simple geometry.
 

Ok I am building a swing in my back yard, a simple A frame design. I am trying to figure out the angles that I need to cut to get what I am looking for.

My sides on the A frame will be 8 feet long, and 4 feet apart at the bottom.

I can not for the life of me remember how to solve for the angles that I need.

Some help would be greatly appreciated, if anyone knows what i'm talking about.

It's sad that a 5th year senior in college can't figure this out.

dola- maybe 4 feet apart at the bottom isn't what i'm looking for.

Anyone have any swing building advice?

I'm thinking 5-1/2 feet at the bottom?

and i found this page after i posted that...

seems to be what i was looking for....but is the angle i'm looking for only 7.5? that seems way to small?

http://www.ajdesigner.com/phptriangl...isector_tb.php

Hey, glad we could help.

Well, if all three sides of the "A" were equal (i.e., 8x8x8) every angle would be 60 degrees, since a triangle must form 180 degrees.

law of cosines

a^2=b^2+c^2-2bccosA

so...
4^2=8^2+8^2-2*8*8*cosA
or
(16-64-64)/-2*8*8 =cos A
112/128=cosA
0.875=cosA
A=28.955 degrees

Or a much simpler way (if I understand the design correctly)...

Your hypoteneus (the diagonal 8' piece) is 8'.

5.5' at the bottom divided by 2 (to make a right angle) = 2.75'.

So to get the bottom angle you simply take the

cos(theta) = adjacent/hyp
cos(theta) = 2.75'/8'
theta = arccos (2.75'/8')
theta = 69.9 degrees

So the angle the 8' piece forms with the ground is 70 degrees.

Or you could use a calculator and ask for the arcsine of 2/8 and then multiply that by 2.

A = 2*ARCSINE(2/8)

:D

............../
............./
..........8'/
.........../
........../.................\
........./..x.............x..\
........_______________
...............5.5'

x = 70 degrees
the other angle is intuitive... 180 - 140 = 40 degrees.

This is what I get for not reading carefully. I was using 4 ft at the bottom instead of 5.5 feet.

I don't know how you could remember that silly law of cosines.

I wish I could, as a lot of my classes require all kinds of force balancing (fortunately as you move up in levels they go from angled forces to horiztonal & vertical forces!:)), but I always revert to drawing a stupid picture.

2/3?

I better remember it! I have to teach it later this semester. That was the first thing I thought of since I knew the 3 sides. using right angle trig is more effective. Just wish that hadf been my first thoughtr instead. :D