Problem 6 - Comprehensive Trigonometry for IIT JEE Main and Advanced Rejaul Makshud MeGraw Hill

Question

A man notices two objects in a striaght line due west of him. After walking a distance \(c\) due north he observes that the objects subtend an angle \(\alpha\) at his eye and after walking a further distance \(c\) due north, an angle \(\beta\). Find the distance between the objects.

Step-by-Step Solution

Answer

The distance x between the two objects is \(2c \cdot \tan(\alpha)\).

Step 1: Draw the problem

From the problem, draw a figure to understand better. Let O1 and O2 be the two objects, P be the original position of the man, Q be the position of the man after walking distance c, and R be the man's position after walking another distance c. Mark angles PO1O2 as alpha and QO2O1 as beta. Now there will be two right triangles PO1O2 and QO1O2 with PO1 = QO1 = c.

Step 2: Use Tangent of angles to derive equations

In the triangle PO1O2, we have \(\tan(\alpha) = \frac{O1O2}{c}\). And in the triangle QO1O2, we have \(\tan(\beta) = \frac{O1O2}{2c}\). Denote O1O2 as x. So we have two equations: \(x = c \cdot \tan(\alpha)\) and \(x = 2c \cdot \tan(\beta)\).

Step 3: Solve for x

We have \(c \cdot \tan(\alpha) = 2c \cdot \tan(\beta)\). This can be simplified to \(\tan(\alpha) = 2 \cdot \tan(\beta)\). By using the identity for the tangent of the sum of two angles, this equation simplifies to \(x = \frac{2c \cdot \tan(\beta)}{1 - \tan^{2}(\beta)} = \frac{2c \cdot \tan(\beta)}{1 - \tan(\alpha)}.\)

Step 4: Find x

From the equation in Step 3, solve for x, which is the distance between the two objects. We can write: \(x = \frac{2c \cdot \tan(\beta) \cdot (1 + \tan(\alpha))}{1 - \tan(\alpha)} = 2c \cdot \tan(\alpha)\). Solve for x to get the distance between the objects:

Key Concepts

Angle of Elevation

The angle of elevation is an important concept in trigonometry. It is the angle formed between the horizontal line and the line of sight to an object that is above the horizontal line. In this exercise, the angle of elevation is used to determine the angles \( \alpha \) and \( \beta \), as the man observes the two objects from a distance.
Understanding the angle of elevation helps in visualizing how the height or distance of an object can be calculated when standing at a certain distance from it. This angle is always measured from the ground upwards to the point of observation, which in geometry helps establish vertical and horizontal relationships.

• Angles \( \alpha \) and \( \beta \) represent the man's perspective as he moves northward.
• These angles help us establish equations using trigonometric functions like tangent.

Recognizing these angles allows us to apply trigonometric identities effectively to calculate distances or heights in real-world contexts.

Tangent Function

The tangent function in trigonometry relates an angle of a right triangle to the ratio of the opposite side over the adjacent side. In the context of this problem, the tangent of angles \( \alpha \) and \( \beta \) are crucial.
The initial equation \( \tan(\alpha) = \frac{O1O2}{c} \) links the length between the objects (\( O1O2 \)) to the distance the man walked north (\( c \)). Similarly, \( \tan(\beta) = \frac{O1O2}{2c} \) forms the second equation after further observation.

• Tangent identities are employed to relate angles and distances in right triangles.
• The equations help determine the relationship to solve for the unknown distance.

By understanding how to use the tangent function, students can solve geometric problems involving height or distance.

Right Triangle Properties

Right triangles form the basis of many trigonometric applications, including this exercise. They are defined by having one angle equal to 90 degrees, forming two legs and a hypotenuse.
Here, two right triangles, \( PO_1O_2 \) and \( QO_1O_2 \), allow us to use the properties of right triangles to move forward in solving the problem.

• The properties of these triangles facilitate the use of the tangent function for solving the angle and distance problems.
• Consistency in the length of one side (\( c \)) helps create symmetrical equations to solve for \( x \).

Studying the properties of right triangles supports understanding how and why certain trigonometric functions are applied to different scenarios and enhances problem-solving skills in geometry.

Distance Calculation

Distance calculation using trigonometry involves relating angles and sides of triangles. In this exercise, the distance \( x \) between the two objects is found by creating and solving equations with tangent functions.
To find \( x \) from the given angles \( \alpha \) and \( \beta \):

• Initially, derive two equations \( x = c \cdot \tan(\alpha) \) and \( x = 2c \cdot \tan(\beta) \).
• Equate the two expressions for \( x \) to isolate \( x \).
• Solve this equation to find the value of \( x \), representing the distance between the objects.

By understanding these steps, students can approach similar problems involving distances measured indirectly through angles and trigonometric equations.