Problem 32 - A Complete Resource Book in Mathematics for JEE Main

Question

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

Step-by-Step Solution

Answer

The height of the pole is \( \frac{7 \sqrt{3}}{2} \times (\sqrt{3} + 1) \) meters (Option B).

Step 1: Initial Condition at Point C

From point C, the angle of elevation to the top of the pole A is given as \(60^\circ\). This creates a right triangle with the height of the pole AB as opposite the angle and the distance BC as the adjacent. Using tangent, \( \tan 60^\circ = \frac{AB}{BC} \).

Step 2: Define Tangent Function at 60°

We know \( \tan 60^\circ = \sqrt{3} \). So we get the equation: \[ \sqrt{3} = \frac{h}{x} \] where \( h \) is the height of the pole and \( x \) is the distance BC. Rearranging gives: \[ h = x \sqrt{3} \]

Step 3: Condition at Point D

When the man moves to point D, the angle of elevation becomes \(45^\circ\). Therefore, \( \tan 45^\circ = \frac{h}{y} \) where \( y \) is the distance from D to the base B which equals \( x + 7 \).

Step 4: Tangent Function at 45°

We know \( \tan 45^\circ = 1 \). This gives the equation: \[ 1 = \frac{h}{x+7} \] simplifying gives: \[ h = x + 7 \]

Step 5: Equate and Solve for x

We now have two equations for h: \( h = x \sqrt{3} \) and \( h = x + 7 \) from Step 4. Setting them equal gives: \[ x \sqrt{3} = x + 7 \]Solving for \( x \), we have:\[ x (\sqrt{3} - 1) = 7 \]Thus, \[ x = \frac{7}{\sqrt{3} - 1} \]

Step 6: Substitute x to find h

Now substitute \( x = \frac{7}{\sqrt{3} - 1} \) into \( h = x \sqrt{3} \) to get:\[ h = \left(\frac{7}{\sqrt{3} - 1}\right) \times \sqrt{3} \]To simplify, multiply numerator and denominator by the conjugate \(\sqrt{3} + 1\):\[ h = \frac{7 \sqrt{3} \times (\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} = \frac{7 \sqrt{3} \times (\sqrt{3} + 1)}{2} \]

Step 7: Final Solution

Rewriting, the height of the pole is given by:\[ h = \frac{7 \sqrt{3}}{2} \times (\sqrt{3} + 1) \text{ meters} \]Thus, the correct answer is option B.

Key Concepts

Angle of Elevation

The angle of elevation is an angle formed by the horizontal plane and a line of sight above this plane. Imagine looking upwards from ground level at an object; this angle helps us determine how steeply we are looking up. In our exercise, the man sights the top of the pole, and two angles of elevation—60° from point C and 45° from point D—are given. These angles lead to forming two different right triangles, which are essential for solving height-related problems.
The angle of elevation is a fundamental part of trigonometry because it allows us to relate distances and heights without needing a direct measurement of either. By knowing one angle and one side length (either opposite or adjacent), you can use trigonometric functions to find the missing dimensions.
Remember, angles of elevation are only applicable when considering observers looking above the horizontal, making them vital in scenarios like this pole-sighting problem.

Tangent Function

The tangent function in trigonometry describes the ratio between the opposite side and the adjacent side in a right triangle. It is one of the basic trigonometric functions, often abbreviated as 'tan.' This function is incredibly useful when solving for unknown variables in right triangles, especially when dealing with angles of elevation.
In the provided exercise, the tangent function is crucial because:

• At point C, with an angle of elevation of 60°, the tangent function translates to \( an 60° = \frac{AB}{BC}\), which simplifies to \( an 60° = \sqrt{3}\). This relationship provides our first equation relating the height of the pole to the distance from the base.
• At point D, with a 45° angle, the relationship is described as \( an 45° = \frac{h}{x+7}\), which equals 1, yielding a second equation.

The solving process involves using these tangent identities to develop equations, allowing us to address unknown values systematically, such as the height of the pole.

Right Triangle Problem Solving

Right triangle problem solving is an essential part of trigonometry, utilizing concepts like angles of elevation and the tangent function. It involves identifying right triangles within a scenario and applying trigonometric principles to solve for missing sides or angles.
This technique is particularly effective when:

• Given one angle (other than the right angle) and a side, like in our problem where angles of 60° and 45° are used to solve for the pole's height.
• Using basic trigonometric identities—sine, cosine, and tangent—to find unknowns with given information. For example, using \( an \theta = \frac{opposite}{adjacent}\).

For the exercise at hand, solving the right triangles involved two steps: Using the tangent equations from both reference points (C and D) to form two equations as outlined in steps 3 and 4. By equating values and manipulating these equations, we found the pole's height, showcasing how right triangle problem-solving bridges the gap between theoretical and real-world applications of trigonometry.