[LAGC]

Yeah, to be honest with you, I wasn't sure I'd ever have any use for any of that trig stuff ever again either. Other than the occasional fun math problem, I certainly haven't used it at all in real life.

Which is why I leapt at the chance to give it a try. But even then, I had to whip out my Precalculus textbook for reference, as I couldn't remember those theorems from heart.

Try that Triangle Calculator out though, that link I posted above. Just plug in 1550 for sides a and b (the two long sides of the Star Destroyer) and 24 for the known angle (γ).

As you can see, it is able to calculate the unknown angles and unknown side (c) as 644.526 meters, which is the widest point we are after in the diagram.

[N/A]

Yeah, using a calculator comes up with the answer with no work involved; just plug and play. Trying to work it out with pencil and paper with limited knowledge of formula is where the fun comes in.

I like to look for ways to solve math problems if you don't have a calculator; ways to make it easier by hand and brain. I can punch in a number into a calculator to get a square root, but I have figured out a way to do small numbers just by using the last number and a set of 4 different tables of using 0 thru 10 in certain sets. Say if you want to know the square root of a seven digit number that ends in 2, then the 2 would tell you which table to go to and then you could find the right set of numbers in that group to tell you what the square root is. I got tired of working on it and haven't been on it for awhile.

[LAGC]

It never hurts to be able to do math problems by hand instead of using computational devices.

After all, you never know when we might be hit with an EMP and all our electronics go bye-bye.

[NAPOTS]

Without knowing that angle, the problem verges on being unsolvable. Anyway, an angle of ∅ degrees is just a straight line. A ∅ degree angle doesn't exist.

That's the Greek letter theta not the number zero, is commonly used to indicate an angle in the general sense as a variable just like x is used in algebra. The fact that both sides were marked theta tells me a) the value of the angle is not disclosed and b) both angles are the same. Also was a hint to me that it wasn't 90 even though it looked like it could be

Not unsolvable by any means without the angle, I explained how I did it and I only needed one trig function sine, and Pythagorean theorem

[N/A]

Yeah, I had to guess if it was theta or zero. I gueszed zero, and it was the wrong guess. Oh well.