(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.
(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.
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Correct Answer: Option n
Explanation:
(a) Total surface area of a sphere = \(4\pi r^{2}\)
Let the TSA of the smaller sphere be \(S_{1}\) with radius r and the bigger sphere be \(S_{2}\) with radius R.
\(\frac{S_{1}}{S_{2}} = \frac{9}{49}\)
\(\frac{9}{49} = \frac{4\pi r^{2}}{4\pi R^{2}}\)
\(\implies \frac{9}{49} = \frac{12^{2}}{R^{2}}\)
\(\frac{49 \times 144}{9} = R^{2}\)
\(R = \sqrt{49 \times 16} = 28 cm\)
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°.
(a) Total surface area of a sphere = \(4\pi r^{2}\)
Let the TSA of the smaller sphere be \(S_{1}\) with radius r and the bigger sphere be \(S_{2}\) with radius R.
\(\frac{S_{1}}{S_{2}} = \frac{9}{49}\)
\(\frac{9}{49} = \frac{4\pi r^{2}}{4\pi R^{2}}\)
\(\implies \frac{9}{49} = \frac{12^{2}}{R^{2}}\)
\(\frac{49 \times 144}{9} = R^{2}\)
\(R = \sqrt{49 \times 16} = 28 cm\)
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°.