# Trig questions

I never took trig in school and now I need to get some quick figures, and help in figuring out some other answers. This is not for school homework.

Q1: Take a square ABCD, bisect one side AB, and draw lines from the bisection point (A’) to the corners of the opposite side, C and D. What is angle DA’C in degrees?

Q2: How can I calculate either the angle subtended by a given line at a given distance or the ratio of the two perpendicular lines for a given angle?

What I’m looking for is a practical explanation of how to get results of the type I’ve described from a scientific calculator, keeping in mind that I know nothing about trig at all. What functions do I need to use, in what order, to get the answers to these (and similar questions.)

FYI, the practical reason for the question is that I need to know the viewing angles for people sitting at various distances from at movie theater screen. IOW: for a given screen width, at what distance does the screen fill 60 degrees of the viewer’s field of view (assuming a seat in the center)? Alternately, if the viewer is x times the screen width from the screen, what angle is his field of view?

Please keep in mind that I’m completely ignorant of trig!

Thanks.

For question 1, the relationship you’re looking for is the law of cosines. Assuming that A is connected to C and B to D by the sides of the square, triangles A’AC and A’BD are congruent right triangles, and therefore angle DA’C is equal to 180[sup]o[/sup] minus twice the value of angle AA’C. Triangle A’AC is a right triangle with side lengths x and 2x, so the hypotenuse length is xsqrt(5). Angle AA’C is opposite the side of length 2x, so by the law of cosines, we have 4x[sup]2[/sup] = x[sup]2[/sup] + 5x[sup]2[/sup] - 2x[sup]2[/sup]sqrt(5)cos(AA’C). Therefore, cos(AA’C) = 1/sqrt(5), and angle AA’C is about 63.4[sup]o[/sup], and angle DA’C is about 53.1[sup]o[/sup].

I don’t understand your second question.

ETA: If you don’t know any trig, you’re going to have to learn some, as it covers a lot more than plugging numbers into a scientific calculator. You’ll also need to remember a lot of high school geometry.

Thanks.

Although I’m all for learning trig one of these days, for the small set of problems I have now, I’m hoping a simple set of calculator or (even better) spreadsheet functions will get me the answers I need in the short term.

Basically, it’s the relationships between angles and distances I outlined in the next-to-last paragraph of the OP: Assuming a unit screen width, if I know the viewing angle, what is the distance to the screen, and if I know the distance, what is the angle? All I need for now are those inverse functions.

Can we put them into a spreadsheet formula?

While the given proof is valid, I’m going to try to dumb it down to simple words in case the #s confuse you. Simply put, a sine, cosine, or tangent is an operator (not a number) that changes an angle into a relationship of sides of a triangle. It’s closer to being a plus sign than, say, the number 4.

Now to your problem. Let’s say you’re labeling the square like you’d read, so A and B are on the top and C and D are on the bottom. A’ is between A and B on the top. Create a new point called M, for “Middle of screen”. It’s directly opposite A’ and sits between C and D. If you draw the lines you’ve described and add in a dotted A’M line, you should see intuitively (you can proveit, but shouldn’t have to) that there are 4 congruent triangles. Now all we need are the measurements of 2 lines of any of those triangles. If you plug in the size of the screen, that gives you the long side. If you halve it, that gives you the short side. Now we just need the trig function:

Tan(Angle)=x/2x=1/2 because Tangent describes the opposite side over the adjacent side. The opposite of the Tangent is the Arctangent. So Arctan(.5)=26.5degrees.

As for question 2, you need to know either 2 sides of the triangle or a side and angle. The rest can be done easily with the functions. If you know the measurements of the 2 adjacent sides, you use ArcCos(short/long)=angle. For the known angle, you do Cos(angle)=ratio.

On the windows calculator, I just punched in values for the angle until i found one close enough to .5. Not sure how to do secant, cosecant, or any arc functions. I missed the edit window for above, but I forgot to mention that the angle you find will need doubled if you’re asking how far it is from left to right. If you just want center to side, don’t double the number.

For secant, calculate cos and then take the reciprocal (1/x). For cosecant, calculate sin and then take the reciprocal. For arcsin, hit the “inv” check box followed by “sin”, etc. Don’t forget that if you want your answer in degrees, the “degrees” radio button must be highlighted (it’s the default).

File this under “lies to children” in that, while it may aid understanding, it’s not actually correct. sin, cos etc are instances of what we call “functions” in math, which is to say, a black box with an input hopper that you drop a number into, and an output hopper that it drops a number into when you crank the handle. For any angle there is one and only one sine, one and only one cosine, etc. There’s actually something rather elegant and beautiful inside the black box, but you don’t need to worry about it for now. But yes, sin, cos and tan for any given angle do reflect the ratio between two sides of a right-angled triangle - and with a bit of jiggery-pokery you can get some useful information about non-right-angled triangles too.

“Sin(x)” is a function. “Sin” isn’t. “x+2” is a function as well, but “+” isn’t. Ok, yeah, if you want to literally define “Sine”, you’d have to say it’s a ratio and that the sine function calculates that ratio. Yet that definition always runs into the problem of someone going “Yeah, but if I imagine ‘sine’, what would it look like? I mean, I get that it yieldsa ratio, but what is it really?” and it always sets off the lightbulb if you tell them to think of it like a plus sign or something like that, not a value.

This really isn’t the thread to get into this, but you’re confused here. sin() and + are both functions. sin(x) is the function sin() evaluated at x , and x + 2 is the function + evaluated at the point (x, 2).

I wrote this post yesterday and thought I had sent it in, but the hamsters must have eaten it.

I’ve found this site, which does exactly what I was looking for: provide the angle for a given distance and vice versa.