1. **Problem 53:** Given $\tan x = \frac{3}{4}$ and $0 < x < 90^\circ$, evaluate $\frac{\cos x}{2 \sin x}$. 2. **Recall the definitions and relationships:** - $\tan x = \frac{\sin x}{\cos x}$ - From $\tan x = \frac{3}{4}$, we can consider a right triangle where the opposite side is 3 and adjacent side is 4. - The hypotenuse $h = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = 5$. - Therefore, $\sin x = \frac{3}{5}$ and $\cos x = \frac{4}{5}$. 3. **Substitute values into the expression:** $$\frac{\cos x}{2 \sin x} = \frac{\frac{4}{5}}{2 \times \frac{3}{5}} = \frac{\frac{4}{5}}{\frac{6}{5}} = \frac{4}{5} \times \frac{5}{6} = \frac{4}{6} = \frac{2}{3}.$$ 4. **Final answer:** $$\boxed{\frac{2}{3}}.$$ --- 5. **Problem 54.1:** The angle of depression from the top $T$ of a building to a point $P$ on the ground is $23.6^\circ$. The horizontal distance $PT = 50$ m. Find the height $h$ of the building. 6. **Use the angle of depression equals angle of elevation rule:** The angle of elevation from $P$ to $T$ is also $23.6^\circ$. 7. **Apply the tangent function:** $$\tan 23.6^\circ = \frac{h}{50} \implies h = 50 \times \tan 23.6^\circ.$$ 8. **Calculate:** $$h = 50 \times 0.436 = 21.8 \approx 22 \text{ m (nearest metre)}.$$ --- 9. **Problem 54.2:** Two ladders lean against opposite sides of a vertical pole at point $L$ which is 9.6 m above ground. Ladder $LB$ is 12 m long. Points $A$ and $B$ are 10 m apart on the ground along the same line as the pole's foot. Find: (i) length of ladder $LA$, (ii) angle $LA$ makes with the ground. 10. **Set up the problem:** Let the pole be at point $O$. $L$ is 9.6 m above ground on the pole. $A$ and $B$ are on opposite sides of $O$, $AB = 10$ m. Ladder $LB$ leans from $B$ to $L$, length $12$ m. Ladder $LA$ leans from $A$ to $L$, length unknown. 11. **Use Pythagoras for $LB$:** $$LB^2 = OL^2 + OB^2 \implies 12^2 = 9.6^2 + OB^2 \implies OB^2 = 144 - 92.16 = 51.84.$$ $$OB = \sqrt{51.84} = 7.2 \text{ m}.$$ 12. **Find $OA$ using $AB = OA + OB = 10$ m:** $$OA = 10 - 7.2 = 2.8 \text{ m}.$$ 13. **Find length $LA$ using Pythagoras:** $$LA = \sqrt{OA^2 + OL^2} = \sqrt{2.8^2 + 9.6^2} = \sqrt{7.84 + 92.16} = \sqrt{100} = 10 \text{ m}.$$ 14. **Find angle $\theta$ between $LA$ and ground:** $$\cos \theta = \frac{OA}{LA} = \frac{2.8}{10} = 0.28 \implies \theta = \cos^{-1}(0.28) \approx 73.7^\circ.$$ 15. **Final answers:** (i) $LA = 10$ m (2 significant figures: $10$ m), (ii) angle $\theta \approx 74^\circ$. --- 16. **Problem 54.3:** From two points on opposite sides of a 33 m high pole, angles of elevation to the top are $23^\circ$ and $67^\circ$. (Problem incomplete in prompt, so no solution provided.)