(a) By how much is the sum of (3frac{2}{3}) and (2frac{1}{5}) less than 7?
(b) The height, h m, of a dock above sea level is given by (h = 6 + 4cos (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 (3frac{2}{3}) and (2frac{1}{5}) be less than 7 by z.
Then z = (7 – (3frac{2}{3} + 2frac{1}{5}))
= (7 – (frac{11}{3} + frac{11}{5}))
= (7 – frac{88}{15})
= (frac{17}{15} = 1frac{2}{15})
Hence, the sum of (3frac{2}{3}) and (2frac{1}{5}) is less than 7 by (1frac{2}{15}).
(b) (h = 6 + 4cos (15p)°) (Given)
(i) When p = 4,
(h = 6 + 4cos (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 + 4cos (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).