Problem 3 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

\(A\) and \(B\) are two points in the horizontal plane through \(O\), the foot of pillar \(O P\) of height \(h\), such that \(\Delta A O B=\theta\). 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 length of AB is \(h \cos \theta \operatorname{cosec} \frac{\theta}{2}\). (Option D)

Step 1: Understand the scenario

We have a point O at the foot of a vertical pillar OP with height \(h\). Points A and B are on the horizontal plane through O. From both points A and B, the angle of elevation to the top of the pillar is \(\theta\). Also, the angle \(\angle AOB = \theta\). We need to find the length of AB.

Step 2: Use the angle of elevation information

For points A and B, the elevation angles to the top of the pillar are \(\theta\). This implies that \(\tan \theta = \frac{h}{AO} = \frac{h}{BO}\). Hence, the horizontal distances \(AO = BO = \frac{h}{\tan \theta}\).

Step 3: Analyze geometry of the triangle AOB

In \(\triangle AOB\), we have \(AO = BO\) as previously determined. \(\angle AOB = \theta\), making \(\triangle AOB\) isosceles with equal legs \(AO\) and \(BO\).

Step 4: Use Law of Cosines

Applying the Law of Cosines in \(\triangle AOB\), we have: \[ AB^2 = AO^2 + BO^2 - 2 \cdot AO \cdot BO \cdot \cos(\angle AOB) \]Since \(AO = BO = \frac{h}{\tan \theta}\) and \(\angle AOB = \theta\), the equation becomes: \[ AB^2 = 2 \left( \frac{h}{\tan \theta} \right)^2 (1 - \cos \theta) \] \[ AB^2 = 2 \left( \frac{h^2}{\tan^2 \theta} \right) (1 - \cos \theta) \]

Step 5: Simplifying using trigonometric identities

Use the identity \(1 - \cos \theta = 2 \sin^2\left(\frac{\theta}{2}\right)\) to simplify: \[ AB^2 = 2 \left( \frac{h^2}{\tan^2 \theta} \right) 2 \sin^2\left(\frac{\theta}{2}\right) \] \[ AB^2 = 4 \frac{h^2 \sin^2\left(\frac{\theta}{2}\right)}{\tan^2 \theta} \] Knowing \(\tan \theta = \frac{\sin \theta}{\cos \theta}\), we have:\[ AB^2 = 4 \frac{h^2 \sin^2\left(\frac{\theta}{2}\right)}{\left( \frac{\sin \theta}{\cos \theta} \right)^2} \] \[ AB^2 = 4 h^2 \sin^2\left(\frac{\theta}{2}\right) \left( \frac{\cos^2 \theta}{\sin^2 \theta} \right) \] \[ AB = 2 h \cos \theta \sin\left(\frac{\theta}{2}\right) \]

Step 6: Identify the solution

Rewrite \(AB\) as: \[ AB = h \cos \theta \left(2 \sin\left(\frac{\theta}{2}\right) \right) \frac{1}{\sin \frac{\theta}{2}} = h \cos \theta \operatorname{cosec} \frac{\theta}{2} \] Thus, the correct answer is choice (D).

Key Concepts

Law of Cosines

The Law of Cosines is a crucial concept in trigonometry that helps in finding unknown sides and angles in any triangle, not just right triangles. It's especially useful when you have an oblique triangle, that is, a triangle with all the angles less than 180° but not necessarily 90°.
In \( riangle AOB\), we apply the Law of Cosines to find the length of the side opposite the known angle. The formula is given by:

• \(c^2 = a^2 + b^2 - 2ab\cos(C)\)

Here, \(a\) and \(b\) are the known lengths of the sides, and \(C\) is the angle opposite side \(c\).
In our solution, since \(AO = BO\) and \(\angle AOB = \theta\), we substitute these into the law:

• \(AB^2 = AO^2 + BO^2 - 2\cdot AO \cdot BO \cdot \cos(\theta)\)

This simplified version uses trigonometric values to compute the length of \(AB\), leading us to solve for the unknown side of the triangle. Understanding the application of the Law of Cosines is essential in scenarios involving non-right triangles.

Angle of Elevation

The concept of the angle of elevation is widely used, especially in scenarios involving heights and distances that are not immediately observable. The angle of elevation is the angle formed between the horizontal line and the line of sight to an object above the horizontal.
In the exercise, both points \(A\) and \(B\) are used to observe the top of the pillar \(O P\), creating an angle of elevation, \(\theta\). This indicates the direction from the observer to the top of a vertical structure is upwards.
Some important aspects to consider:

• It's measured from the horizontal up.
• Often combined with trigonometric ratios like tangent, since \(\tan(\theta) = \frac{opposite}{adjacent}\).

Understanding the angle of elevation links directly to computing distances when heights and angles are involved. In problems where heights and distance affect the outcome, the angle of elevation is a pivotal connecting concept.

Trigonometric Identities

Trigonometric identities are fundamental tools in solving many mathematical problems, especially in simplifying complex trigonometric expressions and equations. They express relationships between the trigonometric functions \( extbf{sin, cos, tan, etc.}\), allowing transformations that make solving problems easier.
In this particular problem, several trigonometric identities come into play:

• The identity \(1 - \cos \theta = 2\sin^2\left(\frac{\theta}{2}\right)\) allows transformation of expressions involving cosine into sine functions, simplifying further calculations.
• \[ \tan \theta = \frac{\sin \theta}{\cos \theta} \Rightarrow \tan^2 \theta = \frac{\sin^2 \theta}{\cos^2 \theta} \]This reveals how to express tangent in terms of sine and cosine, crucial in substituting when simplifying formulas.

Overall, trigonometric identities enable handling of complex expressions and help in achieving simpler equivalent solutions. Mastery of these identities is essential for any student looking to deeply understand trigonometry.