1. **Problem statement:** (a)(i) Explain why $\angle ABC = 96^\circ$. (a)(ii) Write down $\angle ABD$. (b) Calculate the lengths of guy wires $AC$ and $AD$. (c)(i) Show that the total area $S$ of the enclosed space $ABCDEFG$ is $S = 2 \cos \theta + 8 \sin \theta + 4$. 2. **Given:** - Tower $AB = 100$ m vertical. - Hill slope angle $6^\circ$. - Points $C$ and $D$ are $75$ m downhill and uphill from base $B$. - $\angle AEB = \theta$, $\angle AED = 90^\circ$, $AE = BE = 4$ cm, $CE = DE = 2$ cm, $EF = 1$ cm. --- ### (a)(i) Why $\angle ABC = 96^\circ$? - The hill slope is $6^\circ$ from horizontal. - $AB$ is vertical, so $AB$ is perpendicular to horizontal. - $BC$ lies along the hill slope, so $\angle ABC$ is the angle between vertical $AB$ and slope line $BC$. - Since vertical and horizontal are $90^\circ$, and hill slope is $6^\circ$ from horizontal, the angle between vertical and slope is $90^\circ + 6^\circ = 96^\circ$. ### (a)(ii) Write down $\angle ABD$. - $D$ is uphill, so $BD$ is along hill slope but uphill direction. - Angle between vertical $AB$ and uphill slope $BD$ is $90^\circ - 6^\circ = 84^\circ$. - Therefore, $\angle ABD = 84^\circ$. --- ### (b) Calculate lengths of guy wires $AC$ and $AD$. - Use Law of Cosines in triangles $ABC$ and $ABD$. For $AC$: - Triangle $ABC$ with sides $AB=100$ m, $BC=75$ m, and angle $\angle ABC=96^\circ$. - Law of Cosines: $$AC^2 = AB^2 + BC^2 - 2 \times AB \times BC \times \cos(\angle ABC)$$ $$= 100^2 + 75^2 - 2 \times 100 \times 75 \times \cos 96^\circ$$ - Calculate $\cos 96^\circ \approx -0.1045$: $$AC^2 = 10000 + 5625 - 2 \times 100 \times 75 \times (-0.1045)$$ $$= 15625 + 1567.5 = 17292.5$$ $$AC = \sqrt{17292.5} \approx 131.5 \text{ m}$$ For $AD$: - Triangle $ABD$ with sides $AB=100$ m, $BD=75$ m, and angle $\angle ABD=84^\circ$. - Law of Cosines: $$AD^2 = AB^2 + BD^2 - 2 \times AB \times BD \times \cos(84^\circ)$$ - Calculate $\cos 84^\circ \approx 0.1045$: $$AD^2 = 10000 + 5625 - 2 \times 100 \times 75 \times 0.1045$$ $$= 15625 - 1567.5 = 14057.5$$ $$AD = \sqrt{14057.5} \approx 118.5 \text{ m}$$ Rounded to nearest metre: - $AC \approx 132$ m - $AD \approx 119$ m --- ### (c)(i) Show $S = 2 \cos \theta + 8 \sin \theta + 4$ - $S$ is area of rectangle $AEFG$ plus areas of isosceles triangles $ABE$ and $CDE$. - Area of rectangle $AEFG = AE \times EF = 4 \times 1 = 4$ cm$^2$. - Area of triangle $ABE$: - Base $BE = 4$ cm - Height from $A$ perpendicular to $BE$ is $AE \sin \theta$ (since $\angle AEB = \theta$) - Area $= \frac{1}{2} \times BE \times AE \sin \theta = \frac{1}{2} \times 4 \times 4 \sin \theta = 8 \sin \theta$ - Area of triangle $CDE$: - Base $DE = 2$ cm - Height from $C$ perpendicular to $DE$ is $CE \cos \theta$ (since $\angle AED = 90^\circ$ and $AE=BE=4$) - Area $= \frac{1}{2} \times DE \times CE \cos \theta = \frac{1}{2} \times 2 \times 2 \cos \theta = 2 \cos \theta$ - Total area: $$S = 4 + 8 \sin \theta + 2 \cos \theta$$ This matches the required expression. --- **Final answers:** - (a)(i) $\angle ABC = 96^\circ$ because vertical and hill slope differ by $90^\circ + 6^\circ$. - (a)(ii) $\angle ABD = 84^\circ$. - (b) $AC \approx 132$ m, $AD \approx 119$ m. - (c)(i) $S = 2 \cos \theta + 8 \sin \theta + 4$.