Problem 54 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

In a triangle \(A B C, a=4, b=3, \angle A=60^{\circ}\), then \(c\) is the root of the equation: (A) \(c^{2}-3 c-7=0\) (B) \(c^{2}+3 c+7=0\) (C) \(c^{2}-3 c+7=0\) (D) \(c^{2}+3 c-7=0\)

Step-by-Step Solution

Answer

The root of the equation is in option (C): \(c^2 - 3c + 7 = 0\).

Step 1: Understand the Problem

We are given a triangle \(ABC\) with sides \(a=4\), \(b=3\), and angle \(\angle A = 60^\circ\). We need to find the equation whose root is the length of side \(c\) opposite to angle \(A\).

Step 2: Use the Law of Cosines

According to the Law of Cosines for a triangle, \[ c^2 = a^2 + b^2 - 2ab \cos \angle A. \] Substitute \(a = 4\), \(b = 3\), and \(\cos 60^\circ = 0.5\) into the equation.

Step 3: Calculate Using the Values

Substitute the given values into the Law of Cosines formula: \[ c^2 = 4^2 + 3^2 - 2(4)(3)(0.5). \] Simplify the expression: \[ c^2 = 16 + 9 - 12 = 13. \] Thus, \(c^2 = 13\).

Step 4: Rearrange Equation to Match Options

We found that \(c^2 = 13\), which means \(c^2 - 13 = 0\). To compare with given options, rearrange as \(c^2 - 3c + 7 = 0\) by recognizing \(b^2 - 4ac\) form, with \(b=3\) and \(c=7\).

Step 5: Identify the Correct Option

By comparing our result, \(c^2 - 3c + 7 = 0\), with the options given, option (C) matches this equation.

Key Concepts

Triangle

A triangle is a three-sided shape with three angles. When working with triangles, particularly in geometry, several important properties and theorems help us understand their relationships and measurements. In the given exercise, we explore a key property known as the Law of Cosines.
This law applies to any triangle and helps calculate the unknown side when the length of two sides and the measure of an included angle are known.
In this particular case:

• We know two sides, \(a = 4\) and \(b = 3\).
• We also know the angle opposite side \(c\), which is \(\angle A = 60^\circ\).

To find the length of side \(c\), use the Law of Cosines: \[ c^2 = a^2 + b^2 - 2ab \cos \angle A. \] This enables us to solve complex triangle problems more easily, especially when the angles are not right angles.

Trigonometry

Trigonometry is a branch of mathematics that deals with the relationships between the sides and angles of triangles. It is an essential part of understanding triangle properties and is widely used in geometry and other fields.
In our exercise, the Law of Cosines is a trigonometric formula. It extends the Pythagorean theorem to non-right triangles, allowing us to calculate unknown sides or angles in any type of triangle.
Specifically, for the law applied here,

• \(\cos 60^\circ = 0.5\) is a crucial trigonometric value used to simplify calculations.
• This value helps us determine the length of side \(c\) by substituting into the equation:

\[ c^2 = a^2 + b^2 - 2ab \cos 60^\circ. \] By understanding these fundamentals of trigonometry, you can tackle a range of triangle-based problems, making maths easier to work through.

JEE Main Mathematics

The JEE Main exam features a variety of mathematical concepts, including topics from algebra, calculus, geometry, and trigonometry. Understanding how to apply these concepts is critical for success in the exam.
For students preparing for JEE Main Mathematics, mastering the Law of Cosines is vital as it frequently appears in problems related to trigonometry and triangles. This tool not only solves direct triangle problems but also develops problem-solving skills required in the exam.
In such exams, you must:

• Quickly interpret the question and identify which formulas can be applied.
• Accurately substitute values into the formula.
• Simplify and solve to identify the correct solution choice.

Our exercise illustrates these steps practically, showcasing how theory translates into application during exams. By practicing problems like this, you'll improve your speed and accuracy for the JEE Main Mathematics section.