# Math Help - 3D trig

1. ## 3D trig

Help me with this question please

http://i667.photobucket.com/albums/v...0001/wedge.jpg

2. Hello anonymous_maths
Originally Posted by anonymous_maths
Help me with this question please

http://i667.photobucket.com/albums/v...0001/wedge.jpg

I'm assuming that all the angles at $B$ and $C$ are right-angles.

Then, in the diagram I've attached, $N$ is the required point on $FE$; $P$ is the foot of the perpendicular from $N$ to the base; and the required angle, $a = \angle NDP$.

Then, from $\triangle NDP,\, \tan a = \frac{NP}{DP}= \frac{6}{DP}$

$\Rightarrow DP = \frac{6}{\tan a}$

Note now that $FN = CP$, and then consider $\triangle DPC$. By Pythagoras' Theorem: $CP^2 = DP^2 - DC^2$

$\Rightarrow FN = CP = \sqrt{\Big(\frac{6}{\tan a}\Big)^2- 12^2}=\sqrt{\frac{36}{\tan^2a}-144}$