Problem 73 - A Complete Resource Book in Mathematics for JEE Main

Question

\(A B C D\) is a trapezium such that \(A B\) and \(C D\) are parallel and \(B C \perp C D .\) If angle \(A D B=B, B C=P\) and \(C D=q\), then \(A B\) is equal to (A) \(\frac{p^{2}+q^{2} \cos \theta}{p \cos \theta+q \sin \theta}\) (B) \(\frac{p^{2}+q^{2}}{p^{2} \cos \theta+q^{2} \sin \theta}\) (C) \(\frac{\left(p^{2}+q^{2}\right) \sin \theta}{(p \cos \theta+q \sin \theta)^{2}}\) (D) \(\frac{\left(p^{2}+q^{2}\right) \sin \theta}{p \cos \theta+q \sin \theta}\)

Step-by-Step Solution

Answer

Option (A) \(\frac{p^{2} + q^{2} \cos \theta}{p \cos \theta + q \sin \theta}\).

Step 1: Understanding the problem

We are given a trapezium \(ABCD\) where \(AB\) and \(CD\) are parallel, and \(BC\) is perpendicular to \(CD\). We need to find the expression for \(AB\) using the given angles and lengths. The angle \(ADB\) is given as \(\theta\), \(BC\) is \(p\), and \(CD\) is \(q\).

Step 2: Analyzing geometry

Since \(AB \parallel CD\) and \(BC \perp CD\), \(\triangle BCD\) is a right triangle. The side \(BD\) acts as the hypotenuse, and \(CD\) is the adjacent side in the context of angle \(\theta\).

Step 3: Using trigonometry

In the right triangle \(BCD\), we apply trigonometric identities. We use \(\cos \theta = \frac{CD}{BD}\) and \(\sin \theta = \frac{BC}{BD}\). Using Pythagoras theorem, \(BD^2 = BC^2 + CD^2\), so \(BD = \sqrt{p^2 + q^2}\).

Step 4: Express \(AD\) in terms of \(BD\) and angle

For the line \(AD\), we project \(BD\) horizontally toward \(AB\) using \(\cos \theta\): \(AD = BD \cdot \cos \theta = \sqrt{p^2 + q^2} \cdot \cos \theta\).

Step 5: Set up relation for \(AB\) using geometry

Since \(AB \parallel CD\), the length \(AB\) can be derived using the trigonometric projection of \(AD\) along \(AB\). From the diagram, \(AB = AD + DB \cdot \sin \theta = \sqrt{p^2 + q^2} \cdot \cos \theta + q \cdot \sin \theta\).

Step 6: Final expression for \(AB\)

Combine the length expressions to find \(AB\). We simplify and rearrange to match the given options. \(AB = \frac{p^2 + q^2 \cdot \cos \theta}{p \cdot \cos \theta + q \cdot \sin \theta}\).

Step 7: Finding the correct answer

Referring to the given multiple-choice answers, option (A) matches our derived expression \(\frac{p^{2} + q^{2} \cos \theta}{p \cos \theta + q \sin \theta}\). Hence, the correct answer is option (A).

Key Concepts

Trigonometry in Geometry

Trigonometry is a powerful tool used in geometry to find unknown sides and angles within a shape. In the context of a trapezium like the one in our exercise, trigonometry helps us relate angles and lengths. Understanding trigonometric identities such as sine, cosine, and tangent is crucial. In our problem, the cosine of the angle \( \theta \) helps express the relationship between the sides of the right triangle \( BCD \).

• **Cosine Function**: Describes the ratio of the adjacent side to the hypotenuse in a right triangle. For \( \triangle BCD \), \( \cos \theta = \frac{CD}{BD} \).


• **Sine Function**: Describes the ratio of the opposite side to the hypotenuse. In our exercise, \( \sin \theta = \frac{BC}{BD} \).

These functions allow us to derive side lengths from known angles, reinforcing the interplay between trigonometry and geometry.

Right Triangle Properties

Right triangles are fundamental in solving geometric problems because they adhere to specific properties, such as the Pythagorean theorem. In this exercise, the triangle \( \triangle BCD \) is a right triangle because \( BC \) is perpendicular to \( CD \). One key property is the Pythagorean theorem, which states that in any right triangle:\[ BD^{2} = BC^{2} + CD^{2} \]This theorem allows us to find the hypotenuse \( BD \) of the triangle. Given \( BC = p \) and \( CD = q \), we calculate:\[ BD = \sqrt{p^2 + q^2} \] This property is crucial because it connects the perpendicularity of lines \( BC \) and \( CD \) to the calculated length \( BD \). Understanding these simple rules can greatly simplify complex geometric problems.

Parallel Lines in Geometry

Parallel lines are lines in a plane that never meet, and they have the same slope. In the trapezium \( ABCD \), lines \( AB \) and \( CD \) are parallel. This parallelism creates specific geometric relationships and makes certain calculations possible.

• **Segment Addition**: Relating parallel lines can help express line segments in terms of the angles and sides of the intersecting triangles. As in the exercise, knowing \( AD \) allows us to express \( AB \) by combining it with the projected line on \( DB \).


• **Using Trigonometry**: When lines are parallel, trigonometric projections remain constant, allowing us to use functions like cosine and sine to express other parts of the shape.

Identifying and leveraging parallel lines not only helps in theoretical understanding but also in deriving practical solutions to geometric problems.