Calculating internal angles of trapezium

I have to calcute the internal angles of trapezium when I know all sides of that trapezium. How can I do it? I've tryied to solve it but I can't find any solution to this.

Here are the sides of the ABCD trapezium: |AB|=73,6; |BC|=57; |CD|=60; |DA|=58,6; AB//CD

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Re: Calculating internal angles of trapezium

Quote:

Originally Posted by

**Serillan** I have to calcute the internal angles of trapezium when I know all sides of that trapezium. How can I do it? I've tryied to solve it but I can't find any solution to this.

Here are the sides of the ABCD trapezium: |AB|=73,6; |BC|=57; |CD|=60; |DA|=58,6; AB//CD

1. Draw a sketch.

2. I'm referring to my labels now:

$\displaystyle x+y=13.6~\implies~y=13.6-x$

$\displaystyle h^2+x^2=58.6^2$

$\displaystyle h^2+y^2=57^2$

3. Subtract the last 2 equations. You'll get:

$\displaystyle x^2-y^2=58.6^2-57^2$

Replace y by 13.6 - x. You'll get an equation in x.

4. You should com out by x = 13.6 and y = 0.

5. Now use the right triangle at the left to determine the missing angles.

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Re: Calculating internal angles of trapezium

If you drop a line from point D that is perpendiclar to AB, and call the intersection point with line AB 'E'. Let x = the distance from E to B. Do the same with a line from point B to the line CD, and call that intersection point F, and the distance from F to D is also x . From pythagoras you have the height of the figure is:

$\displaystyle 58.6^2 - (73.6-x)^2$

which also equals $\displaystyle 57^2-(60-x)^2$

From this you can solve for x, and then apply standard trig functions to find the angles.

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