How can I work out the angle between a face and base of a triangular prism

Question

I'm struggling with a particular 3d problem - https://i.stack.imgur.com/SmOus.jpg (question 4)

To workout the length of $EM$, forming a right angled triangle from face $EAB$ I get: $$EM =\sqrt{4^2 - 2^2} = 3.46410161514 $$ Making $O$ the middle of a line $MN$ connecting the mid points of both prism faces, I then was trying to form another right angled triangle with the previously acquired length however this is obviously not the slanting height of the prism face and so when I tried to get workout the angle between the face and base by using a trigonometric ratio, I didn't get the right answer.

Any help on this question would be greatly appreciated.

Answer 1

The altitude of face $EAB$ which is segment $EM$ is $2 \sqrt{3}$. Let the required angle between face $EAB$ and the base $ABCD$ be $\theta$, then

$ 2 \sqrt{3} \cos \theta = \dfrac{7 - 5}{2} = 1 $

Therefore,

$ \cos \theta = \dfrac{1}{2 \sqrt{3}} $

And this gives $\theta = 73.22^\circ$

Answer 2

The height $\,EM\,$ of the isosceles triangle is $\sqrt{EB^2-\bigg(\dfrac{AB}{2}\bigg)^2} =\sqrt{4^2-2^2}=\sqrt{12}.\quad$ The perpendicular height from $\,ABCD\,$ to $\,E\,$ is $\sqrt{\big(\sqrt{12}\big)^2-\bigg(\dfrac{7-5}{2}\bigg)} =\sqrt{12-1}=\sqrt{11}.\quad$ The sine of the height one unit in from $\,M\,$ over the height $\,EM\,$ is $\dfrac{\sqrt{11}}{\sqrt{12}}\approx 0.957.\quad$ $\angle\theta\approx\sin^{-1}{0.957}\approx 1.278 \text{ radians} \approx 73.22°.\quad$