10. $\frac{3}{4}$
10. On the unit circle, the points at $-114^{\circ},-42^{\circ}, 30^{\circ}, 102^{\circ}$ and $174^{\circ}$ form a regular pentagon. So the sum of the cosines of these angles is 0 . It follows that
$$
\begin{aligned}
& \left(\cos 42^{\circ}+\cos 102^{\circ}+\cos 114^{\circ}+\cos 174^{\circ}\right)^{2} \\
= & {\left[\cos \left(-114^{\circ}\right)+\cos \left(-42^{\circ}\right)+\cos 102^{\circ}+\cos 174^{\circ}\right]^{2} } \\
= & \left(-\cos 30^{\circ}\right)^{2} \\
= & \frac{3}{4}
\end{aligned}
$$

Alternative Solution
We have
$$
\begin{aligned}
& \left(\cos 42^{\circ}+\cos 102^{\circ}+\cos 114^{\circ}+\cos 174^{\circ}\right)^{2} \\
= & \left(2 \cos 72^{\circ} \cos 30^{\circ}+2 \cos 144^{\circ} \cos 30^{\circ}\right)^{2} \\
= & 3\left(\cos 72^{\circ}+\cos 144^{\circ}\right)^{2}
\end{aligned}
$$

Now
$$
\begin{aligned}
& 2 \sin 36^{\circ}\left(\cos 72^{\circ}+\cos 144^{\circ}\right) \\
= & 2 \sin 36^{\circ} \cos 72^{\circ}+2 \sin 36^{\circ} \cos 144^{\circ} \\
= & \sin 108^{\circ}-\sin 36^{\circ}+\sin 180^{\circ}-\sin 108^{\circ} \\
= & -\sin 36^{\circ}
\end{aligned}
$$
so that $\cos 72^{\circ}+\cos 144^{\circ}=-\frac{1}{2}$. It follows that the answer is $\frac{3}{4}$.
[As one could expect, there are many other ways of obtaining the answer, notably by solving $\cos 36^{\circ}$ in surd form. The above solutions are surd-free in the entire process.]
10. Simplify $\left(\cos 42^{\circ}+\cos 102^{\circ}+\cos 114^{\circ}+\cos 174^{\circ}\right)^{2}$ into a rational number.
10. 把 $\left(\cos 42^{\circ}+\cos 102^{\circ}+\cos 114^{\circ}+\cos 174^{\circ}\right)^{2}$ 化簡成一個有理數。