[dhYTvlAv]

Been lurking on this forum for many years (pre-board days) and just recently thought of a practical use I might put it to, which is not say it hasn't been useful to me in the past.
Some time ago I passed the half century and took stock of a great deal of water that had gone by the bridge and decided that there were things amiss I should put right. Having had a fine education, I had been reasonably successful in my chosen field but found there were certain capacities that, under-utilised in my youth, had atrophied and dropped off. The single most irritating one was basic trigonometry. I can recall enduring the whole rigmarole and becoming reasonably proficient at it but (because I knew better) it was something I consigned to the dustbin of the mind and no matter what I have tried in the last several years I have been unable to pick up that knack again.
In the course of my employ I became used to manipulating spreadsheets and creating simple formulas to achieve simple results. I tumbled to the fact that you could use the functions of Excel to write a spreadsheet that would calculate the angles of a triangle from the given lengths of it's elements and even wasted a not inconsiderable number of hours trying to construe such a thing by myself.
Yet it came to nothing. I'm still an ignoramus. Moreover, I feel that this shortcoming is symptomatic of something deeper and I'd like to reacquaint myself again with the more sophisticated levels of high-school mathematics that have now seized up with the rust of passing time. Without returning to high school can anyone recommend some sort of real-world/online course that I could undertake that would act as a suitable refresher? Failing that can anyone help me with the design of the spreadsheet?

[voksveta]

Familiar with the cosine rule?

[voksveta]

Anyway, let a, b, c be the side lengths.
Let A, B, C be the angles opposite each of those sides, respectively.

We have:
a^2 = b^2 + c^2 - 2bccosA
b^2 = c^2 + a^2 - 2cacosB
c^2 = a^2 + b^2 - 2abcosC
.

[Vikonbarius]

a^2 = b^2 + c^2 - 2bccosA
b^2 = c^2 + a^2 - 2cacosB
c^2 = a^2 + b^2 - 2abcosC Row 1:
Angles of a traingle given the three side lengths
Row 2:
a b c A B C
Row 3 under headings A, B, C:
=ACOS(($B3^2+$C3^2-$A3^2)/(2*$B3*$C3))*180/PI()
=ACOS(($A3^2+$C3^2-$B3^2)/(2*$A3*$C3))*180/PI()
=ACOS(($A3^2+$B3^2-$C3^2)/(2*$A3*$B3))*180/PI()
Then input side lengths under headings a, b,c.

> Failing that can anyone help me with the design of the spreadsheet?

There are some spreadsheet experts on the forum. I'm not one of them.

You may be interested that I had a trigonometry problem to solve today. Point A is at (a,0,0), point B is at (0,b,0), point C is at (0,0,c). Find the area of the triangle ABC. It turned out that I could use the cross product to do it. Vector AB is (-a,b,0), Vector AC is (-a,0,c). The cross product ABxBC is (bc,ac,ab) with magnitude sqrt((bc)2+(ac)2+(ab)2) so the area of the triangle is half this.

Patted self on back figuring this out.

[RorieSorNearop]

A+B+C=180 degrees