Problem 67 - A Complete Resource Book in Mathematics for JEE Main

Question

\(\mathrm{AB}\) is a vertical pole with \(\mathrm{B}\) at the ground level and A at the top. A man finds that the angle of elevation of the point A from a certain point \(\mathrm{C}\) on the ground is \(60^{\circ} .\) He moves away from the pole along the line BC to a point \(\mathrm{D}\) such that \(\mathrm{CD}=7 \mathrm{~m}\). From D the angle of elevation of the point \(\mathrm{A}\) is \(45^{\circ}\). Then the height of the pole is \(\quad\) (A) \(\frac{7 \sqrt{3}}{2} \cdot \frac{1}{\sqrt{3}-1} m\) (B) \(\frac{7 \sqrt{3}}{2} \cdot(\sqrt{3}+1) m\) (C) \(\frac{7 \sqrt{3}}{2} \cdot(\sqrt{3}-1) m\) (D) \(\frac{7 \sqrt{3}}{2} \cdot \frac{1}{\sqrt{3}+1}\)

Step-by-Step Solution

Answer

The height of the pole is (A) \( \frac{7 \sqrt{3}}{2} \cdot \frac{1}{\sqrt{3}-1} \) m.

Step 1: Draw the Triangle and Name Variables

Visualize the scenario with a right triangle where \( B \) is at ground level, \( A \) is the top of the pole, and \( C \) and \( D \) are points on the ground. Let \( h \) represent the height of the pole \( AB \), \( x \) represent the distance \( BC \), and use the fact that \( CD = 7 \) m.

Step 2: Use Trigonometry at Point C

Apply the tangent function at point \( C \) where the angle of elevation is \( 60^{\circ} \). We have \( \tan(60^{\circ}) = \frac{h}{x} \). Since \( \tan(60^{\circ}) = \sqrt{3} \), this gives the equation \( h = x \sqrt{3} \).

Step 3: Use Trigonometry at Point D

Now, apply the tangent function at point \( D \) where the angle of elevation is \( 45^{\circ} \). In this case, \( \tan(45^{\circ}) = \frac{h}{x + 7} \). Since \( \tan(45^{\circ}) = 1 \), we have \( h = x + 7 \).

Step 4: Set Up an Equation System

From Steps 2 and 3, we have two equations: \( h = x \sqrt{3} \) and \( h = x + 7 \). Equating both expressions for \( h \), we get \( x \sqrt{3} = x + 7 \).

Step 5: Solve for x

Rearrange the equation \( x \sqrt{3} = x + 7 \) to isolate \( x \). This gives \( x(\sqrt{3} - 1) = 7 \). Solve for \( x \) to deduce that \( x = \frac{7}{\sqrt{3} - 1} \).

Step 6: Calculate the Height h

Substitute \( x = \frac{7}{\sqrt{3} - 1} \) back into the equation for \( h \), \( h = x \sqrt{3} \). Thus, \( h = \frac{7 \sqrt{3}}{\sqrt{3} - 1} \). Simplify to find \( h = \frac{7 \sqrt{3}}{2} \cdot \frac{1}{\sqrt{3} - 1} \).

Step 7: Simplify the Expression

The expression for the height \( h \) corresponds to option (A) in the given problem statement, as \( \frac{7 \sqrt{3}}{2} \cdot \frac{1}{\sqrt{3} - 1} \) matches the option provided.

Key Concepts

Angle of Elevation

The concept of the **angle of elevation** is important in trigonometry when solving real-world problems involving heights. The **angle of elevation** is the angle between the line of sight of an observer and a horizontal line drawn from the observer's position. This angle is measured from the ground up towards an object that the observer is looking at. For example, if you are observing the top of a pole, the angle of elevation will be the angle formed from your line of sight upwards to the top of the pole.

In problems, the **angle of elevation** often helps determine heights or distances that cannot be easily measured directly. When an observer moves further from the object, this angle decreases, which is a key point in such exercises. It creates different scenarios for applying trigonometric functions like the tangent function, as the distance from the object changes your viewpoint and thus the angle of elevation.

Tangent Function

The **tangent function** in trigonometry is a useful tool for finding unknown sides in right triangles. Generally, the tangent of an angle is defined as the ratio of the opposite side to the adjacent side. In mathematical terms, for a triangle with angle \( \theta \), the tangent function is expressed as \( \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} \).

In our problem, the tangent function is used at two points, C and D, with angles of elevation of \(60^{\circ}\) and \(45^{\circ}\) respectively:

• At point C, \( \tan(60^{\circ}) = \sqrt{3} \), leading to the equation \( h = x\sqrt{3} \).
• At point D, \( \tan(45^{\circ}) = 1 \), forming the equation \( h = x + 7 \).

These equations form the foundation for setting up a relationship between the height of the pole (h) and the horizontal distances (x and x+7 meters) from each reference point.

Right Triangle

A **right triangle** is a triangle in which one of the angles is exactly \(90^{\circ}\). This implies that the other two angles are complementary, adding up to \(90^{\circ}\) as well. The defining feature of a right triangle makes it an essential shape in trigonometry because trigonometric ratios like sine, cosine, and tangent are defined based on the sides of a right triangle.

In the given exercise, the scenario can be visualized through two separate right triangles:

• One triangle at point C, where the base is \(BC\), height is \(h\), and the angle of elevation is \(60^{\circ}\).
• Another triangle at point D, where we further extend the base by \(CD = 7\) meters. Here, the angle of elevation is \(45^{\circ}\).

These right triangles are crucial for applying trigonometric functions, which help solve for unknown lengths when given an angle. Each triangle's properties, such as its angles and sides, allow us to establish the direct relationships needed to solve the height of the pole accurately.