1. Problem 1: Find the bearing from F to G on the square grid map. 2. To find the bearing, first determine the horizontal and vertical distances between points F and G using the scale. 3. Suppose F is at coordinates (x_F, y_F) and G at (x_G, y_G). Calculate the difference: $\Delta x = x_G - x_F$, $\Delta y = y_G - y_F$. 4. The bearing is measured clockwise from north. Calculate the angle $\theta$ from the north direction using: $$\theta = \arctan\left(\frac{|\Delta x|}{|\Delta y|}\right)$$ 5. Adjust $\theta$ depending on the quadrant to get the bearing from north clockwise. 6. Round the bearing to the nearest degree. --- 1. Problem 2: Work out the area of quadrilateral EFGH with given sides and right angles. 2. The quadrilateral has two right angles at H and F, and a diagonal EG. 3. Split the quadrilateral into two triangles EHG and EFG by diagonal EG. 4. Calculate length EG using Pythagoras in triangle FGH: $$EG = \sqrt{(FG)^2 + (HG)^2} = \sqrt{4^2 + 11^2} = \sqrt{16 + 121} = \sqrt{137} \approx 11.7\text{ cm}$$ 5. Calculate area of triangle EHG: $$\text{Area}_{EHG} = \frac{1}{2} \times EH \times HG = \frac{1}{2} \times 8 \times 11 = 44\text{ cm}^2$$ 6. Calculate area of triangle EFG using base FG and height (from E to FG), which equals EH = 8 cm: $$\text{Area}_{EFG} = \frac{1}{2} \times FG \times EH = \frac{1}{2} \times 4 \times 8 = 16\text{ cm}^2$$ 7. Total area of quadrilateral EFGH: $$44 + 16 = 60\text{ cm}^2$$ --- 1. Problem 3: Find length $d$ in the right triangle with angle $\theta$ and base segments 6.2 cm and 3.8 cm. 2. Write two equations involving $\cos \theta$ and $d$: - From the triangle, $\cos \theta = \frac{6.2}{d}$ - The total base length is $6.2 + 3.8 = 10$ cm. 3. Using the right angle, the vertical height $h$ can be expressed as: $$h = d \sin \theta$$ 4. Using Pythagoras theorem on the right triangle: $$d^2 = h^2 + 10^2$$ 5. Substitute $h = d \sin \theta$: $$d^2 = (d \sin \theta)^2 + 10^2 = d^2 \sin^2 \theta + 100$$ 6. Rearrange: $$d^2 - d^2 \sin^2 \theta = 100$$ $$d^2 (1 - \sin^2 \theta) = 100$$ 7. Use identity $1 - \sin^2 \theta = \cos^2 \theta$: $$d^2 \cos^2 \theta = 100$$ 8. Substitute $\cos \theta = \frac{6.2}{d}$: $$d^2 \left(\frac{6.2}{d}\right)^2 = 100$$ $$d^2 \frac{38.44}{d^2} = 100$$ $$38.44 = 100$$ This is a contradiction, so re-examine the problem setup. 9. Alternatively, use the triangle with base 6.2 cm and hypotenuse $d$: $$\cos \theta = \frac{6.2}{d}$$ 10. The vertical height is: $$h = d \sin \theta$$ 11. The other segment 3.8 cm is the horizontal distance from the right angle vertex to the base end. 12. Using Pythagoras on the smaller right triangle with base 3.8 cm and height $h$: $$d^2 = h^2 + (6.2 + 3.8)^2$$ 13. Since $6.2 + 3.8 = 10$, and $h = d \sin \theta$, then: $$d^2 = (d \sin \theta)^2 + 10^2 = d^2 \sin^2 \theta + 100$$ 14. Rearranged: $$d^2 - d^2 \sin^2 \theta = 100$$ $$d^2 (1 - \sin^2 \theta) = 100$$ $$d^2 \cos^2 \theta = 100$$ 15. Substitute $\cos \theta = \frac{6.2}{d}$: $$d^2 \left(\frac{6.2}{d}\right)^2 = 100$$ $$38.44 = 100$$ 16. This contradiction suggests the problem requires numerical solving or more information. 17. Assuming $\cos \theta = \frac{6.2}{d}$, solve for $d$: $$d = \frac{6.2}{\cos \theta}$$ 18. Without $\theta$ value, $d$ cannot be numerically found. Final answers: - Bearing from F to G: depends on coordinates, calculate $\theta = \arctan\left(\frac{\Delta x}{\Delta y}\right)$ and convert to bearing. - Area of quadrilateral EFGH: $60.0$ cm$^2$ - Length $d$: $d = \frac{6.2}{\cos \theta}$ (exact formula), numerical value requires $\theta$.