C2. (a) Let $A B C D$ be an isosceles trapezoid with $A B=C D=5, B C=2, A D=8$. Find the height of the trapezoid and the length of its diagonals.
(b) For the trapezoid introduced in (a), find the exact value of $\cos \angle A B C$.
(c) In triangle $K L M$, let points $G$ and $E$ be on segment $L M$ so that $\angle M K G=\angle G K E=$ $\angle E K L=\alpha$. Let point $F$ be on segment $K L$ so that $G F$ is parallel to $K M$. Given that $K F E G$ is an isosceles trapezoid and that $\angle K L M=84^{\circ}$, determine $\alpha$.
Solution for (a): Let BE be the height of the trapezoid. Thus, BEA is a right triangle with hypotenuse $\mathrm{AB}=5$ and $\mathrm{AE}=(8-2) / 2=3$. Therefore, the height $\mathrm{BE}=4$.
Let $\mathrm{AF}$ be the height. From the right triangle $A F C$, where $A F=4, F C=F B+B C=3+2=5$ we find $A C=\sqrt{16+25}=\sqrt{41}$.
Answer: 4 and $\sqrt{41}$.
Solution for $(b)$ :
From ABC, using the Law of Cosines we find $41=4+25-20 \cos \angle A B C$, so $\cos \angle A B C=-0.6$.
Alternatively, since $F B=A E=3, \mathrm{AB}=5, \cos \angle F B A=3 / 5$. Then
$$
\cos \angle A B C=\cos (180-\angle F B A)=-\cos \angle F B A=-3 / 5=-0.6 \text {. }
$$

Answer: -0.6 .
Solution for (c): Since $K M \| F G$ and $K F E G$ is an isosceles trapezoid, we have that $\angle K M L=\angle F G E=\angle F K E$, so $\angle L K M=3 \angle K M L$. Since the two angles sum up to 96 degrees, we have that $\angle L K M=72^{\circ}=3 \alpha$. Thus, $\alpha=24^{\circ}$.
Alternatively, since $K F E G$ is an isosceles trapezoid, due to symmetry $K L G$ is an isosceles triangle. Then $2 \alpha+2 \alpha+84^{\circ}=180^{\circ}$, so $\alpha=24^{\circ}$.
Answer: $24^{\circ}$.