Problem 55 - A Complete Resource Book in Mathematics for JEE Main

Question

In a cubical hall abcdpqrs with each side \(10 \mathrm{~m}, \mathrm{G}\) is the centre of the wall berq and \(\mathrm{T}\) is the mid point of the side \(\mathrm{AB}\). The angle of elevation of \(\mathrm{G}\) at the point \(\mathrm{T}\) is (A) \(\sin ^{-1} \frac{1}{\sqrt{3}}\) (B) \(\cos ^{-1} \frac{1}{\sqrt{3}}\) (C) \(\tan ^{-1} \frac{1}{\sqrt{3}}\) (D) \(\cot ^{-1} \frac{1}{\sqrt{3}}\)

Step-by-Step Solution

Answer

The correct answer is (C) \( \tan^{-1}\left(\frac{1}{\sqrt{3}}\right).\)

Step 1: Understand the Problem and Diagram the Cube

Visualize the cube with sides of length 10 m. Let the vertices be labeled such that points \( A, B, C, D \) are one face, \( P, Q, R, S \) are another face.\( G \) is the center of the wall \( BERQ \), and \( T \) is the midpoint of \( AB \). To solve for the angle of elevation from \( T \) to \( G \), we'll establish coordinates for these points and then find the angle of elevation.

Step 2: Assign Coordinates to Known Points

Let's place the cube in a 3D coordinate system:- \( A = (0, 0, 0) \)- \( B = (10, 0, 0) \)- \( T = (5, 0, 0) \) because it is the midpoint of \( AB \) - \( E = (10, 10, 0) \)- \( R = (10, 10, 10) \)- \( Q = (0, 10, 10) \)- \( G = (5, 10, 5) \) because it is the midpoint of the face \( BERQ \).

Step 3: Calculate the Distance Between T and G

\( \text{The vector from } T \text{ to } G \text{ is } (5, 10, 5). \)The length of this vector, \( TG \), can be calculated as follows:\[ TG = \sqrt{(5-5)^2 + (10-0)^2 + (5-0)^2} = \sqrt{0 +100 + 25} = \sqrt{125} = 5\sqrt{5}. \]

Step 4: Calculate the Angle of Elevation from T to G

The angle of elevation \( \theta \) is found using the vertical and horizontal distances between \( T \) and \( G \). The vertical distance is 5 (\( z \) component) and the horizontal distance is the projection of \( TG \) onto the x-y plane, which is 10:\[ \tan \theta = \frac{\text{Opposite (vertical distance)}}{\text{Adjacent (horizontal distance)}} = \frac{5}{\sqrt{5^2 + 10^2}} = \frac{5}{\sqrt{125}} = \frac{1}{\sqrt{3}}. \]Thus, \( \theta = \tan^{-1}\left(\frac{1}{\sqrt{3}}\right).\)

Step 5: Choose the Correct Option Based on Calculations

Based on our calculation, the correct angle of elevation is represented by the inverse tangent function. Thus, the correct answer is option (C) \( \tan^{-1}\left(\frac{1}{\sqrt{3}}\right). \)

Key Concepts

Angle of Elevation

The angle of elevation refers to the angle formed between the horizontal plane and the line of sight to a point above the horizontal plane. In three-dimensional Euclidean geometry, understanding the angle of elevation is crucial, especially when it comes to visualizing and solving problems involving heights and distances.
For example, consider standing at point T and looking up at point G. The line of sight from T to G creates the angle of elevation with a horizontal line at your eye level. When you know the vertical height and horizontal distance between two points, you can calculate this angle using trigonometric functions such as tangent. The formula to find the angle of elevation, when given the vertical height (opposite side) and the horizontal distance (adjacent side), is the inverse tangent function:

• \( \tan \theta = \frac{\text{Vertical height}}{\text{Horizontal distance}} \)

In this exercise, the angle from point T to point G is calculated using this principle, leading you to the correct choice of \( \theta = \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) \). Simplifying these computations and using inverse trigonometric functions will help you find the angle easily.

Coordinate System

A coordinate system in 3D Euclidean geometry provides a framework to pinpoint the location of points in space using an ordered set of numbers. In this exercise, the points of the cube are established in a three-dimensional Cartesian coordinate system.
Each vertex of the cube is assigned a set of coordinates, such as \( A = (0, 0, 0) \), where each number corresponds to a position along the x, y, and z axes. This helps in calculating distances and angles between points by providing precise positions in space. Specifically, this exercise places:

• T = (5, 0, 0), middle of AB, on the x-axis.
• G = (5, 10, 5), center of the face BERQ, within the cube.

By placing these points onto a math-based grid, you can easily compute vectors, distances, and angles using simple arithmetic and geometric equations. The application of a 3D coordinate system is foundational in visually and numerically solving problems involving spatial orientation and measurement.

Vector Analysis

Vector analysis is a powerful mathematical tool used to deal with quantities having both magnitude and direction. In the context of 3D Euclidean geometry, vectors help describe the relative positions and movements within space.
In this exercise, the vector from point T to point G is crucial. Identified as \((5, 10, 5)\), this vector outlines the path one would take to travel from one point to another in the cube. Vectors allow us to solve problems like finding lengths, angles, and projections. Steps involve:

• Determining the vector coordinates by subtracting the coordinates of point T from those of point G.
• Calculating the magnitude of the vector using the formula \( \| {\mathbf{v}} \| = \sqrt{x^2 + y^2 + z^2} \).

For instance, the length of the vector TG, known as the magnitude, represents the straight-line distance between T and G, which simplifies geometric computations in 3D space. Understanding vectors and how to manipulate them is essential for various fields, including physics, engineering, and computer graphics.