Answer
$$\theta\approx18.4^\circ$$
Work Step by Step
$$(l_1):x+y=9\hspace{2cm}(l_2):2x+y=-1$$
$$(l_1):y=-x+9\hspace{2cm}(l_2):y=-2x-1$$
Let's call the angle between the pair of lines $\theta$.
We need to apply the formula built in Exercise 74
$$\tan\theta=\frac{m_2-m_1}{1+m_1m_2}$$
in which $m_1$ is the slope of $l_1$ and $m_2$ is the slope of $l_2$.
However, since $\tan\theta\gt0$, so $$\tan\theta=\Big|\frac{m_2-m_1}{1+m_1m_2}\Big|$$
The slope of line $(l_1)$: $m_1=-1$
The slope of line $(l_2)$: $m_2=-2$
Therefore,
$$\tan\theta=\Big|\frac{-2-(-1)}{1+(-2)(-1)}\Big|$$
$$\tan\theta=\Big|-\frac{1}{3}\Big|$$
$$\tan\theta=\frac{1}{3}$$
So, $$\theta\approx18.4^\circ$$