Problem 67 - A Complete Resource Book in Mathematics for JEE Main 2017

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 \(\mathrm{BC}\) 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 (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 option (B): \(\frac{7 \sqrt{3}}{2} \cdot(\sqrt{3}+1)\) m.

Step 1: Understand the Problem

When the man is at point C, he sees the top of pole A at an angle of elevation of \(60^{\circ}\) and then moves 7 m further to point D where the angle changes to \(45^{\circ}\). The task is to find the height of the pole AB.

Step 2: Setting the Scene with a Diagram

Visualizing the problem can be helpful. Imagine a right triangle where AB is the height, BC is the horizontal distance to point C, and AC is the hypotenuse. Similarly, AD is the hypotenuse of the triangle with \(45^{\circ}\).

Step 3: Use Trigonometry for First Triangle at Point C

For the first position at C, the tangent of \(60^{\circ}\) can be used: \(\tan(60^{\circ}) = \frac{AB}{BC} = \sqrt{3}\). Thus, \(AB = BC \cdot \sqrt{3}\).

Step 4: Use Trigonometry for Second Triangle at Point D

From point D, we use \(\tan(45^{\circ}) = \frac{AB}{BD} = 1\), implying \(AB = BD\). This shows that the height from D to A is the same as the distance BD.

Step 5: Express BD in Terms of BC

Since CD = 7 m and D lies on the same line as BC, then \(BD = BC + 7.\) Substituting \(AB = BD\) from Step 4, we have \(AB = BC + 7.\)

Step 6: Solve the System of Equations

You now have \(AB = BC \cdot \sqrt{3}\) and \(AB = BC + 7\). Set \(BC \cdot \sqrt{3} = BC + 7\). Rearrange to find BC: \(BC(\sqrt{3} - 1) = 7\), giving \(BC = \frac{7}{\sqrt{3} - 1}\).

Step 7: Calculate the Height of the Pole

Substitute \(BC\) back into the expression for \(AB\): \(AB = \frac{7\sqrt{3}}{\sqrt{3} - 1}\). Simplifying by rationalizing the denominator, \(AB = \frac{7\sqrt{3}}{\sqrt{3} - 1} \cdot \frac{\sqrt{3} + 1}{\sqrt{3} + 1}\), gives \(AB = \frac{7\sqrt{3}(\sqrt{3} + 1)}{2}= \frac{7\sqrt{3}}{2}(\sqrt{3}+1)\).

Step 8: Match the Solution to Given Choices

Compare the simplified expression \(\frac{7\sqrt{3}}{2}(\sqrt{3}+1)\) with the provided choices. It matches option (B).

Key Concepts

Angle of Elevation

The angle of elevation is an essential concept in trigonometry, especially in solving real-world problems involving heights and distances. When an observer looks at an object that is above the horizontal line of sight, the angle formed between the line of sight and the horizontal is known as the angle of elevation. It is always measured from the horizontal upwards.

In the given exercise, the man observes the top of the pole (point A) from two different points (C and D). Initially, the angle of elevation from point C is given as \(60^\circ\), and then from point D, it decreases to \(45^\circ\). These angles of elevation help us form right triangles and set up trigonometric equations to find unknown distances and heights.

Understanding how to use angles of elevation is key to solving problems involving right triangles and indirect measurement. Be sure to observe which side or segment of a triangle relates to the angle given for correct equation formation.

Right Triangle

Right triangles are a fundamental element in trigonometry. They comprise one \(90^\circ\) angle and two other angles that add up to \(90^\circ\). Their special properties, such as the Pythagorean theorem and trigonometric ratios, make them incredibly useful in solving problems.

In the problem given, two right triangles are formed as the man observes the pole from different points. In both triangles, the height of the pole represents the vertical leg, while the base corresponds to the horizontal distance from the observer to the point directly beneath the object being observed.

Using the right triangle formed at point C, with an angle of elevation \(60^\circ\), we apply the tangent function to form the equation \( \tan(60^\circ) = \frac{AB}{BC} \). Similarly, in the second triangle, from point D, \( \tan(45^\circ) = \frac{AB}{BD} \). These trigonometric ratios are crucial for relating the different sides of the triangle when angles are known.

Problem Solving

Problem solving in trigonometry often involves understanding the scenario, identifying right triangles, and applying relevant formulas to find missing measurements. The step-by-step process demonstrated in the solution illustrates how to approach a typical exercise.

The key steps include:

• Comprehending the problem by visualizing the scenario and drawing diagrams to map out the situation.
• Identifying known and unknown values in the setup.
• Utilizing trigonometric identities, such as the tangent function, which relates angles to the opposite and adjacent sides of a right triangle.
• Creating equations from these identities and solving for unknowns through algebraic manipulation.

In this specific problem, by setting equations for the heights and solving the system, we systematically found the height of the pole. Problem solving in this context requires a clear understanding of trigonometric concepts and the ability to apply them in sequence.