1. **Problem 1: Find height of the tree.** Given: $d=20$ m, angle of elevation $\theta=30^\circ$. Formula: $\tan \theta = \frac{h}{d}$. Calculation: $$\tan 30^\circ = \frac{h}{20}$$ $$h = 20 \times \tan 30^\circ = 20 \times \frac{1}{\sqrt{3}}$$ $$h \approx 20 \times 0.5774 = 11.55\text{ m}$$ 2. **Problem 2: Find angle of elevation of the sun.** Given: Opposite side = 10 m, Adjacent side = $10\sqrt{3}$ m. Formula: $\tan \theta = \frac{\text{Opposite}}{\text{Adjacent}}$. Calculation: $$\tan \theta = \frac{10}{10\sqrt{3}} = \frac{1}{\sqrt{3}}$$ $$\theta = \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) = 30^\circ$$ 3. **Problem 3: Find height of the window (equilateral triangle).** Given: side $s=1.6$ m. Formula: $h = \frac{\sqrt{3}}{2} s$ or use Pythagoras. Calculation: $$h^2 + (0.8)^2 = (1.6)^2$$ $$h^2 = 2.56 - 0.64 = 1.92$$ $$h = \sqrt{1.92} \approx 1.39\text{ m}$$ 4. **Problem 4: Find length of the slide.** Given: height = 5 m, angle of depression = $30^\circ$. Formula: $\sin \theta = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{5}{L}$. Calculation: $$\sin 30^\circ = \frac{5}{L}$$ $$L = \frac{5}{0.5} = 10\text{ m}$$ 5. **Problem 5: Find height of each pillar and position of point P.** Given: Two pillars height $h$, distance between pillars = 120 m, angles $60^\circ$ and $30^\circ$. Setup: $$\tan 60^\circ = \frac{h}{x} \Rightarrow h = x\sqrt{3}$$ $$\tan 30^\circ = \frac{h}{120 - x} \Rightarrow h = \frac{120 - x}{\sqrt{3}}$$ Equate: $$x\sqrt{3} = \frac{120 - x}{\sqrt{3}}$$ Multiply both sides by $\sqrt{3}$: $$3x = 120 - x$$ $$4x = 120 \Rightarrow x = 30\text{ m}$$ Calculate $h$: $$h = 30 \times \sqrt{3} \approx 51.96\text{ m}$$ Answer: Height of each pillar $\approx 51.96$ m. Position of point P is 30 m from $60^\circ$ pillar and 90 m from $30^\circ$ pillar. 6. **Problem 6: Find distance between the boats.** Given: Tower height = 120 m, angles of depression $60^\circ$ and $45^\circ$. Setup: $$\tan 60^\circ = \frac{120}{x} \Rightarrow x = \frac{120}{\sqrt{3}} = 40\sqrt{3}$$ $$\tan 45^\circ = \frac{120}{y} \Rightarrow y = 120$$ Distance between boats: $$d = y - x = 120 - 40\sqrt{3} \approx 120 - 69.28 = 50.72\text{ m}$$ 7. **Problem 7: Find distance between the men.** Given: Tree height = 15 m, angles of elevation $30^\circ$ and $60^\circ$. Setup: $$\tan 30^\circ = \frac{15}{x} \Rightarrow x = \frac{15}{\tan 30^\circ} = 15\sqrt{3}$$ $$\tan 60^\circ = \frac{15}{y} \Rightarrow y = \frac{15}{\tan 60^\circ} = \frac{15}{\sqrt{3}} = 5\sqrt{3}$$ Total distance: $$d = x + y = 15\sqrt{3} + 5\sqrt{3} = 20\sqrt{3} \approx 34.64\text{ m}$$ 8. **Problem 8: Find height of building and distance between tree and building.** Given: Tree height $H=12$ m, angles of elevation $30^\circ$ and depression $45^\circ$. Setup: From angle of depression: $$\tan 45^\circ = \frac{12}{d} \Rightarrow d = 12$$ From angle of elevation: $$\tan 30^\circ = \frac{h_2}{d} \Rightarrow h_2 = 12 \times \frac{1}{\sqrt{3}} = 4\sqrt{3} \approx 6.928$$ Total building height: $$h = 12 + 6.928 = 18.93\text{ m}$$ Answer: Distance = 12 m, Building height $\approx 18.93$ m. 9. **Problem 9: Find height of tower and value of $\beta$.** Given: Distance = 200 m, $\tan \alpha = \frac{2}{5}$, point B is 80 m nearer. Step 1: Tower height: $$\tan \alpha = \frac{h}{200} = \frac{2}{5} \Rightarrow h = \frac{2 \times 200}{5} = 80\text{ m}$$ Step 2: Distance to point B: $$d_B = 200 - 80 = 120\text{ m}$$ Step 3: Calculate $\tan \beta$: $$\tan \beta = \frac{80}{120} = \frac{2}{3}$$ Step 4: Calculate $\beta$: $$\beta = \tan^{-1}\left(\frac{2}{3}\right) \approx 33.69^\circ$$ 10. **Problem 10: Find speed of boat.** Given: Lighthouse height = 206.6 m, time = 120 s, angle changes from $60^\circ$ to $45^\circ$. Step 1: Initial distance: $$d_1 = \frac{206.6}{\sqrt{3}} \approx 119.29\text{ m}$$ Step 2: Final distance: $$d_2 = 206.6\text{ m}$$ Step 3: Distance traveled: $$\Delta d = d_2 - d_1 = 206.6 - 119.29 = 87.31\text{ m}$$ Step 4: Speed: $$\text{Speed} = \frac{87.31}{120} \approx 0.728\text{ m/s}$$