1. **Problem statement:** Find the length of the side marked $x$ in each right triangle given the angle and one side. 2. **Key formulas and rules:** - In a 30°-60°-90° triangle, sides are in ratio $1 : \sqrt{3} : 2$ (opposite 30°, opposite 60°, hypotenuse respectively). - In a 45°-45°-90° triangle, sides are in ratio $1 : 1 : \sqrt{2}$ (legs : hypotenuse). 3. **Solve each part:** (a) Angle 60°, side opposite 60° is 15, side adjacent (x) is opposite 30°. - Using ratio, side opposite 30° = $\frac{1}{\sqrt{3}}$ times side opposite 60°. - So, $x = \frac{15}{\sqrt{3}}$. - Rationalize denominator: $$x = \frac{15}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{15\sqrt{3}}{3} = 5\sqrt{3}$$ (b) Angle 60°, base side 12 (opposite 30°), hypotenuse $x$. - Hypotenuse = 2 times side opposite 30°. - So, $x = 2 \times 12 = 24$ (c) Angle 45°, base side 4 (leg), hypotenuse $x$. - Hypotenuse = leg $\times \sqrt{2}$. - So, $x = 4 \sqrt{2}$ (d) Angle 45°, hypotenuse 20, perpendicular side $x$ (leg). - Leg = $\frac{\text{hypotenuse}}{\sqrt{2}}$. - So, $x = \frac{20}{\sqrt{2}}$. - Rationalize denominator: $$x = \frac{20}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{20\sqrt{2}}{2} = 10\sqrt{2}$$ (e) Angle 30°, hypotenuse $x$, perpendicular side 8 (opposite 60°). - Side opposite 60° = $\sqrt{3}$ times side opposite 30°. - Let side opposite 30° be $s$, then $8 = s \sqrt{3}$. - So, $s = \frac{8}{\sqrt{3}}$. - Hypotenuse $x = 2s = 2 \times \frac{8}{\sqrt{3}} = \frac{16}{\sqrt{3}}$. - Rationalize denominator: $$x = \frac{16}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{16\sqrt{3}}{3}$$ (f) Angle 30°, hypotenuse 22, perpendicular side $x$ (opposite 60°). - Side opposite 60° = $\sqrt{3}$ times side opposite 30°. - Side opposite 30° = half hypotenuse = $\frac{22}{2} = 11$. - So, $x = 11 \sqrt{3}$ **Final answers:** (a) $5\sqrt{3}$ (b) $24$ (c) $4\sqrt{2}$ (d) $10\sqrt{2}$ (e) $\frac{16\sqrt{3}}{3}$ (f) $11\sqrt{3}$