Problem 55 - A Complete Resource Book in Mathematics for JEE Main 2017

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

None of the options correctly match \(\tan^{-1}(2)\). Double-check the problem statement.

Step 1: Identify Key Points and Distances

Since the hall is cubical with sides of 10 m, we first note the coordinates of points in the 3D space. Assume point A is at (0,0,0), B at (10, 0, 0), etc. Since T is the midpoint of AB, its coordinates are (5, 0, 0). G, being the center of the face BERQ, has coordinates (5, 5, 10).

Step 2: Determine the Vector GT

Find the vector \(\overrightarrow{GT}\) by subtracting the coordinates of T from G: \(\overrightarrow{GT} = (5, 5, 10) - (5, 0, 0) = (0, 5, 10)\). This vector gives the direction from T to G.

Step 3: Calculate the Magnitude of Vector GT

Use the formula for the magnitude of a vector: \(\|\overrightarrow{GT}\| = \sqrt{0^2 + 5^2 + 10^2} = \sqrt{125}\). This simplifies to \(5\sqrt{5}\).

Step 4: Find Vertical Component Height

Note that the vertical component of the vector is the difference in z-coordinates from T to G, which is 10 m.

Step 5: Compute Angle of Elevation Using Trigonometry

The angle of elevation, \(\theta\), is found using \( \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} = \frac{10}{5} = 2\). Therefore, \(\theta = \tan^{-1}(2)\).

Step 6: Correctly Identify the Given Angle Options

The task is to match this angle \(\theta = \tan^{-1}(2)\) with the given options. Note that none directly match \(\tan^{-1}(2)\). Instead, note if there was any mistake or double-check calculations.

Key Concepts

Angle of Elevation

The angle of elevation is the angle formed between the horizontal plane and a line of sight pointing upward towards an object. It is a fundamental concept often used in real-world applications and problems involving heights and distances.

Here's how it works:

• Imagine you are standing on the ground, and you look up at the top of a building. The angle your line of sight makes with the level ground is called the angle of elevation.
• To calculate the angle of elevation, you commonly use the tangent function from trigonometry: \( \tan(\theta) = \frac{\text{opposite side}}{\text{adjacent side}} \), where the opposite side is the height of the object and the adjacent side is the distance from the object.
• In 3D geometry, this concept becomes crucial when determining perspectives or constructing architectural designs, much like the cubical hall problem where we calculated \( \theta = \tan^{-1}(2) \).

Recognizing and calculating this angle accurately is vital for solving problems that deal with distances and elevations in a spatial context.

Vector Magnitude

Vector magnitude is a measure of a vector's length. It is calculated by taking the square root of the sum of the squares of its components.

Vectors have both magnitude (length) and direction, and they are essential in representing quantities in physics and engineering that aren't just confined to a linear path.

For a vector \(\overrightarrow{v} = (x, y, z)\):

• The magnitude is calculated as \( \|\overrightarrow{v}\| = \sqrt{x^2 + y^2 + z^2} \).
• This calculation tells us how strong or how far a vector reaches in its space, similar to how we found the magnitude of \( \overrightarrow{GT} \) to be \( 5\sqrt{5} \) in the cubical hall problem.
• Vector magnitude is frequently used in force calculations, motion analysis, and any context where directional quantities need resolution.

Understanding vector magnitude strengthens your ability to manipulate and interpret vectors in multidimensional contexts.

Trigonometric Concepts

Trigonometry involves the study of relationships between the angles and sides of triangles, often extending into 3D applications in geometry and physics.

This field of mathematics uses specific functions to relate angles with side ratios:

• Sine, Cosine, and Tangent: Three primary trigonometric functions. They help relate angles to side lengths in right triangles: for an angle \(\theta\), \( \sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}} \), \( \cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}} \), and \( \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} \).
• In 3D scenarios, these functions help derive angles of elevation and depression, as they help translate positional points into meaningful angles when viewing objects from different spots.
• Through these studies, we could identify the angle of elevation using \( \tan(\theta) = \frac{10}{5} = 2 \), situationally solving how the viewer's line of sight ascends towards a point in space.

Grasping trigonometric concepts is crucial for solving spatial problems, and they serve as a bridge between geometric interpretations and algebraic solutions.