1. **State the problem:** Brogan wants to position a coin so that it completely blocks her view of the Sun. We know: - Diameter of coin $d_c = 28$ mm - Diameter of Sun $d_s = 1.4$ million km = $1.4 \times 10^6$ km - Distance to Sun $D_s = 150$ million km = $1.5 \times 10^8$ km We need to find the maximum distance $D_c$ from Brogan's eye to the coin so that the coin fully covers the Sun. 2. **Formula and concept:** The coin blocks the Sun if the angular size of the coin as seen from the eye equals the angular size of the Sun. Angular size $\theta = \frac{\text{diameter}}{\text{distance}}$ (small angle approximation). So, $$\frac{d_c}{D_c} = \frac{d_s}{D_s}$$ 3. **Convert units:** - Coin diameter $d_c = 28$ mm = $0.028$ m - Sun diameter $d_s = 1.4 \times 10^6$ km = $1.4 \times 10^9$ m - Sun distance $D_s = 1.5 \times 10^8$ km = $1.5 \times 10^{11}$ m 4. **Solve for $D_c$:** $$D_c = d_c \times \frac{D_s}{d_s}$$ Substitute values: $$D_c = 0.028 \times \frac{1.5 \times 10^{11}}{1.4 \times 10^9}$$ 5. **Simplify:** $$D_c = 0.028 \times \frac{1.5}{1.4} \times 10^{11-9} = 0.028 \times 1.0714 \times 10^2$$ 6. **Calculate:** $$D_c = 0.028 \times 107.14 = 3.0$$ So, the furthest distance Brogan can position the coin is approximately $3.0$ metres. **Final answer:** $$\boxed{3.0 \text{ metres}}$$