# Determine length of side of right triangle for given angle?

Discussion in 'Math' started by spinnaker, Oct 29, 2012.

1. ### spinnaker Thread Starter AAC Fanatic!

Oct 29, 2009
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I have a solar panel mounted on a fence post. I need to adjust the panel angle throughout the year.

Instead of requiring a protractor, I would like to make a graduated scale marked with the months. To make the job easier, I would like to measure one of the sides of the triangle to set the required angle.

Example below. How would I calculate the length of side B for 26°?

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2. ### MrChips Moderator

Oct 2, 2009
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I believe, in this application A will be changing.
You want to use the fixed length C.
sin(angle) = B/C
where the angle is in radians, i.e.

angle in radians = (angle in degrees) x 3.14159 / 180

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3. ### MrChips Moderator

Oct 2, 2009
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The correct equation is

B = A * tan(angle)

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4. ### Wendy Moderator

Mar 24, 2008
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Are you familiar with the law of sines? It is one of those formula I hang on to, because it tends to fill in for so many problems.

A / Sine Angle A = B / Sine Angle B = C / Sine Angle C

The Angles are opposite the sides. I'll stick in an illustration when I have more time.

Last edited: Oct 29, 2012
5. ### blah2222 Well-Known Member

May 3, 2010
554
33
This is correct, something seems to be in my water. Sorry about that.

6. ### spinnaker Thread Starter AAC Fanatic!

Oct 29, 2009
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I take back my thanks.

7. ### imasoundhound New Member

Oct 28, 2012
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0
you can't.

you have to have the measure of at least one side of the triangle in order to use any of the trigonometric functions.

assuming side C is the solar panel, and therefore of constant length, any change made to the angle formed by sides A and C will change the lengths of sides A and B (remember C doesn't change). If the measure of the angle increases, B must get longer while A gets shorter. If the measure of the angle decreases, B will get shorter while A gets longer.

if C is of constant length, then:

C*sin (26 degrees) = B

and

C*cos (26 degrees) = A

Last edited: Oct 30, 2012