thanks

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- Dec 14th 2008, 04:29 AMhelloyingtrapezoid
thanks

- Dec 14th 2008, 07:23 AMearboth
Cut off thetwo right triangles of the trapezium (see attachment)

Then you know

**(A)**$\displaystyle \dfrac hx=\tan(72^\circ)$ ........ and ........**(B)**$\displaystyle \dfrac hy = \tan(68^\circ)$

According to my sketch you can see that

$\displaystyle x+y=8~\implies~y=8-x$

Solve the equations (A) and (B) for h and substitute (8-x) instead of y:

$\displaystyle x\tan(72^\circ) = (8-x) \tan(68^\circ)$

Solve for x. I've got $\displaystyle x \approx 3.5659$ and consequently $\displaystyle y \approx 4.4341$ and $\displaystyle h \approx 10.9747$

I'll leave the rest for you.