Two straight lines L_1 and L_2 have slopes -2 and 3 , respectively....

The angle of inclination of a straight line always lies in the interval $[0, \pi)$.

For line $L_1$, the slope is $\tan\theta = -2$. Since $\tan\theta$ is negative and $\theta \in [0, \pi)$, $\theta$ must lie in the $\text{II}^{\text{nd}}$ quadrant \left(\dfrac{\pi}{2} < \theta < \pi\right)$.<br>Thus,$\sin\theta = \dfrac{2}{\sqrt{5}}$and$\cos\theta = -\dfrac{1}{\sqrt{5}}$.<br><br>For line$L_2$, the slope is$\tan\phi = 3$. Since$\tan\phi$is positive and$\phi \in [0, \pi)$,$\phi$must lie in the$\text{I}^{\text{st}}$quadrant \left(0 < \phi < \dfrac{\pi}{2}\right)$.
Thus, $\sin\phi = \dfrac{3}{\sqrt{10}}$ and $\cos\phi = \dfrac{1}{\sqrt{10}}$.

Now, using the sine addition formula:
$\sin(\theta+\phi) = \sin\theta\cos\phi + \cos\theta\sin\phi$
$\Rightarrow \sin(\theta+\phi) = \left(\dfrac{2}{\sqrt{5}}\right)\left(\dfrac{1}{\sqrt{10}}\right) + \left(-\dfrac{1}{\sqrt{5}}\right)\left(\dfrac{3}{\sqrt{10}}\right)$
$\Rightarrow \sin(\theta+\phi) = \dfrac{2}{\sqrt{50}} - \dfrac{3}{\sqrt{50}} = -\dfrac{1}{\sqrt{50}} = -\dfrac{1}{5\sqrt{2}}$.

To determine the quadrant of $\theta+\phi$:
We know $\dfrac{\pi}{2} < \theta < \pi$ and $0 < \phi < \dfrac{\pi}{2}$.
Adding these inequalities gives $\dfrac{\pi}{2} < \theta+\phi < \dfrac{3\pi}{2}$.
This means $\theta+\phi$ lies in either the $\text{II}^{\text{nd}}$ or $\text{III}^{\text{rd}}$ quadrant.
Since $\sin(\theta+\phi)$ is negative, $\theta+\phi$ must lie in the $\text{III}^{\text{rd}}$ quadrant.

Answer: $-\dfrac{1}{5\sqrt{2}}$ and $\text{III}^{\text{rd}}$ quadrant