Macro help for right angle diagram.

Jamtart · #1 08-26-2012, 04:09 PM

I am using macros to show the components of a right angle diagram. So far I can input an angle and generate the cosine of the angle, then enter a value for the Hponenous and I can get the total of the horizontal but then I have to use Pythagorean Theorem to solve for the side opposite. Im sure there is an other way to do the side opposite.
Here is my script thus far:
Private Sub angle_Change()
If Not IsNumeric(Me.angle) And Len(Me.angle) > 0 Then Me.angle.Text = Left(Me.angle, Len(Me.angle) - 1)

If Val(Me.angle) < 0 Then Me.angle.Text = "0"
If Val(Me.angle) > 90 Then Me.angle.Text = "90"
End Sub
Private Sub angle_KeyDown(ByVal KeyCode As MSForms.ReturnInteger, ByVal Shift As Integer)
Me.pfactor.Text = Cos(Val(Me.angle) / 57.29578)
pfactor = Round(pfactor, 3)
End Sub
Private Sub ind_Change()
If Not IsNumeric(Me.ind) And Len(Me.ind) > 0 Then Me.ind.Text = Left(Me.ind, Len(Me.ind) - 1)
If Val(Me.ind) < 0.0001 Then Me.ind.Text = ".0001"
If Val(Me.ind) > 1000 Then Me.ind.Text = "1000"
End Sub
Private Sub ind_KeyDown(ByVal KeyCode As MSForms.ReturnInteger, ByVal Shift As Integer)
Me.indx.Text = 2 * 3.14159265358979 * 60 * Val(Me.ind)
indx = Round(indx, 3)
Me.indh.Text = Val(Me.indx) * Val(Me.pfactor)
indh = Round(indh, 3)
Me.indv.Text = Sqr((Val(Me.indx ^ 2)) - (Val(Me.indh ^ 2)))
End Sub

Any suggestions would be greatly appreciated.

Jamtart · #2 08-26-2012, 07:08 PM

Ok, cruising right along. Cant get one more item at all. I have created a rt angle triangle, I divided the side adjacent by the Hypontenous, which gives you the Cosine. How can I convert this into an angle? For example, a cosine of .866 is 30 degrees.

JohnWilson · #3 08-27-2012, 01:51 AM

PowerPoint doesn't have an ACos function (unlike Excel) which makes it difficult to get the Angle from a Cosine

You can write a User defined function though (hope your math is good!).

Code:

Function GetAngle(ThisCos As Double) As Double
Const PI As Double = 3.141592
Dim Acos As Double
Acos = Atn(-ThisCos / Sqr(-ThisCos * ThisCos + 1)) + 2 * Atn(1)
GetAngle = Round(Acos * 180 / PI, 2)
End Function

From your normal code you would use this :

Code:

Dim theAngle As Double
theAngle = GetAngle(0.866)
MsgBox theAngle

Jamtart · #4 08-27-2012, 06:35 AM

Works perfect John. Would have been so much easier if PPT had the ACOS like excel, found that out in my readings trying to find a solution.
So I guess using Pythagorean Theorem (Me.indv.Text = Sqr((Val(Me.indx ^ 2)) - (Val(Me.indh ^ 2)))) in my original post was indeed the easiest way to solve for the side opposite (Vertical component of the triangle) after all.

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