ISASP High School Geometry. Practice it free.

The angles sum to $180°$: $(2x+5)+(3x+5)+(x+50)=180$, so $6x+60=180$ and $x=20$. The angles are $45°$, $65°$, and $70°$, so the largest is $70°$.

Set up the proportion $\dfrac{8\text{ pounds}}{1\text{ cubic foot}} = \dfrac{1{,}920\text{ pounds}}{x\text{ cubic feet}}\textsf{.}$ Cross-multiplying gives $8x = 1{,}920\textsf{,}$ so $x = 240\text{ ft}^{3}\textsf{.}$

$\cos\alpha = \dfrac{\sin\alpha}{\tan\alpha} = \dfrac{\dfrac{4}{5}}{\dfrac{4}{3}} = \dfrac{4}{5} \cdot \dfrac{3}{4} = \dfrac{3}{5}$.

The short leg is half the hypotenuse: $\dfrac{14}{2} = 7$.

Consecutive angles are supplementary: $(4x+6)+(2x+24)=180$, so $6x+30=180$ and $x=25$. Then $m\angle P=4(25)+6=106°$.

Set up the proportion $\dfrac{x\text{ packages}}{1\text{ cubic foot}} = \dfrac{4{,}800\text{ packages}}{160\text{ cubic feet}}\textsf{.}$ Cross-multiplying gives $160x = 4{,}800\textsf{,}$ so $x = 30\text{ packages per ft}^{3}\textsf{.}$

$\sin\theta = \tan\theta \cdot \cos\theta = \dfrac{5}{12} \cdot \dfrac{12}{13} = \dfrac{5}{13}$.

Both legs of a $45$-$45$-$90$ triangle are equal, so the other leg is $12$.