Problem 6 - A Complete Resource Book in Mathematics for JEE Main

Question

A lamp post standing at a point \(A\) on a circular path of radius \(\mathrm{r}\) subtends an angle \(\alpha\) at some point \(B\) on the path, and \(A B\) subtends an angle of \(45^{\circ}\) at any other point on the path, then height of the lampost is (A) \(\sqrt{2} r \cot \alpha\) (B) \((r / \sqrt{2}) \tan \alpha\) (C) \(\sqrt{2} r \tan \alpha\) (D) \((r / \sqrt{2}) \cot \alpha\)

Step-by-Step Solution

Answer

The height of the lamp post is \(\sqrt{2} r \tan \alpha\), which is option (C).

Step 1: Understanding the Problem

We have a lamp post at point \( A \) on a circular path with radius \( r \). The lamp post subtends an angle \( \alpha \) at point \( B \) on the path, and the segment \( AB \) subtends an angle of \( 45^\circ \) at any other point on the path. Our goal is to find the height of the lamp post using the given information.

Step 2: Exploring Geometry of Points A, B, and O

In a circle, if a line subtends an angle \( \theta \) at the circumference, it subtends an angle \( 2\theta \) at the center. Point \( O \) is the center of the circle, so \( \angle AOB = 2 \times 45^\circ = 90^\circ \). This implies that \( AOB \) forms a right-angled triangle with \( AO = r = BO \).

Step 3: Checking Right Triangle AOB

In right triangle \( AOB \), where \( \angle AOB = 90^\circ \) and \( AO = BO = r \), use the fact that the line from the center to the tangent at the circumference is perpendicular to the radius. Thus, \( \triangle AOB \) is an isosceles right triangle with \( AB = r\sqrt{2} \).

Step 4: Subtended Angle \(\alpha\) by Lamp Post

At point \( B \) on the circle, the lamp post \( AB \) subtends an angle \( \alpha \). Assuming the height of the lamp post is \( h \), by considering the geometry of the circle and the tangent, \( \tan(\alpha) = \frac{h}{AB} \).

Step 5: Calculating Height of the Lamp Post

Substitute \( AB = r\sqrt{2} \) into \( \tan(\alpha) = \frac{h}{AB} \). Solving for \( h \), we have:\[\tan(\alpha) = \frac{h}{r\sqrt{2}} \implies h = r\sqrt{2} \tan(\alpha)\]

Step 6: Determining the Correct Option

From previous calculations, the expression for the height \( h = r\sqrt{2} \tan(\alpha) \) matches option (C). Thus, the correct answer is (C) \( \sqrt{2} r \tan \alpha \).

Key Concepts

Right Triangle Properties

Right triangle properties are fundamental in understanding geometric problems involving circles. In our problem, we establish that triangle \( \triangle AOB \) is a right-angled triangle, with a right angle at point \( O \). This particular triangle not only has a right angle but is also isosceles since \( AO = BO = r \). The right-angled property comes into play because any line subtending an angle at the circle's circumference subtends twice that angle at the circle’s center.
So, when \( AB \) subtends a \( 45^\circ \) angle at the circle, at the center \( O \), it subtends an angle of \( 2 \times 45^\circ = 90^\circ \). These properties aid in defining that in \( \triangle AOB \), the relationship of sides is such that the hypotenuse is \( r \sqrt{2} \).

• Right triangles have one angle that is \( 90^\circ \).
• In an isosceles right triangle, the hypotenuse is \( \sqrt{2} \times \) the length of each equal side.

Understanding these properties helps us identify the hypotenuse as \( AB = r\sqrt{2} \), which is crucial in further calculations.

Trigonometric Functions

Trigonometric functions come into play in calculating the height of the lamp post, denoted as \( h \). When we know that a given side subtends an angle \( \alpha \) at a point on the circumference, trigonometry provides us with the formula \( \tan(\alpha) = \frac{h}{AB} \).
In this formula, \( \tan(\alpha) \) establishes the ratio of the opposite side (the height \( h \)) to the adjacent side \( AB \), which in our problem is \( r \sqrt{2} \). Subsequently, solving \( \tan(\alpha) = \frac{h}{r\sqrt{2}} \) for \( h \), we discover:

• \( h = r \sqrt{2} \tan(\alpha) \)

Trigonometric functions like tangent are crucial for solving problems where we need to find unknown sides or angles using known measurements on right triangles.

Circular Path

The circular path presents a unique setup that intertwines geometry and trigonometry. When we talk about points such as \( A \), \( B \), and the center \( O \) on a circular path, they form a cyclic geometry which has specific properties.
In our scenario, a full circle surrounds the problem, and features such as:

• Perpendicular radii and tangent properties create right angles.
• The segment \( AB \) subtends equal angles along numerous points on the path, showcasing a symmetrical behavior.

In circle geometry, when a line subtends angles at various points, using known angles and trigonometric relationships, we can find unknown values related to the circle, such as the height of the lamp post. The circular path isn't just a geometrical constraint but a source of symmetric relationships exploited in solving the problem.