MO MAP Geometry Similarity, Right Triangles, and Trigonometry. Practice it free.

$\tan\alpha = \dfrac{\sin\alpha}{\cos\alpha} = \dfrac{\dfrac{7}{25}}{\dfrac{24}{25}} = \dfrac{7}{25} \cdot \dfrac{25}{24} = \dfrac{7}{24}$.

In a $30^{\circ}$-$60^{\circ}$-$90^{\circ}$ triangle, the short leg is half the hypotenuse: $\dfrac{16}{2} = 8\textsf{.}$

Both legs of a $45$-$45$-$90$ triangle are equal, so the other leg is $7$.

Points $5$ units from the origin satisfy $x^{2} + y^{2} = 25$. Being exactly $3$ units from $x = 0$ means $|x| = 3$, so $x = 3$ or $x = -3$. Substituting gives $y^{2} = 16$, so $y = 4$ or $y = -4$ for each $x$-value. That yields $(3, 4)$, $(3, -4)$, $(-3, 4)$, and $(-3, -4)$: $4$ points.

Set up $\dfrac{7}{14} = \dfrac{h}{20}$. Cross-multiply: $14h = 140$, so $h = 10$ feet.

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

In a $30^{\circ}$-$60^{\circ}$-$90^{\circ}$ triangle, the hypotenuse is twice the short leg: $2 \times 10 = 20\textsf{.}$

In a $45^{\circ}$-$45^{\circ}$-$90^{\circ}$ triangle, the hypotenuse is a leg times $\sqrt{2}\textsf{:}$ $12\sqrt{2}\textsf{.}$