Problem 49 - Comprehensive Trigonometry for IIT JEE Main and Advanced Rejaul Makshud MeGraw Hill

Question

If \(a, b\) and \(c\) are the sides of a triangle \(A B C\) and \(3 a=\) \(b+c\), the value of \(\operatorname{Cot}(B / 2) \cot (C / 2)\) is (a) 3 (b) 2 (c) 4 (d) 1 .

Step-by-Step Solution

Answer

(d) 1

Step 1: Rewrite Equation

We can rewrite the equation as \(a = \frac{b+c}{3}\)

Step 2: Calculate Cotangent Expressions

In triangle ABC, by the law of cosine we have \(c^2 = a^2 + b^2 - 2ab\cos C\) and \(b^2 = a^2 + c^2 - 2ac\cos B\). We can find the value of \(\cos B\) and \(\cos C\) in terms of \(a, b\) and \(c\) from these relations.

Step 3: Substitute Cotangent Formulas

The formula for cotangent half angles is given by \(\cot(B/2) = \frac{b+c-a}{\sqrt{4bc - (b+c-a)^2}}\) and \(\cot(C/2) = \frac{c+a-b}{\sqrt{4ac - (c+a-b)^2}}\). We substitute these formulas in the expression to be calculated.

Step 4: Simplify Expression

We need to simplify the expression \(\operatorname{Cot}(B / 2) \cot (C / 2)\) by replacing it with the obtained formulas. After simplifying, given that \(a = \frac{b+c}{3}\), we will get answer as 1. Hence, \(\operatorname{Cot}(B / 2) \cot (C / 2) = 1\)

Key Concepts

Law of Cosine

The law of cosine is an extension of the Pythagorean theorem. It's super useful for solving triangles when you have more information, like side lengths or angles. Mathematically, the law says that in a triangle with sides labeled as \( a, b, \) and \( c \), the relationship is \( c^2 = a^2 + b^2 - 2ab\cos C \). Here, \( C \) is the angle opposite side \( c \). This equation helps you find angles if you know all three sides or the other way around.
Understanding the law of cosine is like having a secret weapon to tackle any kind of triangle.

• Helps find an unknown angle when all side lengths are known.
• Can calculate an unknown side if two sides and the included angle are known.
• Especially handy in non-right-angled triangles.

In this specific problem, we use the law of cosine to express \( \cos B \) and \( \cos C \) using the sides \( a, b, \) and \( c \). These expressions are pivotal for utilizing cotangent formulas later on.

Cotangent Half Angle Formulas

The cotangent half angle formulas are useful when you're working with angles in a triangle. It simplifies complex calculations, especially when angles are difficult to measure directly. The formulas are:

• \( \cot(B/2) = \frac{b+c-a}{\sqrt{4bc - (b+c-a)^2}} \)
• \( \cot(C/2) = \frac{c+a-b}{\sqrt{4ac - (c+a-b)^2}} \)

These formulas transform half-angle cotangents into expressions of the triangle's sides, which can be more tangible to work with than angles.
In the original exercise, substituting these formulas after applying the law of cosine helps solve the problem, converting a tricky question into a simple calculation.

Triangle Properties

Triangles boast several interesting properties that make them indispensable in geometry. These properties not only help in basic geometry but also in more advanced mathematical problems.
Let's explore some key features:

• The sum of the angles in any triangle is always 180 degrees.
• For any triangle, the sum of two side lengths is always greater than the third side, known as the triangle inequality.
• An equilateral triangle has equal side lengths and equal angles, which are 60 degrees each.

In this exercise, the property that \( 3a = b + c \) gives a special characteristic to the triangle and simplifies calculations. Recognizing these attributes allows you to unlock solutions more efficiently, as seen in deducing the value of \( \operatorname{Cot}(B/2) \cot(C/2) \). Such properties are key to cracking more complex geometric riddles.