hi KerimF

Apologies for not responding to this sooner. I had an idea when I read your first post but it has knocked around in my brain for some days before immerging as a method. It's got lots of steps and, if you wanted an actual formula, it could be done but it would be messy. Instead I'll outline an algorithm that will work.

If you just knew the three angles you could draw a triangle with those. But there are an infinite number of such triangles, all similar in shape, each scalable to scale up its size, but only one will have the right perimeter.

There's a formula, called Heron's formula after it's discoverer, that will calculate the area of a triangle if you know its perimeter. The semi perimeter = P/2 , sidea a, b, and c, and the formula is:

This Wiki page has an article about it https://en.wikipedia.org/wiki/Heron%27s_formula but there P already stands for the semi perimeter.

So how to make use of this.

Choose a random value for a; I'll call it p.

You can then use the sine rule to work out the other two sides, q and r.

Work out P' the perimeter of this triangle by P' = p + q + r

If P' = P then you're done because you had a really lucky choice for a. Unlikely though. But what you have got is the sides of a similar triangle that is either a scaled down or scaled up version of the correct triangle.

So you can work out the scale factor P/P' for converting your triangle to the correct one.

So now you know a, b, and c and so can proceed with Heron's formula.

Bob

Very nice work, thank you.

As expected, there are more than one way to solve it.

Here is how I did it:

(back from step 1, to show below the complete solution)

The area's formula of a triangle ABC is:

S = base * height / 2

As a start, I chose the base AB. In this case:

height = CM (from C to AB, M on AB)

And,

S = AB * CM / 2

Now, we have two right triangles ACM and BCM (right angle at M)

From them, let us find:

AB = f(CM,A,B)

In the triangle ACM

AM = CM/tan(A)

And, in the triangle BCM

MB = CM/tan(B)

So AB=f(CM,A,B) is:

AB = AM+MB = CM*[1/tan(A) + 1/tan(B)]

===

Again, in the triangle ACM

AC=CM/sin(A)

And, in the triangle BCM

BC=CM/sin(B)

So,

P = AB+AC+BC = CM*[1/tan(A)+1/tan(B)+1/sin(A)+1/sin(B)]

P = CM*[cos(A)/sin(A)+cos(B)/sin(B)+1/sin(A)+1/sin(B)]

P = CM*{[1+cos(A)]/sin(A) + [1+cos(B)]/sin(B)}

===

Now, it is time to substitude CM with S from the area's formula:

S=AB*CM/2 ==> CM=2*S/AB

P = 2*S/AB*{[1+cos(A)]/sin(A) + [1+cos(B)]/sin(B)}

And we get:

AB = 2*S/P*{[1+cos(A)]/sin(A) + [1+cos(B)]/sin(B)}

===

Similarly, we can say:

AC = 2*S/P*{[1+cos(A)]/sin(A) + [1+cos(C)]/sin(C)}

CB = 2*S/P*{[1+cos(C)]/sin(C) + [1+cos(B)]/sin(B)}

===

By adding the above three formulas of AB, AC and CB

P = AB+AC+CB = 2*S/P*{2*[1+cos(A)]/sin(A) + 2*[1+cos(B)]/sin(B) + 2*[1+cos(C)]/sin(C)}

P = 4*S/P*{[1+cos(A)]/sin(A) + [1+cos(B)]/sin(B) + [1+cos(C)]/sin(C)}

And the required formula/function is:

S = f(P,A,B,C) = P^2 / {[1+cos(A)]/sin(A) + [1+cos(B)]/sin(B) + [1+cos(C)]/sin(C)} / 4

Kerim

]]>Apologies for not responding to this sooner. I had an idea when I read your first post but it has knocked around in my brain for some days before immerging as a method. It's got lots of steps and, if you wanted an actual formula, it could be done but it would be messy. Instead I'll outline an algorithm that will work.

If you just knew the three angles you could draw a triangle with those. But there are an infinite number of such triangles, all similar in shape, each scalable to scale up its size, but only one will have the right perimeter.

There's a formula, called Heron's formula after it's discoverer, that will calculate the area of a triangle if you know its perimeter. The semi perimeter = P/2 , sidea a, b, and c, and the formula is:

This Wiki page has an article about it https://en.wikipedia.org/wiki/Heron%27s_formula but there P already stands for the semi perimeter.

So how to make use of this.

Choose a random value for a; I'll call it p.

You can then use the sine rule to work out the other two sides, q and r.

Work out P' the perimeter of this triangle by P' = p + q + r

If P' = P then you're done because you had a really lucky choice for a. Unlikely though. But what you have got is the sides of a similar triangle that is either a scaled down or scaled up version of the correct triangle.

So you can work out the scale factor P/P' for converting your triangle to the correct one.

So now you know a, b, and c and so can proceed with Heron's formula.

Bob

]]>My first step was to write the area's formula of a triangle ABC.

S = base * height / 2

I chose:

base = AB

height = CM (from C to AB, M on AB)

So,

S = AB * CM / 2

Now, we have two rectangular triangles ACM and BCM (or CMA and CMB, if you like).

The second step was to find at its end:

AB = f(CM,A,B)

...

]]>KerimF wrote:The perimeter (P) of a triangle ABC is known, also its three angles (A), (B) and (C).

Find its surface (S).

S = f(P, A, B, C)a triangle is 2D

surface is 3D

do you maybe mean 'area'?

Sorry, I was French educated at school. You are right, I meant by surface the area of the triangle.

]]>The perimeter (P) of a triangle ABC is known, also its three angles (A), (B) and (C).

Find its surface (S).

S = f(P, A, B, C)

a triangle is 2D

surface is 3D

do you maybe mean 'area'?

Find its area (S).

S = f(P, A, B, C)]]>