(a) The total surface area of two spheres are in the ratio 9 : 49. If the radius of the smaller sphere is 12 cm, find, correct to the nearest (cm^{3}), the volume of the bigger sphere.
(b) A cyclist starts from a point X and rides 3 km due West to a point Y. At Y, he changes direction and rides 5 km North- West to a point Z.
(i) How far is he from the starting point, correct to the nearest km? ; (ii) Find the bearing of Z from X, to the nearest degree.
Explanation
(a) Total surface area of a sphere = (4pi r^{2})
Let the TSA of the smaller sphere be (S_{1}) with radius r and the bigger sphere be (S_{2}) with radius R.
(frac{S_{1}}{S_{2}} = frac{9}{49})
(frac{9}{49} = frac{4pi r^{2}}{4pi R^{2}})
(implies frac{9}{49} = frac{12^{2}}{R^{2}})
(frac{49 times 144}{9} = R^{2})
(R = sqrt{49 times 16} = 28 cm)
Volume of bigger sphere = (frac{4}{3} pi r^{3})
= (frac{4}{3} times frac{22}{7} times 28 times 28 times 28)
= (frac{275,968}{3})
= (91989.33 cm^{3})
(approxeq 91989 cm^{3}) (nearest whole number).
(b)
(|ZX|^{2} = |YZ|^{2} + |YX|^{2} – 2|YZ||YX| cos Y)
= (5^{2} + 3^{2} – 2(5)(3) cos 135)
= (25 + 9 – 30(-0.7071))
= (34 + 21.213)
(|ZX|^{2} = 55.213)
(|ZX| = 7.43 km approxeq 7 km)
(ii) Using sine rule,
(frac{sin theta}{5} = frac{sin 135}{7.43})
(sin theta = frac{5 times sin 135}{7.43})
= (frac{3.5355}{7.43})
(sin theta = 0.4758)
(theta = sin^{-1} (0.4758))
= (28.41° approxeq 28°) (to the nearest degree).
Bearing of Z from X = 270° + 28.41° = 298.41°
(approxeq) 298°.