Associated Topics || Dr. Math Home || Search Dr. Math

### Compound Angles

```
Date: 01/14/2001 at 02:37:08
From: Aaron Mangwiro
Subject: Compound Angles

I am installing a Ranch Rail vinyl fence with my Dad, and we have
posts (5 x 5) with rails (2 x 3 1/2). The posts have 2.020 X 3.520
holes routed into them, and the rails fit into these holes perfectly.

But we have one odd one. We have a situation where we have a post but
the rail does not go into the post horizontally, but has a rise angle
of 30 degrees and a side angle of 39 degrees. The hole to be routed
needs to be a parallelogram instead of a square. I have tried to
calculate the size of this parallelogram now for three weeks and
desperate.

Thanks.
Aaron
```

```
Date: 01/15/2001 at 09:51:02
From: Doctor Mitteldorf
Subject: Re: Compound Angles

Dear Aaron,

Here's a formula that you can use: cosA*cosB = cosC. It applies to the
three angles on planes that meet at a point, where two of the planes
meet at right angles.

Specifically: take a piece of cardboard, draw angle A, and cut it out.
Take a second piece of cardboard and draw angle B. Now tape the pieces
of cardboard together so the two vertices coincide, and one line of
each angle is in common, and adjust the tape hinge so the cardboard
planes are at right angles to one another. The third angle formed by
the free edges of the cardboard is angle C, which obeys the formula
above.

(You can can convince yourself that this is true with some heavy-duty
visualization, or by building a physical model.  If you know vector
algebra, the vector derivation is much easier.)

Now picture the rail where it joins the plane of the post. Take as
your vertex the point where the longest edge of the rail meets the
post. Take angles A and B to be on the sides of the rail; they are
(90-39) and (90-30) degrees. The third angle (C) is one angle of the
parallelogram at the mouth of your hole in the post. By the formula
above, cosC = cos(90-39)*cos(90-30), so C = 71.66 degrees, the acute
angle in the parallelogram.

- Doctor Mitteldorf, The Math Forum
http://mathforum.org/dr.math/
```

```
Date: 01/21/2001 at 13:12:22
From: Doctor Mitteldorf
Subject: Re: Compound Angles

Dear Aaron,

Ever since you wrote this question to me, I've been going over it in
my spare time and working on it. I gave you the wrong answer last
week, and I think I have the right answer now.

In the answer I gave you before, I assumed that the angles 30 and 39
degrees were line angles between the axis of the rail and the line
orthogonal to the post. I don't think that's what you intended - I
think the given angles are probably plane angles: first between the
top face of the rail and the horizontal, second between the vertical
face of the rail and the vertical plane perpendicular to the post. So
I've re-done the analysis under that assumption.

Last time, I started with the formula I call the hinge formula. As
foundation, I want to clarify this and generalize it.

Start with a hinge. On one arm of the hinge, draw an angle with the
axis of the hinge as one side. Call this angle alpha. On the other arm
of the hinge, draw angle beta, again with the hinge axis as one side.
You now have two angles in different planes, sharing a common ray.

Open the hinge to a right angle. Measure the angle gamma between the
outer rays of angles alpha and beta. Then the three angles are related
by cos(gamma) = cos(alpha)*cos(beta). This is the hinge formula for
right angle opening.

In general: Suppose the hinge is open to an arbitrary angle zeta.
Then the formula becomes

cos(gamma) = cos(alpha)*cos(beta) + sin(alpha)*sin(beta)*cos(zeta)

This is the general hinge formula for arbitrary opening angle zeta.

Now imagine three planes that come together at a point. They form
three plane angles and three line angles, and you can apply the hinge
formula in three different ways. Just as you can write cos(gamma) in
terms of alpha and beta, you can also write cos(alpha) in terms of
beta and gamma, and you can write cos(beta) in terms of gamma and
alpha. In fact, as the hinge formula is set up, it gives you one line
angle in terms of the other two line angles and the opposite plane
angle; but any three of the angles are sufficient to solve the system.

In your case, I believe you have started with a rail aligned along a
horizontal axis, with its top face horizontal and its side faces
vertical and perpendicular to the rail. You've turned the rail
horizontally through an angle of 39 degrees, and then tilted it down
through an angle of 30 degrees. So consider the point where the
longest edge of the rail meets the rail. There are three faces coming
together at that point: two of them are faces of the rail, and the
third is the vertical face of the post. The two rail faces, of course,
meet each other at a right angle. The horizontal face of the rail
meets the vertical face of the post at an angle 90-30 = 60 degrees,
and the vertical face of the rail meets the vertical face of the post
at an angle 90-39 = 51 degrees.

So in this case, you have the three plane angles, and you'd like to
find the line angles - in particular, the angle between the lines in
the parallelogram, which you need to rout in the vertical face of the
post.

So we go back to these three hinge formulas, specialized where one
face angle is a right angle. We can solve the three equations for the
three line angles. Call the three line angles alpha, beta, and gamma;
call the two plane angles theta and phi - the third is 90 degrees.
The three hinge formulas are:

1) cos(alpha)=cos(gamma)*cos(beta) + sin(gamma)*sin(beta)*cos(theta)
2) cos(beta)=cos(gamma)*cos(alpha) + sin(gamma)*sin(alpha)*cos(phi)
3) cos(gamma)=cos(alpha)*cos(beta)

You can solve this system for alpha, beta, and gamma, given theta and
phi. Here are some hints how to do this:

1) Write x for cos(alpha), y for cos(beta) and z for cos(gamma).
Then sin(alpha)=sqrt(1-x^2), etc.

2) Substitute equation (3) for z into equations (1) and (2) to give
two equations in x and y. Solve either one, and substitute into the
other.

Some fortunate simplifications are available along the way, and the
result is surprisingly simple:

cos(alpha) = cos(theta)/sin(phi)
cos(beta) = cos(phi)/sin(theta)

gamma is the angle in the parallelogram that you're looking for.

cos(alpha) = cos(51)/sin(60)  ==>  alpha = 43.4 deg
cos(beta) = cos(60)/sin(51)  ==>  beta = 50.0 deg

cos(gamma) = cos(alpha)*cos(beta)=0.468   ==>  gamma = 62.1 deg

- Doctor Mitteldorf, the Math Forum
```

```
Date: 01/24/2001 at 13:04:10
From: Dennis Mangwiro
Subject: Re: Compound Angles

Thank you very much - that was amazing.

I have one more question. Will the width of the parallelogram be the
same, taking into consideration the wall thickness of 0.150 inch, as
the width of the the widened rectangle due to the side angle only?
Similarly, will the length of the parallelogram side be the same as
the length of the side of the rectangle, enlarged, due to the rise
angle only?

Thank you.
Aaron
```

```
Date: 01/25/2001 at 06:52:12
From: Doctor Mitteldorf
Subject: Re: Compound Angles

Dear Aaron,

Last time, I described a method for calculating the three line angles
where three planes meet. The three planes we're looking at here are
the top and side of the rail and the vertical face of the post.

When you look at the point where the longest edge of the rail meets
the post, one of the line angles we calculated gives you the acute
angle in the parallelogram traced on the face of the post. The other
two line angles, which we calculated as alpha and beta, appear on the
top and the vertical side of the rail. In fact, you can draw a picture
for yourself and see that the length and width of the rectangular
cross-section of the rail are elongated by factors of 1/cos(alpha) and
1/cos(beta).

Does alpha go with the length and beta with the width or vice versa?
I'll leave that to you to figure out.

- Doctor Mitteldorf, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
College Higher-Dimensional Geometry
High School Higher-Dimensional Geometry

Search the Dr. Math Library:

 Find items containing (put spaces between keywords):   Click only once for faster results: [ Choose "whole words" when searching for a word like age.] all keywords, in any order at least one, that exact phrase parts of words whole words

Submit your own question to Dr. Math
Math Forum Home || Math Library || Quick Reference || Math Forum Search