Question

Angles of elevation of the top of a tower from three points (collinear) A, B and C on a road leading to the foot of the tower are 30\(^{\circ}\), 45\(^{\circ}\) and 60\(^{\circ}\) respectively. The ratio of AB and BC is

• A. \( \sqrt{3}:1 \)
• B. \( \sqrt{3}:2 \)
• C. 1:2
• D. \( 2:\sqrt{3} \)

Hint:

In height and distance problems, it's often easier to work with `cot` instead of `tan`. The distances from the base are `h cot(30)`, `h cot(45)`, and `h cot(60)`, which are `h√3`, `h`, and `h/√3`. The differences `AB` and `BC` and their ratio can then be quickly calculated.

Answer 1

The Correct Option is A

Let the height of the tower be \( h \), and let the foot of the tower be O. The points A, B, C are on a line leading to O. Let the distances of A, B, and C from the foot of the tower be \( OA, OB, OC \). From the angles of elevation, we have: \[\begin{array}{rl} \bullet & \text{From point A (angle 30\(^{\circ}\)): \( \tan(30^{\circ}) = \frac{h}{OA} \Rightarrow OA = \frac{h}{\tan(30^{\circ})} = h\sqrt{3} \)} \\ \bullet & \text{From point B (angle 45\(^{\circ}\)): \( \tan(45^{\circ}) = \frac{h}{OB} \Rightarrow OB = \frac{h}{\tan(45^{\circ})} = h \)} \\ \bullet & \text{From point C (angle 60\(^{\circ}\)): \( \tan(60^{\circ}) = \frac{h}{OC} \Rightarrow OC = \frac{h}{\tan(60^{\circ})} = \frac{h}{\sqrt{3}} \)} \\ \end{array}\] Since the angles of elevation are increasing (30, 45, 60), the points are getting closer to the tower. So the order is A, B, C, O. The distances between the points are: \[ AB = OA - OB = h\sqrt{3} - h = h(\sqrt{3} - 1) \] \[ BC = OB - OC = h - \frac{h}{\sqrt{3}} = h\left(1 - \frac{1}{\sqrt{3}}\right) = h\left(\frac{\sqrt{3} - 1}{\sqrt{3}}\right) \] Now we find the ratio of AB to BC: \[ \frac{AB}{BC} = \frac{h(\sqrt{3} - 1)}{h\left(\frac{\sqrt{3} - 1}{\sqrt{3}}\right)} \] \[ \frac{AB}{BC} = \frac{1}{1/\sqrt{3}} = \sqrt{3} \] So, the ratio \( AB:BC \) is \( \sqrt{3}:1 \).