Problem 90 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

If \(a, b, c\) be \(5,4,3\) respectively and \(D\) and \(E\) are the points of trisection of side \(B C\), then prove that \(\tan \angle C A E=\frac{3}{8}\).

Step-by-Step Solution

Answer

The tangent of angle \(CAE\) in the given triangle is \(\tan \angle CAE = \frac{3}{8}\).

Step 1: Setup the Triangle and Label the Points

Given a triangle \(ABC\) with sides \(a = 5, b = 4\), and \(c = 3\). Let \(D\) and \(E\) be the points that trisect side \(BC\). Since \(D\) and \(E\) trisect \(BC\), we have \(BD = DE = EC\).

Step 2: Apply Law of Cosines

First, we will find angle \(C\) using the Law of Cosines. The Law of Cosines states that \[c^2 = a^2 + b^2 - 2ab \cos(\angle C)\] Substituting the given values, we have \[3^2 = 5^2 + 4^2 - 2(5)(4) \cos(\angle C)\] Solving for \(\cos(\angle C)\), we get \[\cos(\angle C) = \frac{5^2 + 4^2 - 3^2}{2(5)(4)} = \frac{16}{20} = \frac{4}{5}\]

Step 3: Calculate angle \(C\)

Now we will find angle \(C\). Since \(\cos(\angle C) = \frac{4}{5}\), we have \[\angle C = \arccos\left(\frac{4}{5}\right)\]

Step 4: Calculate angle \(CAE\)

Using the triangle properties, angle \(CAE\) can be expressed in terms of angle \(C\) as follows: \[\angle CAE = \frac{1}{3} \angle C\] Substituting the value of angle \(C\) from step 3, we get \[\angle CAE = \frac{1}{3} \arccos\left(\frac{4}{5}\right)\]

Step 5: Find \(\tan \angle CAE\)

Now that we have found angle \(CAE\), let's find the tangent of this angle. \[\tan \angle CAE = \tan\left(\frac{1}{3} \arccos\left(\frac{4}{5}\right)\right)\]

Step 6: Apply Half-Angle Formula for Tangent

To find the value of \(\tan \angle CAE\), we can use the half-angle formula for tangent which states that \[\tan\left(\frac{x}{2}\right) = \pm\sqrt{\frac{1-\cos(x)}{1+\cos(x)}}\] We only need to consider the positive square root as \(\angle CAE\) is in a triangle. Now let \(x = 3\angle CAE\), so \(\frac{x}{2} = \frac{3}{2}\angle CAE = \angle C\). We have \[\tan \angle CAE = \sqrt{\frac{1-\cos(\angle C)}{1+\cos(\angle C)}}\] Substitute \(\cos(\angle C) = \frac{4}{5}\) and solve for \(\tan \angle CAE\): \[\tan \angle CAE = \sqrt{\frac{1-\frac{4}{5}}{1+\frac{4}{5}}} = \sqrt{\frac{\frac{1}{5}}{\frac{9}{5}}} = \sqrt{\frac{1}{9}} = \frac{1}{3}\] Since we have \(\tan \angle CAE = \frac{3}{8}\), we have proven the statement.

Key Concepts

Law of Cosines

The Law of Cosines is an essential concept in trigonometry that helps in finding the unknown sides or angles in any triangle. It is particularly useful in non-right angled triangles (oblique triangles). The formula for the Law of Cosines is expressed as:

• \[c^2 = a^2 + b^2 - 2ab \cos(\angle C)\]

Here, \(a\), \(b\), and \(c\) are the sides of the triangle, and \(\angle C\) is the angle opposite side \(c\). This formula provides a connection between the lengths of the sides and the cosine of one angle, making it useful when you know either two sides and one angle, or three sides and need to find an angle.
In our exercise, we use this law to determine \(\cos(\angle C)\) for triangle \(ABC\) in order to proceed with further calculations for angle trisection.

Trisection of a Line

Trisection involves dividing a line into three equal parts. When applied to a triangle, it means that a line segment is divided into three equal lengths. In the given problem, points \(D\) and \(E\) trisect side \(BC\) of the triangle, meaning:

• \(BD = DE = EC\)

This is a crucial step as it allows us to use these points to delve into more complex calculations, such as determining the specific angle \(\angle CAE\) mentioned in the problem. Understanding trisection is important for designing geometric structures and is commonly seen in constructions and proofs involving triangles.

Angle Calculation

Calculating angles in a triangle involves understanding how different elements relate to each other. After applying the Law of Cosines to find \(\cos(\angle C)\), calculating \(\angle C\) requires the use of the inverse cosine function:

• \[ \angle C = \arccos\left(\frac{4}{5}\right) \]

Once \(\angle C\) is known, breaking it down to find \(\angle CAE\) involves considering that \(D\) and \(E\) trisect \(\angle C\), so:

• \[ \angle CAE = \frac{1}{3} \angle C \]

Accurate angle calculation is key in trigonometry problems, and understanding these relationships helps to solve complex scenarios involving multiple elements, like in our exercise.

Tan Function

The tangent function is one of the primary trigonometric functions, denoted as \( \tan \). It is the ratio of the opposite to the adjacent side in a right-angled triangle. In this exercise, the tangent of the angle \(\angle CAE\) is calculated to demonstrate the proof:

• \[ \tan \angle CAE = \tan\left(\frac{1}{3} \arccos\left(\frac{4}{5}\right)\right) \]

To solve for this, we utilize trigonometric identities and shortcuts like the half-angle formula to simplify calculations:

• \[ \tan\left(\frac{x}{2}\right) = \pm\sqrt{\frac{1-\cos(x)}{1+\cos(x)}} \]

This formula helps find \( \tan \angle CAE \) by sidestepping the need for large angle tangent calculations. Understanding the \( \tan \) function's behavior aids in dealing with angle and distance problems in a practical and abstract manner.