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### Calculating the Angle of a Plank

```
Date: 08/22/2001 at 09:36:50
From: nick pounder
Subject: Calculating the angle of a plank.

I've been through all the weird and wonderful high school trig
formulas, tried all sorts of substitutions and simultaneous equations,
but I have not been able to crack this. On the face of it it looks
very simple and is a very practical problem:

a|.
|   .
b|.     .
|   .     .
|      .     .
|_________.___ .
0          c    d

A hypothetical plank bounded by the dots is leaning against a wall.
All I want to know is the angle the plank makes.  The problem is: I
know the height a and I also know the length c, the first point
of contact with the floor. However, I don't know length d, or height b
for that matter.

I do know that the plank is z width. If I knew the length d I could
calculate the angle easily; however, I don't know how much farther on
from c that d is, because I don't know the angle.

Are there any kind of simultaneous trig equations that could be used
to solve this for a plank of known width? And if the solution is

The diagram above isn't very good as it shows the plank already cut to
the correct angle; in practice my new plank has 90-degree corners.  I
want to calculate the angle without further measurements, so I know
what angle to cut the plank so it rests smoothly against both the wall
and floor.

Thanks very much!
```

```
Date: 08/23/2001 at 10:27:43
From: Doctor Rick
Subject: Re: calculating the angle of a plank.

Hi, Nick, thanks for writing to Ask Dr. Math.

This is not a trivial problem, no apologies are needed!

Let's solve the problem using coordinate geometry; you have already

a *-
/|  -
/ |    -
z/th|       -
/   |          -
(x0, *    |            -
y0)   -  |               -
-|                 -
b *-                   -
|  -                    -
|     -                   -
|       -                    -
|          -                   -
|             -                   -
|               -                   -
|                  -                   -
|                     -                   -
|                        -                  -
|                          -                   -
|                             -                  -
|                                -                  -
|                           theta  -           theta   -
+-------------------------------------*------------------*
0                                      c                  d

You know a, c, and z. In terms of the unknown angle theta, we can find
the coordinates of the point (x0,y0) that I added to the figure:

x0 = -z*sin(theta)
y0 = a - z*cos(theta)

the slope of the line marking the bottom edge of the plank is

m = -tan(theta)

Thus the equation of this line is

y = m(x - x0) + y0

y = -tan(theta)(x + z*sin(theta)) + a - z*cos(theta)

This line passes through the point (c,0), so

0 = -tan(theta)(c + z*sin(theta)) + a - z*cos(theta)

We want to solve this equation for theta. To do this, we rewrite
tan(theta) and cos(theta) in terms of sin(theta):

cos(theta) = sqrt(1 - sin^2(theta))

tan(theta) = sin(theta)/cos(theta)
= sin(theta)/sqrt(1 - sin^2(theta))

Call s = sin(theta) to simplify the equations:

cos(theta) = sqrt(1 - s^2)
tan(theta) = s/sqrt(1 - s^2)

0 = -s/sqrt(1 - s^2)(c + zs) + a - z*sqrt(1 - s^2)

Multiply through by sqrt(1 - s^2):

0 = -s(c + zs) + a*sqrt(1 - s^2) - z(1 - s^2)

Simplifying, the s^2 term cancels out:

sc + z = a*sqrt(1 - s^2)

Square both sides (we'll need to check that we don't introduce
spurious solutions at this step):

(sc + z)^2 = a^2(1 - s^2)

(a^2+c^2)s^2 + 2zcs + (z^2-a^2) = 0

Solve this quadratic equation for s:

s = (-2zc +or- sqrt(4z^2c^2-4(a^2+c^2)(z^2-a^2)))/(2(a^2+c^2))
= (-zc +or- sqrt(z^2c^2-(a^2+c^2)(z^2-a^2)))/(a^2+c^2)
= (-zc +or- sqrt(a^4+a^2c^2-a^2z^2))/(a^2+c^2)
= (-zc +or- a*sqrt(a^2+c^2-z^2))/(a^2+c^2)

The negative solution is spurious, since we know that theta is in the
first quadrant, so sin(theta) is between 0 and 1. Thus, setting
sin(theta) equal to s, we have our solution:

+----------------------------------------------------+
| theta = sin^-1((a*sqrt(a^2+c^2-z^2)-zc)/(a^2+c^2)) |
+----------------------------------------------------+

Let's check the formula by plugging in the numbers for a simple case.
If z = 1 and theta = 45 degrees, then we can see from the figure that
a = b+sqrt(2) and c = b. Make these substitutions:

sin(theta) = ((b+sqrt(2))*sqrt((b+sqrt(2))^2+b^2-1)-b)
/((b+sqrt(2))^2+b^2)

= ((b+sqrt(2))*sqrt(2b^2+2b*sqrt(2)+1)-b)/(2b^2+2b*
sqrt(2)+2)

= ((b+sqrt(2))*(sqrt(2)*b+1)-b)/(2b^2+2b*sqrt(2)+2)

= (b^2*sqrt(2)+2b+sqrt(2))/(2b^2+2b*sqrt(2)+2)

= sqrt(2)(b^2+b*sqrt(2)+1)/(2(b^2+b*sqrt(2)+1)

= sqrt(2)/2

This is indeed the sin of 45 degrees, so the formula checks out in
this case.

I hope this is helpful to you. I'm assuming you can finish the work,
finding the length of at least one edge of the plank so you can cut it
to fit.

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

```
Date: 01/04/2016 at 16:27:53
From: nick
Subject: Hullo!  This is a question on a trig solution 2001 !

Hi there.

Way back in 2001, I posted the above question on Dr. Math. And a
solution -- albeit a very convoluted one -- was kindly provided back
then by Doctor Rick.

So fifteen years later, I have started looking at recreational math and I
decided to revisit this question. On this second look, the solution is
really straightforward and was a case of me not seeing the wood for the
trees. So, for others seeking a nice easy solution ...

Referring to Doctor Rick's diagram, all you need to do is join point a to
point c to make a new triangle, and work out the hypotenuse. Use basic
trig to determine the angle with the horizontal; call this theta-1. Next,
consider the new triangle that this line has made "within the plank." The
hypotenuse is the same as the big new triangle you have just created.

Then, armed with this hypotenuse and the length of the thickness of the
plank (the "opposite"), you can now work out the angle at the horizontal
of this little triangle by applying a little more basic trig. To get the
final answer, just subtract the small angle from the big one :-)

Best regards -- and thanks for replying all those years ago.

```

```
Date: 01/04/2016 at 17:34:46
From: Doctor Rick
Subject: Re: Hullo!  This is a question on a trig solution 2001 !

Hi, Nick.

You're right! That's a simpler solution method, for sure.

The formula it produces for theta in terms of a, c, and the
thickness z, is:

theta = arctan(a/c) - arcsin(z/sqrt(a^2 + c^2))

How does that compare to my original formula?

theta = arcsin((a*sqrt(a^2 + c^2 - z^2) - zc)/(a^2 + c^2))

Rather than try to prove that this formula is equivalent, I used graphing
software to plot both functions of z, for fixed values of a and c. They
look identical for z > -c; and they only have physical significance for
positive values of z (and a and c) -- so we can regard them as
equivalent formulas.

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

```
Associated Topics:
High School Coordinate Plane Geometry
High School Geometry
High School Practical Geometry
High School Trigonometry

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