(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.

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(a) The total surface area of two spheres are in the ratio 9 : 49. If the radius of the ...

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(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

c m^{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 =

4 π 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 π r^{2}}{4 π R^{2}}

⟹ \frac{9}{49} = \frac{12^{2}}{R^{2}}

\frac{49 \times 144}{9} = R^{2}

R = \sqrt{49 \times 16} = 28 c m

Volume of bigger sphere =

\frac{4}{3} π r^{3}

\frac{4}{3} \times \frac{22}{7} \times 28 \times 28 \times 28

\frac{275, 968}{3}

91989.33 c m^{3}

≊ 91989 c m^{3}

(nearest whole number).
(b)

| Z X |^{2} = | Y Z |^{2} + | Y X |^{2} - 2 | Y Z | | Y X | \cos Y

5^{2} + 3^{2} - 2 (5) (3) \cos 135

25 + 9 - 30 (- 0.7071)

34 + 21.213

| Z X |^{2} = 55.213

| Z X | = 7.43 k m ≊ 7 k m

(ii) Using sine rule,

\frac{\sin θ}{5} = \frac{\sin 135}{7.43}

\sin θ = \frac{5 \times \sin 135}{7.43}

\frac{3.5355}{7.43}

\sin θ = 0.4758

θ = \sin^{- 1} (0.4758)

28.41 ° ≊ 28 °

(to the nearest degree).
Bearing of Z from X = 270° + 28.41° = 298.41°

≊

298°.

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