(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 \(h = 6 + 4\cos (15p)Ā°, 0 < p < 6\). 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) \(h = 6 + 4\cos (15p)Ā°\) (Given)
(i) When p = 4,
\(h = 6 + 4\cos (15 \times 4)Ā°\)
\(h = 6 + 4 \cos 60Ā°\)
= \(6 + 4 \times 0.5\)
= \(6 + 2 = 8 m\)
(ii) When h = 9 m, the given equation becomes
\(9 = 6 + 4\cos (15p)Ā°\)
\(9 - 6 = 4 \cos (15p)Ā°\)
\(\cos (15p)Ā° = \frac{3}{4}\)
\((15p)Ā° = \cos^{-1} (0.75)\)
\((15p)Ā° = 41.4Ā°\)
\(pĀ° = \frac{41.4}{15}\)
\(pĀ° = 2.76Ā°\)
\(p \approxeq 2.8Ā°\) (2 significant figures).