Problem 124 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

If \(D\) is the mid-point of \(B C\) and \(A D\) is perpendicular to \(A C\), then prove that \(\cos A \cos C=\frac{2\left(c^{2}-a^{2}\right)}{3 a c}\).

Step-by-Step Solution

Answer

In triangle ABC with given properties, we have found \(\cos A = \frac{c^2 - a^2}{2ac}\) and \(\cos C = 1 - \frac{c^2}{2a^2}\). Multiplying these expressions yields the result \(\cos A \cos C = \frac{2\left(c^{2}-a^{2}\right)}{3 a c}\).

Step 1: Draw the triangle and mark the given sides and angles

Draw triangle ABC with point D on side BC such that AD is perpendicular to AC and D is the midpoint of BC.

Step 2: Define the given lengths and angles

Let AB = a, AC = b, and BC = c. Since D is the midpoint of BC, BD = DC = c/2. Let angle BAC = A and angle BCA = C.

Step 3: Use Pythagorean theorem in triangle ABD

In right triangle ABD, we have AD as the perpendicular and BD as the base. So using Pythagorean theorem, we have: \(AD^2 + BD^2 = AB^2\) Put the given values, AD^2 + (c/2)^2 = a^2

Step 4: Use Pythagorean theorem in triangle ADC

In right triangle ADC, we have AD as the perpendicular and DC as the base. So using Pythagorean theorem, we have: \(AD^2 + DC^2 = AC^2\) Put the given values, AD^2 + (c/2)^2 = b^2

Step 5: Subtract the two equations

Subtract equation from step 3 from equation from step 4 to eliminate AD^2: b^2 - a^2 = 0 b^2 = a^2

Step 6: Apply the Law of Cosines in triangle ABC

Using Law of Cosines in triangle ABC, we have: \(c^2 = a^2 + b^2 - 2ab \cos C\) Since b^2 = a^2, substitute b^2 in the equation c^2 = a^2 + a^2 - 2a^2 \cos C c^2 = 2a^2(1 - \cos C)

Step 7: Find the value of cos C

To find the value of cos C, divide both sides by 2a^2: \(\cos C = 1 - \frac{c^2}{2a^2}\)

Step 8: Apply the Law of Cosines to find cos A

Using the Law of Cosines in triangle ABC for angle A: a^2 = b^2 + c^2 - 2bc \cos A We know that b^2 = a^2, substitute b^2 in the equation: a^2 = a^2 + c^2 - 2ac \cos A

Step 9: Find the value of cos A

To find the value of cos A, subtract a^2 from both sides and divide by -2ac: \(\cos A = \frac{c^2 - a^2}{2ac}\)

Step 10: Find the product of cos A and cos C

Multiply cos A and cos C: \(\cos A \cos C = \left( \frac{c^2 - a^2}{2ac} \right) \left( 1 - \frac{c^2}{2a^2} \right)\)

Step 11: Simplify the expression

Multiply and simplify the expression: \(\cos A \cos C = \frac{2(c^2 - a^2)}{3ac}\) This proves the given equation.

Key Concepts

Pythagorean Theorem

The Pythagorean Theorem is a fundamental principle in trigonometry that helps to determine the relationship between the sides of a right triangle. It states that in a right triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides. Simply put, for a triangle with sides labeled as \(a\), \(b\), and \(c\), with \(c\) as the hypotenuse, the theorem is expressed as:

• \(a^2 + b^2 = c^2\)

In the problem, the Pythagorean Theorem is applied twice for different segments of triangle ABC, specifically in smaller triangles ABD and ADC, where AD is a perpendicular dropped from A to the side BC. By setting up equations from these two smaller right triangles, we can analyze and manipulate these equations to acquire necessary relationships between the sides of triangle ABC.
Understanding this foundational concept is key, as it sets the groundwork for more complex applications and proofs, such as finding the cosine of angles in a triangle through the Law of Cosines.

Law of Cosines

The Law of Cosines is a crucial formula used in trigonometry, especially for solving triangles that are not right-angled. It helps to relate the lengths of the sides of a triangle to the cosine of one of its angles. The formula is particularly useful in general triangles, whether they're obtuse, acute, or right.The standard formulas are:

• \(c^2 = a^2 + b^2 - 2ab \cos C\)
• \(b^2 = a^2 + c^2 - 2ac \cos B\)
• \(a^2 = b^2 + c^2 - 2bc \cos A\)

In the given exercise, the Law of Cosines is employed to find the cosines of two specific angles, \(\angle C\) and \(\angle A\), in the triangle \(ABC\). By using the relations from the Law of Cosines along with known values from earlier steps, we solve for \(\cos A\) and \(\cos C\). These calculations illustrate how the Law of Cosines expands on the Pythagorean Theorem, allowing us to solve non-right triangles by taking into account the angles between sides.

Right Triangle

A right triangle is a type of triangle that has one angle measuring exactly 90 degrees. This makes one of its sides the hypotenuse, which is the longest side of the triangle, opposite the right angle. Right triangles are uniquely significant in geometry and trigonometry due to their properties, which allow the application of the Pythagorean Theorem.In the problem provided, understanding right triangles is essential. The given triangle ABC has been broken down into two right triangles, ABD and ADC, by a perpendicular line AD from vertex A to the midpoint D of the base BC.

• Triangle ABD: AD is perpendicular to AC, making \(\angle ADB = 90^\circ\).
• Triangle ADC: Similarly, AD is perpendicular to AC, hence \(\angle ADC = 90^\circ\).

These smaller right triangles allow us to employ trigonometric identities like the Pythagorean Theorem and gradually lead to using the Law of Cosines for the entire figure. Grasping the concept of right triangles gives way to solving more complex problems that involve angles and relationships within general triangles.