(A). The angles of depression of two ships from the top of a lighthouse are 60° and 45° towards the east. If the ships are 300 meter apart, the height of the lighthouse is \(150(3+\sqrt{3})\)
(B). If the surface area of a cube is 726 m2, then its volume shall be 1331 m3
(C) If the ratio of diameters of two spheres is 3 : 5, then the ratio of their surface area shall be 9 : 25
Determine as to which of the statements given above are correct.
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A), (B) & (C)
(A). The angles of depression of two ships from the top of a lighthouse are 60° and 45° towards the east. If the ships are 300 meter apart, the height of the lighthouse is \(150(3+\sqrt{3})\) Correct. The angle of depression from the top equals the angle of elevation from the ship (alternate angles), and both ships are on the same side, so the nearer ship has the larger angle, 60°.
Nearer ship, at distance \(x\) from the foot: \(\tan60^\circ=\frac{h}{x}\), so \(\sqrt{3}=\frac{h}{x}\) and \(x=\frac{h}{\sqrt{3}}\) (1)
Farther ship, 300 m beyond it: \(\tan45^\circ=\frac{h}{x+300}\), so \(1=\frac{h}{x+300}\) and \(h=x+300\) (2)
Putting \(x\) from (1) into (2) gives one equation in \(h\): \(\displaystyle h-\frac{h}{\sqrt{3}}=300\), so \(\displaystyle h\left(\frac{\sqrt{3}-1}{\sqrt{3}}\right)=300\)
\(\displaystyle h=\frac{300\sqrt{3}}{\sqrt{3}-1}=\frac{300\sqrt{3}(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)}=\frac{300(3+\sqrt{3})}{2}=150(3+\sqrt{3})\) m
(B). If the surface area of a cube is 726 m2, then its volume shall be 1331 m3 Correct.
Six faces of area \(a^2\) each: \(6a^2=726\), so \(a^2=121\) and \(a=11\) m
Volume \(=11^3=1331\,\mathrm{m^3}\)
(C) If the ratio of diameters of two spheres is 3 : 5, then the ratio of their surface area shall be 9 : 25 Correct. The surface area of a sphere is \(4\pi r^2\), so surface areas are in the ratio of the squares of the diameters: \(3^2:5^2=9:25\).