Problem 3 - A Complete Resource Book in Mathematics for JEE Main

Question

\(A\) and \(B\) are two points in the horizontal plane through \(O\), the foot of pillar \(O P\) of height \(h\), such that \(\triangle A O B=\theta\). If the elevation of the top of the pilar from \(A\) and \(B\) are also equal to \(\theta\), then \(A B\) is equal to (A) \(h \cot \theta\) (B) \(h \cos \theta \sec \frac{\theta}{2}\) (C) \(h \cot \theta \sin \frac{\theta}{2}\) (D) \(h \cos \theta \operatorname{cosec} \frac{\theta}{2}\)

Step-by-Step Solution

Answer

The correct option is (C) \(h \cot \theta \sin \frac{\theta}{2}\).

Step 1: Understand the problem

We have a vertical pillar OP of height \( h \). Two points, \( A \) and \( B \), are on the horizontal ground plane through \( O \). The elevation angle of the top of the pillar from both \( A \) and \( B \) is \( \theta \), and the angle \( AOB \) is also \( \theta \). We need to find the distance between \( A \) and \( B \).

Step 2: Use trigonometric relations for elevation

For the point \( A \), the elevation angle to the pillar is \( \theta \), so we can express the tangent function as \( \tan \theta = \frac{h}{OA} \). Thus, \( OA = \frac{h}{\tan \theta} \). Similarly, for point \( B \), \( OB = \frac{h}{\tan \theta} \).

Step 3: Consider triangle AOB

In the triangle \( AOB \), since \( \angle AOB = \theta \), and we have \( OA = OB = \frac{h}{\tan \theta} \). This makes \( \triangle AOB \) an isosceles triangle with equal side lengths.

Step 4: Calculate AB using the law of cosines

In \( \triangle AOB \), using the law of cosines, we have:\[ AB^2 = OA^2 + OB^2 - 2 \cdot OA \cdot OB \cdot \cos(\theta) \]Substitute \( OA = OB = \frac{h}{\tan \theta} \):\[ AB^2 = 2 \left(\frac{h}{\tan \theta}\right)^2 (1 - \cos \theta) \]

Step 5: Simplify the expression for AB

Simplifying, we have:\[ AB = \sqrt{2} \cdot \frac{h}{\tan \theta} \sqrt{1 - \cos \theta} \]Using the identity \( 1 - \cos \theta = 2 \sin^2 \frac{\theta}{2} \), it becomes:\[ AB = \sqrt{2} \cdot \frac{h}{\tan \theta} \cdot \sqrt{2} \cdot \sin \frac{\theta}{2} \]\[ = \frac{2h}{\tan \theta} \cdot \sin \frac{\theta}{2} \]\( \cot \theta = \frac{1}{\tan \theta} \), so:\[ AB = 2h \cdot \cot \theta \cdot \sin \frac{\theta}{2} \]

Step 6: Choose the correct option

Among the given options, the expression that matches is (C): \( h \cot \theta \sin \frac{\theta}{2} \).

Key Concepts

Law of Cosines

The Law of Cosines is a vital tool in trigonometry, particularly when dealing with any kind of triangle. It helps to find unknown sides or angles by relating the lengths of the sides with the cosine of one of the angles. This comes particularly handy in non-right triangles.

For any triangle with sides of lengths \(a\), \(b\), and \(c\), and an angle \(C\) opposite side \(c\), the law of cosines is expressed as:

• \[ c^2 = a^2 + b^2 - 2ab\cos(C) \]

This formula is useful when you know two sides of a triangle and the included angle, making it possible to solve for the third side.

In the problem, we used the Law of Cosines in triangle \(AOB\) to find the length of \(AB\). We knew the sides \(OA\) and \(OB\) and the angle \(\theta\) between them. The law allowed us to set up an equation to solve for \(AB\), which simplifies the process when direct measurement isn't possible.

This principle is not just theoretical; it finds real-world application in various fields such as engineering, navigation, and architecture. Whenever precise measurements of angles and distances are necessary, the Law of Cosines can be an indispensable ally.

Elevation Angle

The concept of an elevation angle is commonly encountered in problems involving heights and distances. It is the angle formed by the line of sight of an observer looking at an object above or below them and the horizontal plane.

Consider a scenario where you're standing on the ground looking at the top of a pillar. The line of sight forms an angle with the horizontal plane running along the ground. This is the angle of elevation.

• In mathematical terms, if\( OP\) is the height, and `\( OA\)` is the distance along the ground: \( \tan(\theta) = \frac{height}{base}\)

In the exercise, both points \(A\) and \(B\) had equal elevation angles \(\theta\) relative to the same pillar:\[ \tan \theta = \frac{h}{OA} \] and \[ \tan \theta = \frac{h}{OB} \]Thus, \( OA \) and \( OB \) became easily computable as the same length, employing the tangent function to relate them to the pillar's height \( h \).

Elevation angles play a pivotal role in surveying, astronomy, and various engineering applications, where it is often crucial to determine positions and angles accurately.

Isosceles Triangle

An isosceles triangle is one of the simpler yet fascinating kinds of triangles. By definition, it has at least two sides of equal length, which also means it has two equal angles opposite those sides.

For example, if a triangle has two sides \(x\) and \(x\), the triangle is classified as isosceles.

• The property of having equal sides makes calculations involving isosceles triangles straightforward because you can use symmetry, which often simplifies problems.

In the problem, \( \triangle AOB\) was an isosceles triangle. Both sides \(OA\) and \(OB\) were equal due to the equal elevation angles, making calculations for the unknown side \(AB\) easier.

Understanding isosceles triangles is crucial, as they appear frequently in various geometric proofs and real-life structures. Whether in constructing stable bases or in tiling designs, recognizing isosceles triangles helps one appreciate symmetry and balance, advantageous in scientific and artistic endeavors alike.