Finding the internal angles of a polygon

• Aug 15th 2010, 12:59 AM
Tomska
Finding the internal angles of a polygon
Hi there,

This might seem a simple question, but I suck at maths. I need to cut some ply to fill a gap behind my boat kitchen and I have measured the lengths of the sides. I'd like to have the angles to help me draw a decent template to cut from, but I'm not sure how.

Attachment 18582

A = 1055mm
B = 399mm
C = 860mm
D = 73mm

And the angle between D and A is a right angle.

Any help welcome.

Thanks.
• Aug 15th 2010, 01:03 AM
Prove It
You'll need to draw a line from the vertex between B and C, to the vertex between A and D.

You will then need to use a combination of the sine and cosine rules to find the missing sides and angles...
• Aug 15th 2010, 01:27 AM
Educated
It would help if you are working with triangles.

First draw a line from angle AB to angle CD. This will divide the shape into 2 triangles.
Solve this unknown length (call this E) by using the Pythagorus Theorum (Because it has a right angle):

$A^2 + D^2 = E^2$ (I substituted your letters into it)

$1055^2 + 73^2 = E^2$

$E \approx 1057.52\mbox{mm}$

Now we find the angle of BC by using the cosine rule:

$\mbox{angle BC} = cos^{-1}\left(\dfrac{B^2+C^2-E^2}{2BC}\right)$

$\mbox{angle BC} = cos^{-1}\left(\dfrac{399^2+860^2-1057.52^2}{2 * 399 * 860}\right)$

$\mbox{angle BC} \approx 108.657^{\circ}$

To solve the angle AB, we do the same thing, but must divide it into 2 parts.

$\mbox{angle AE} = cos^{-1}\left(\dfrac{A^2+E^2-D^2}{2AE}\right)$

$\mbox{angle AE} = cos^{-1}\left(\dfrac{1055^2+1057.52^2-73^2}{2 * 1055 * 1057.52}\right)$

$\mbox{angle AE} \approx 3.958^{\circ}$

$\mbox{angle EB} = cos^{-1}\left(\dfrac{E^2+B^2-C^2}{2EB}\right)$

$\mbox{angle EB} = cos^{-1}\left(\dfrac{1057.52^2+399^2-860^2}{2 * 1057.52 * 399}\right)$

$\mbox{angle EB} \approx 50.3976^{\circ}$

$\mbox{angle AB} = \mbox{angle AE} + \mbox{angle EB}$

$\mbox{angle AB} = 3.958^{\circ} + 50.3976^{\circ}$

$\mbox{angle AB} \approx 54.3556^{\circ}$

To find angle CD, we must subtract the other 3 angles by 360.

$\mbox{angle CD} = 360^{\circ} - 54.3556^{\circ} - 108.657^{\circ} - 90^{\circ}$

$\mbox{angle CD} \approx 106.9874^{\circ}$
• Aug 15th 2010, 02:22 AM
Tomska
That's fantastic. Thank you very much.