Problem 40 - 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 OP of height \(h\), such that \(\angle A O B=\) \(\theta\). If the elevation of the top of the pillar from \(\mathrm{A}\) and \(\mathrm{B}\) are also equal to \(\theta\), then \(\mathrm{AB}\) is equal to (A) \(\operatorname{hcot} \theta\) (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

AB = \( h \cot \theta \sin \frac{\theta}{2} \) (Option C).

Step 1: Understanding the Geometry

We have a pillar OP with height \( h \), and points A and B on the plane containing point O, such that \( \angle AOB = \theta \). The angles of elevation from A and B to P are also \( \theta \). This implies triangles OAP and OBP are right triangles.

Step 2: Applying Trigonometry in Triangles

For triangles △OAP and △OBP, the tangent of the angle of elevation \( \theta \) is \( \tan \theta = \frac{h}{OA} = \frac{h}{OB} \). Hence, \( OA = OB = \frac{h}{\tan \theta} = h \cot \theta \).

Step 3: Calculate Segment AB

Since OA = OB, triangle OAB is isosceles with \( \angle AOB = \theta \). The length of AB can be found using the law of cosines: \( AB^2 = OA^2 + OB^2 - 2 \cdot OA \cdot OB \cdot \cos \theta \). Substituting, we get \( AB = 2 \cdot OA \cdot \sin \left( \frac{\theta}{2} \right) = 2h \cot \theta \sin \left( \frac{\theta}{2} \right) \).

Step 4: Select the Correct Option

We derived that \( AB = 2h \cot \theta \sin \left( \frac{\theta}{2} \right) \). Referring to the options, we have none that directly matches this result. Hence, explore simplifications or known identities to match this or verify against similar identities with \( \sin\, 2\theta \), if needed for cross-verification.

Key Concepts

Angles of Elevation

In trigonometry, the angle of elevation is a very useful concept when dealing with right triangles and heights. It is the angle formed by the line of sight when looking upward from a horizontal plane to a higher point. In the exercise, this angle is represented as \( \theta \). Both points \(A\) and \(B\) have the same angle of elevation when looking at the top of the pillar \(OP\).

When you see problems involving an angle of elevation:

• Visualize the problem with points \( A \) and \( B \) on the ground and point \( P \) at the top of the pillar.
• The angle of elevation \( \theta \) can be used to determine distances and heights using trigonometric ratios like \( \tan \theta \).
• In the solution, \( \tan \theta = \frac{h}{OA} \), which rearranges to give \( OA = \frac{h}{\tan \theta} \), simplifying calculations.

Understanding angles of elevation is crucial for solving problems involving right triangles and heights in trigonometry.

Law of Cosines

The Law of Cosines is a powerful tool in trigonometry used to find missing lengths or angles in a triangle. For any triangle with sides \(a\), \(b\), and \(c\), opposite angles \(A\), \(B\), and \(C\), the law states: \[ c^2 = a^2 + b^2 - 2ab \cdot \cos(C) \]

In the exercise involving triangle \( \bigtriangleup OAB \), the Law of Cosines helps calculate side \( AB \) given:

• The points \( A \) and \( B \) are at the same distance from point \( O \), making \( \bigtriangleup OAB \) isosceles.
• This symmetry in \( \angle AOB = \theta \) allows us to substitute and find that \( AB^2 = 2 \cdot OA^2 \cdot \left( 1 - \cos \theta \right) \).
• The law becomes handy because it makes it possible to figure out side lengths even when some angles or other lengths are not directly given.

Utilizing the Law of Cosines means we can solve non-right triangles with more ease, extending our problem-solving capabilities.

Right Triangles

Right triangles are fundamental in trigonometry and appear in numerous applications, such as the problem at hand. A right triangle has one angle equal to 90 degrees. This feature simplifies calculating various trigonometric ratios for sides such as sine, cosine, and tangent.

When looking at the problem:

• Triangles \( \triangle OAP \) and \( \triangle OBP \) are right triangles because the line of sight upwards forms 90 degrees with the ground below.
• The tangent function helps find distances such as \( OA \) or \( OB \) with relationships like \( \tan \theta = \frac{h}{OA} \).
• These relationships are vital for breaking down complex geometric figures into simpler right triangles for calculations.

Recognizing and using right triangles allows students to apply trigonometric functions effectively, simplifying various problems by reducing them to known identities and easier calculations.