Fitting a Triangle

"my aim is to calculate the distance of the maximum peak from the line joining origin and first minimum" - may I ask why?

I think you need to explain what you're really trying to do.

Paraphrasing your question:-

Can I draw a triangle by joining three dots. Can I then measure the height of one dot above the base formed by the other two.

Maybe he is trying to design a new fractal antenna.

I can't see the connection with sine waves, Mildred. You need to explain the problem to me a bit more.

You say you want to calculate the length from a point to a line. At what point of the line do you want the distance to be measured from?

If I understand the word picture you describe correctly, I do not see why you cannot use a triangle vs. arithmetic with x & y coordinates on your ordinate system.

Either way works.

It sounds like you have a graph similar to what is plotted below. The answer is 'Yes' you can solve for the sides, angles and area of the triangle - BUT you need to be working in 'pure' numbers.

Visually, the way you plot the graph can yield very different looking triangles, as you can see in the two examples below. But if you stick to pure numbers - algebra and trigonometry - you will get the same values, regardless of what the triangle looks like.

It looks harder than it is. Imagine a rectangular grid of 12 uniformly spaced points, 4 wide by 3 tall. Number them 1 to 4 (top), 5 to 8 (middle), and 9 to 12 (bottom). The lines go from #5 to #2, #5 to #12, and #2 to #12. If the vertical spacing is dy (the amplitude of the sine wave) and the horizontal spacing is dx (1/4 wavelength), then the line lengths are sqrt(dx^2 + dy^2), sqrt(9dx^2+dy^2), and sqrt(4dx^2+4dy^2). Once you know the lengths of the sides, you can calculate each angle from the length of the three sides using the law of cosines:

C=sqrt(A^2 + B^2 -2ABcos(c)) , where c is the angle opposite side C.

If my graph describes what the OP is asking, then the lengths and angles are given by the following, using pi since pure numbers are required.

h1 = √[1² + (pi/2)²]

h2 = √[1² + (3 pi/2)²]

h3 = √[2² + (pi)²]

A1 = arcsin(1/h1) + arcsin(1/h2)

A2 = pi - 2 arcsin(1/h1)

A3 = pi - A1 - A2

The area can be given by: AREA = ½ (1*pi) + ½ (1 * 3 pi/2) - ½ (1 * pi/2); AREA = pi.

GA; only answer so far that shows (h) what the OP asked.

Yes, Tornado 13, in the sense that a value for 'h' has been given, but that is just one possible answer because it could be any length from 'a' to 'c' depending on where the point is on the line AC.

No answer to my question in post 9 has been given.

The point used by Randall 12 for 'h' might be right, but it would be useful to know if this is the one the OP had in mind.

Surely the distance from a point to a line only has any practical meaning if it refers to the shortest distance?

Easily done using co-ordinate geometry.

' my aim is to calculate the distance of the maximum peak from the line joining origin and first minimum'

Do you mean a point 'anywhere' on the line? the mid point of the line, or one that is vertical ?

Intuitively I feel the answer can be derived in terms of frequency and amplitude.

The mathematicians can help here.

Yes it could fit. If your interest is to approximate the area bounded by it, say Volt* Seconds or Amperes* Seconds, why not use real sine function and integrate it?