50. a) The height of the tree Step 1: Identify the known values in the right-angled triangle BFC. The distance from the foot of the tree (F) to the chameleon (C) is $FC = 14.4 \text{ m}$. The angle of elevation of the bird (B) from the chameleon (C) is $\angle BCF = 41.4^\circ$. Step 2: Use the tangent ratio to find the height of the bird (BF). $$\tan(\angle BCF) = \frac{BF}{FC}$$ $$\tan(41.4^\circ) = \frac{BF}{14.4 \text{ m}}$$ $$BF = 14.4 \text{ m} \times \tan(41.4^\circ)$$ $$BF = 14.4 \text{ m} \times 0.88009$$ $$BF \approx 12.6733 \text{ m}$$ Step 3: Calculate the height of the tree (TF). Since B is the midpoint of FT, the height of the tree TF is twice the height of the bird BF. $$TF = 2 \times BF$$ $$TF = 2 \times 12.6733 \text{ m}$$ $$TF \approx 25.3466 \text{ m}$$ Step 4: Round the height to 2 decimal places. $$TF \approx 25.35 \text{ m}$$ The height of the tree is $\boxed{25.35 \text{ m}}$. b) The angle of depression of the chameleon from point T Step 1: Identify the known values in the right-angled triangle TFC. The height of the tree $TF \approx 25.3466 \text{ m}$ (from part a). The distance from the foot of the tree (F) to the chameleon (C) is $FC = 14.4 \text{ m}$. Step 2: The angle of depression of the chameleon from point T is equal to the angle of elevation of T from C, which is $\angle TCF$. Use the tangent ratio. $$\tan(\angle TCF) = \frac{TF}{FC}$$ $$\tan(\angle TCF) = \frac{25.3466 \text{ m}}{14.4 \text{ m}}$$ $$\tan(\angle TCF) \approx 1.76018$$ Step 3: Calculate the angle $\angle TCF$. $$\angle TCF = \arctan(1.76018)$$ $$\angle TCF \approx 60.399^\circ$$ Step 4: Round the angle to 2 decimal places. $$\angle TCF \approx 60.40^\circ$$ The angle of depression of the chameleon from point T is $\boxed{60.40^\circ}$. 51. Step 1: Convert the side length of the cubical stone from cm to m. $$s = 40 \text{ cm} = 0.4 \text{ m}$$ Step 2: Calculate the volume of the cubical stone in m$^3$. $$V = s^3$$ $$V = (0.4 \text{ m})^3$$ $$V = 0.064 \text{ m}^3$$ Step 3: Use the given mass and calculated volume to determine the density. $$m = 12 \text{ kg}$$ $$\rho = \frac{m}{V}$$ $$\rho = \frac{12 \text{ kg}}{0.064 \text{ m}^3}$$ $$\rho = 187.5 \text{ kg/m}^3$$ The density of the stone is $\boxed{187.5 \text{ kg/m}^3}$. 52. Step 1: Use the area formula for a triangle to find $\angle PQR$. Given Area $= 30\sqrt{3} \text{ cm}^2$, $PQ = 10 \text{ cm}$, $QR = 12 \text{ cm}$. $$\text{Area} = \frac{1}{2} \times PQ \times QR \times \sin(\angle PQR)$$ $$30\sqrt{3} = \frac{1}{2} \times 10 \times 12 \times \sin(\angle PQR)$$ $$30\sqrt{3} = 60 \times \sin(\angle PQR)$$ $$\sin(\angle PQR) = \frac{30\sqrt{3}}{60}$$ $$\sin(\angle PQR) = \frac{\sqrt{3}}{2}$$ Step 2: Determine the obtuse angle $\angle PQR$. Since $\sin(\angle PQR) = \frac{\sqrt{3}}{2}$, the acute angle is $60^\circ$. As $\angle PQR$ is obtuse, $$\angle PQR = 180^\circ - 60^\circ = 120^\circ$$ Step 3: Use the Cosine Rule to find the length of PR. $$PR^2 = PQ^2 + QR^2 - 2 \times PQ \times QR \times \cos(\angle PQR)$$ $$PR^2 = (10 \text{ cm})^2 + (12 \text{ cm})^2 - 2 \times 10 \text{ cm} \times 12 \text{ cm} \times \cos(120^\circ)$$ $$PR^2 = 100 \text{ cm}^2 + 144 \text{ cm}^2 - 240 \text{ cm}^2 \times \left(-\frac{1}{2}\right)$$ $$PR^2 = 244 \text{ cm}^2 + 120 \text{ cm}^2$$ $$PR^2 = 364 \text{ cm}^2$$ $$PR = \sqrt{364} \text{ cm}$$ $$PR \approx 19.0787 \text{ cm}$$ Step 4: Calculate the perimeter of the triangle. $$\text{Perimeter} = PQ + QR + PR$$ $$\text{Perimeter} = 10 \text{ cm} + 12 \text{ cm} + 19.0787 \text{ cm}$$ $$\text{Perimeter} = 41.0787 \text{ cm}$$ Step 5: Round the perimeter to 2 decimal places. $$\text{Perimeter} \approx 41.08 \text{ cm}$$ The perimeter of the triangle is $\boxed{41.08 \text{ cm}}$. 53. Step 1: Calculate the total time elapsed from Monday 6:00 a.m. to Friday 6:00 a.m. From Monday 6:00 a.m. to Friday 6:00 a.m. is 4 full days. Total hours = $4 \text{ days} \times 24 \text{ hours/day} = 96 \text{ hours}$. Step