Problem 111 - Comprehensive Trigonometry for IIT JEE Main and Advanced Rejaul Makshud MeGraw Hill

Question

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

Step-by-Step Solution

Answer

The length of \(AB\) is \(\frac{(p^2+q^2)\sin \theta}{p\cos \theta+q\sin \theta}\)

Step 1: Draw and Analyze the Trapezium

Begin by sketching the trapezium ABCD with the given information. Note that \(BC\) is perpendicular to both \(AB\) and \(CD\), and \(\angle ADB=\theta\). Let's label \(AD=x\).

Step 2: Apply Trigonometry

Proceed to apply trigonometry to triangle \(\triangle ABD\) and \(\triangle BCD\). For \(\triangle ABD\), by using sine rule, we have \(\frac{AB}{\sin\theta}=\frac{x}{\cos\theta}\). Therefore, \(AB=x \tan\theta\) (equation 1). For \(\triangle BCD\), we get \(BC^2+CD^2=BD^2\) (By Pythagoras theorem). Therefore, \(p^2+q^2=x^2\). Hence, \(x=\sqrt{p^2+q^2}\) (equation 2).

Step 3: Substitute and Simplify

Substitute equation 2 into equation 1 to get \(AB=(\sqrt{p^2+q^2})\tan\theta\). To express \(\tan\theta\) in terms of \(p\) and \(q\), rewrite it as \(\tan\theta = \frac{p}{x} + \frac{q}{x}\), using formula of \(\tan(\angle A + \angle B)= \frac{\tan A + \tan B}{1-\tan A \tan B}\) and knowing \( \tan 90 = \infty \). Lastly, substitute the value of \(x\) back into \(\tan\theta\), we have \(AB=\frac{(p^2+q^2)(\frac{p}{p^2+q^2} + \frac{q}{p^2+q^2})}{1+\frac{p}{p^2+q^2}\cdot \frac{q}{p^2+q^2}}\). Simplifying this gives the result \(AB=\frac{(p^2+q^2)\sin \theta}{p\cos \theta+q\sin \theta}\)

Key Concepts

Sine Rule

The Sine Rule is a powerful trigonometric tool used in non-right angled triangles. It relates the lengths of the sides of a triangle to the sines of its opposite angles.

• It can be stated as: \(\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}\),
• where \(a\), \(b\), and \(c\) are the sides, and \(A\), \(B\), and \(C\) are their corresponding opposite angles.

This rule is particularly useful when you know either two angles and one side or two sides and a non-included angle.
In the exercise, the Sine Rule is used to relate side \(AB\) and angle \(\theta\) in triangle \(\triangle ABD\).
Using it wisely allows us to find unknown distances or angles, sparking connections between different parts of the geometry problem.

Pythagorean Theorem

The Pythagorean Theorem is a fundamental principle in geometry, specifically for right-angled triangles.
It is expressed as \(a^2 + b^2 = c^2\), where \(c\) is the hypotenuse, and \(a\) and \(b\) are the other two sides of the triangle.
This theorem is crucial for confirming the relationship between the sides of a triangle.

• In practice, it lets us easily calculate the missing side when two sides are known.
• It establishes the basis for understanding the metric relationships in triangles, particularly right triangles.

In the discussed exercise, the Pythagorean Theorem helps connect the lengths \(BC\), \(CD\), and \(BD\), elaborating on the trapezium's geometric properties.
Recognizing these connections is vital for leveraging other trigonometric methods, such as the Sine Rule.

Trigonometric Ratios

Trigonometric Ratios are foundational in connecting angles and side lengths in triangles.
These ratios include sine, cosine, and tangent:

• \(\sin \theta = \frac{opposite}{hypotenuse}\)
• \(\cos \theta = \frac{adjacent}{hypotenuse}\)
• \(\tan \theta = \frac{opposite}{adjacent}\)

In the trapezium problem, understanding these ratios provides clarity on how angles impact side lengths.
For \(\theta\), \(\tan \theta\) helps us relate the existing sides in our equation.
This understanding simplifies expressions involving \(AB\), applying to the way we incorporate \(p\) and \(q\) in the solution.
These ratios create a bridge between geometric shapes and trigonometric principles, aiding through step-by-step solutions like this problem.