Problem 68 - A Complete Resource Book in Mathematics for JEE Main

Question

A bird is sitting on the top of a vertical pole \(20 \mathrm{~m}\) high which makes an angle of elevation \(45^{\circ}\) from a point \(O\) on the ground. It flies off horizontally straight away from the point \(O\). After one second, the elevation of the bird from \(O\) is reduced to \(30^{\circ}\). Then the speed (in \(\mathrm{m} / \mathrm{s}\) ) of the bird is (A) \(40(\sqrt{2}-1)\) (B) \(40(\sqrt{3}-\sqrt{2})\) (C) \(20 \sqrt{2}\) (D) \(20(\sqrt{3}-1)\)

Step-by-Step Solution

Answer

The speed of the bird is (D) 20(\sqrt{3}-1) m/s.

Step 1: Initial Setup and Understanding

We begin by identifying the given information. The bird starts at the top of a pole that is 20 meters high. Initially, the angle of elevation from the point O to the bird is \(45^\circ\). This scenario is a right triangle with the pole as the opposite side, the distance from point O to the base of the pole as the adjacent side, and the line of sight as the hypotenuse.

Step 2: Calculating Initial Distance from O

Using the tangent of the initial angle (\(45^\circ\)), we have \(\tan(45^\circ) = \frac{20}{x}\), where x is the horizontal distance from point O to the base of the pole. Since \(\tan(45^\circ) = 1\), we have \(x = 20\), so the initial distance is 20 meters.

Step 3: Calculate New Distance Using 30° Elevation

After 1 second, the angle of elevation changes to \(30^\circ\). Using \(\tan(30^\circ) = \frac{20}{x'}\), where \(x'\) is the new horizontal distance, we have \(\frac{1}{\sqrt{3}} = \frac{20}{x'}\). Solving for \(x'\), we get \(x' = 20\sqrt{3}\).

Step 4: Calculating Bird's Speed

The horizontal distance traveled by the bird in one second is \(x' - x = 20\sqrt{3} - 20\). Therefore, the speed of the bird is \(20(\sqrt{3} - 1)\) meters per second.

Key Concepts

Angle of Elevation

In trigonometry, the angle of elevation is the angle formed between the horizontal level and the line of sight when an observer looks at an object above the horizontal. This concept is particularly useful in navigating right triangle problems, such as when calculating distances or heights. For example, if you are standing at point O and looking at the top of a pole, the angle formed between your line of sight and the horizontal ground is the angle of elevation.
In this exercise, the bird's initial angle of elevation is given as \(45^{\circ}\), which subsequently reduces to \(30^{\circ}\) after one second. These angles help us establish relationships between the known heights and unknown distances.

Right Triangle Calculations

Right triangle calculations are pivotal in solving trigonometry problems involving angles of elevation. In a right triangle, one angle is always \(90^{\circ}\). The other two angles add up to \(90^{\circ}\). For this exercise, two different triangles are formed due to the change in the bird's position.
The first triangle is formed with the bird at the top of the pole, the base of the pole as the horizontal side, and the line of sight as the hypotenuse. Using this setup, we can utilize trigonometric functions to find unknown side lengths. Typically, the tangent function is used when dealing only with the opposite and adjacent sides, as seen in our exercise.

Tangent Function

The tangent function in trigonometry is defined as the ratio of the length of the opposite side to the adjacent side for a given angle in a right triangle. It's given by the formula: \(\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}}\). This function is particularly useful in our problem as we know the height of the pole (opposite) and need to calculate the distances (adjacent).
Initially, with an angle of \(45^{\circ}\), its tangent is \(1\), leading to the adjacent side being \(20\, \text{m}\). After one second, with a new angle of \(30^{\circ}\), the tangent becomes \(\frac{1}{\sqrt{3}}\), allowing us to solve for the new horizontal distance \(x'\), which is calculated as \(20\sqrt{3}\).

Speed Calculation

Speed calculation in this context involves determining how fast the bird travels horizontally, given the change in the angle of elevation. Since speed is defined as distance traveled over time, we calculate the bird's speed based on how far it moves horizontally in one second.
From the step-by-step solution, we found that the bird moves from a horizontal distance of \(20\) meters to \(20\sqrt{3}\) meters in one second. Thus, the speed is the difference between these two distances, \(20(\sqrt{3} - 1)\), divided by the time interval, which is one second. This gives us the final speed of the bird as \(20(\sqrt{3} - 1) \; \text{m/s}\). This method provides a straightforward yet effective way to compute the speed using trigonometric relationships.