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Determining an Angle From Side Views

Date: 07/24/2002 at 23:41:51
From: Bob Frost-Stevenson
Subject: Compound angles

I have searched everywhere, and can't come up with the answer. I need 
to know, what is the simple angle of two adjoining angles? 

I can probably explain a bit better. In the end elevation of a 
drawing, a rod rises from point A at 32 degrees. In the side 
elevation, the same rod is seen rising from point A at 48 degrees. 

How do I work out at what angle to cut the end of the rod? Is it
simply adding the angles and dividing by two? 

Thank you very much.

Date: 07/25/2002 at 09:15:26
From: Doctor Rick
Subject: Re: Compound angles

Hi, Bob.

It isn't as simple as that. The easiest way to derive the formula is 
to use analytic geometry. Let the rod start from the origin of a 
cartesian coordinate system, with the end elevation being the 
projection of the rod onto the y-z plane and the side elevation being 
the projection of the rod onto the x-z plane. I'm not sure whether 
your angles are the angles from the vertical or from the horizontal, 
but it turns out that the formula is simplest if all angles are from 
the vertical, so that's how I'll define the angles.

The point on the rod at height 1 unit above the x-y plane has some 
coordinates (x,y,1). In the end elevation, this point has coordinates 
(y,1) and the tangent of the angle from the vertical, which I'll call 
alpha, is y/1. In the side elevation, the point has coordinates (x,1) 
and the tangent of the angle from the vertical (I'll call it beta) is 

The compound angle theta is the angle between the rod and the z axis. 
The distance between the point (x,y,1) and the z axis is 

  r = sqrt(x^2+y^2), 


  tan(theta) = r/1

             = sqrt(x^2 + y^2)

             = sqrt(tan^2(alpha) + tan^2(beta))

and our formula for the compound angle is

  theta = arctan(sqrt(tan^2(alpha) + tan^2(beta))

If the angles from the vertical in the two elevations are 32 degrees 
and 48 degrees, then

  alpha = 32 degrees; tan(alpha) = 0.624869

  beta  = 48 degrees; tan(beta)  = 1.110613

  tan^2(alpha) + tan^2(beta) = 0.624869*0.624869 + 1.110613*1.110613

                             = 1.623922

  sqrt(tan^2(alpha) + tan^2(beta)) = 1.274332

  theta = arctan(1.274332)

        = 51.87 degrees

I hope this does the trick for you!

- Doctor Rick, The Math Forum 
Associated Topics:
High School Geometry
High School Higher-Dimensional Geometry
High School Practical Geometry
High School Trigonometry

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