(a) By how much is the sum of \(3\frac{2}{3}\) and \(2\frac{1}{5}\) less than 7? (b) The height, h m, of a dock above sea level is given by \(h = 6 + 4\cos (15p)°, 0 < p < 6\). Find : (i) the value of h when p = 4 ; (ii) correct to two significant figures, the value of p when h = 9 m.
Explanation
(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) \(h = 6 + 4\cos (15p)°\) (Given) (i) When p = 4, \(h = 6 + 4\cos (15 \times 4)°\) \(h = 6 + 4 \cos 60°\) = \(6 + 4 \times 0.5\) = \(6 + 2 = 8 m\) (ii) When h = 9 m, the given equation becomes \(9 = 6 + 4\cos (15p)°\) \(9 - 6 = 4 \cos (15p)°\) \(\cos (15p)° = \frac{3}{4}\) \((15p)° = \cos^{-1} (0.75)\) \((15p)° = 41.4°\) \(p° = \frac{41.4}{15}\) \(p° = 2.76°\) \(p \approxeq 2.8°\) (2 significant figures).