(a) Let the sum of \(3\frac{2}{3}\) and \(2\frac{1}{5}\) be less than 7 by z. Then z = \(7 - (3\frac{2}{3} + 2\frac{1}{5})\) = \(7 - (\frac{11}{3} + \frac{11}{5})\) = \(7 - \frac{88}{15}\) = \(\frac{17}{15} = 1\frac{2}{15}\) Hence, the sum of \(3\frac{2}{3}\) and \(2\frac{1}{5}\) is less than 7 by \(1\frac{2}{15}\). (b) \(h = 6 + 4\cos (15p)^\circ\) (Given) (i) When p = 4, \(h = 6 + 4\cos (15 \times 4)^\circ\) \(h = 6 + 4 \cos 60^\circ\) = \(6 + 4 \times 0.5\) = \(6 + 2 = 8 m\) (ii) When h = 9 m, the given equation becomes \(9 = 6 + 4\cos (15p)^\circ\) \(9 - 6 = 4 \cos (15p)^\circ\) \(\cos (15p)^\circ = \frac{3}{4}\) \((15p)^\circ = \cos^{-1} (0.75)\) \((15p)^\circ = 41.4^\circ\) \(p^\circ = \frac{41.4}{15}\) \(p^\circ = 2.76^\circ\) \(p \approxeq 2.8^\circ\) (2 significant figures).