O-Level E-Math 2016 Paper 2, Question 8: Cosine rule for an angle

(a) Apply the cosine rule in triangle \(ABC\) with \(BC\) opposite \(A\): \(\cos(\angle BAC) = \dfrac{AB^2 + AC^2 - BC^2}{2 \cdot AB \cdot AC} = \dfrac{7^2 + 3^2 - 6^2}{2(7)(3)} = \dfrac{49 + 9 - 36}{42} = \dfrac{22}{42} = 0.52381\). So \(\angle BAC = \cos^{-1}(0.52381) = 58.41^{\circ} = 58.4^{\circ}\) (1 d.p.). (shown)

(b) The prism has two triangular ends and three rectangular faces. Using the angle found in part (a), the area of one triangular end is \(\tfrac12 \cdot AB \cdot AC \cdot \sin(\angle BAC) = \tfrac12(7)(3)\sin 58.4^{\circ} = 10.5 \times 0.8517 = 8.94 \text{ cm}^2\). The three rectangles have widths \(7, 3, 6\) and common length \(10\) (the prism length \(BD\)), so their total area \(= (7 + 3 + 6) \times 10 = 160 \text{ cm}^2\). Surface area \(= 2(8.94) + 160 = 17.89 + 160 = 177.9 \text{ cm}^2\) (to 1 d.p.).

(c) Treating \(AB\) as the base, the vertical distance of \(C\) above \(AB\) is the perpendicular height \(h\) of triangle \(ABC\) from \(C\) to \(AB\). Using area \(= \tfrac{1}{2} \cdot AB \cdot h\): \(8.944 = \tfrac{1}{2}(7)h \Rightarrow h = \dfrac{2(8.944)}{7} = 2.556 \text{ cm} \approx 2.56 \text{ cm}\).

(d) \(E\) is the apex of the far triangular end, directly "across" the prism from \(C\): it sits at the same height \(h = 2.556\) cm above the base, but a horizontal distance of 10 cm (the prism length) plus the in-base offset away from \(A\). Let \(F\) be the foot of \(C\) on \(AB\); then \(AF = \sqrt{AC^2 - h^2} = \sqrt{3^2 - 2.556^2} = \sqrt{9 - 6.533} = 1.571 \text{ cm}\). The foot \(G\) of \(E\) is 10 cm from \(F\) along the prism (perpendicular to the cross-section), so the horizontal distance \(AG = \sqrt{AF^2 + 10^2} = \sqrt{1.571^2 + 100} = \sqrt{102.47} = 10.12 \text{ cm}\). The angle of elevation of \(E\) from \(A\) is \(\tan^{-1}\dfrac{h}{AG} = \tan^{-1}\dfrac{2.556}{10.12} = \tan^{-1}(0.2526) = 14.2^{\circ}\) (1 d.p.).