Problem 13 - Comprehensive Trigonometry for IIT JEE Main and Advanced Rejaul Makshud MeGraw Hill

Question

In a \(\Delta A B C\), the value of \(\frac{a \cos A+b \cos B+c \cos C}{a+b+c}\) is (a) \(\frac{R}{r}\) (b) \(\frac{R}{2 r}\) (c) \(\frac{r}{R}\) (d) \(\frac{2 r}{R}\)

Step-by-Step Solution

Answer

The correct answer is (a) \(\frac{R}{r}\).

Step 1: Express a, b, c using the law of cosines

Consider \(\Delta A B C\), we can express the sides of the triangle a, b, c in terms of the angles using the law of cosines. Therefore, \(a = 2R\sin{A}\), \(b = 2R\sin{B}\), and \(c = 2R\sin{C}\), where R is the circumradius of the triangle.

Step 2: Substitute these expressions back into the original equation

Substitute these expressions back into the original equation: \[\frac{a \cos A+b \cos B+c \cos C}{a+b+c} = \frac{(2R\sin{A}\cos{A})+(2R\sin{B}\cos{B})+(2R\sin{C}\cos{C})}{2R(\sin{A}+\sin{B}+\sin{C})}\] By applying trigonometric identity, \(\sin{2\theta} = 2\sin{\theta}\cos{\theta}\), the equation becomes \[\frac{R\sin{2A}+R\sin{2B}+R\sin{2C}}{2R(\sin{A}+\sin{B}+\sin{C})}\]

Step 3: Use formulas for the circle radii

According to properties of triangle, the numerator is four times the area of the triangle, which is \(4R^2 \sin{A}\sin{B}\sin{C}\) and the denominator is the semi-perimeter of the triangle, \(s = \frac{a+b+c}{2} = R(\sin{A} + \sin{B} +\sin{C})\).We also know that the area of the triangle can be expressed in terms of the incircle radius r as \(\Delta = rs\). Thus, we substitute these expressions to get\[\frac {4R^2 \sin{A}\sin{B}\sin{C}}{2R(R(\sin{A} + \sin{B} +\sin{C}))} = \frac{R}{r}\]

Step 4: Comparing the result with the options

Now, it can be determined by comparison that the correct option is (a) \(\frac{R}{r}\).

Key Concepts

Law of Cosines

The Law of Cosines is a fundamental rule in trigonometry that helps relate the angles and sides of a triangle. It can be particularly useful in any triangle - not just right ones. This law provides a way to calculate the length of one side of a triangle if the lengths of the other two sides and the measure of the included angle are known. The Law of Cosines states that for any triangle with sides \(a\), \(b\), \(c\) and the opposite angles \(A\), \(B\), \(C\) respectively, the following formula holds:

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

The formula resembles the Pythagorean theorem, which makes sense since the Pythagorean theorem is a special case of the Law of Cosines when the angle is a right angle (90 degrees), making \(\cos 90^\circ = 0\). This law is invaluable when solving triangles that aren’t right-angled as it links all three sides with one angle.

Circumradius

The circumradius is a central concept in trigonometry concerning triangles. It refers to the radius of the circumcircle, which is the circle that passes through all the vertices of a triangle. The circumradius \(R\) can be calculated using the formula: \(R = \frac{abc}{4\Delta}\), where \(a\), \(b\), \(c\) are the sides of the triangle and \(\Delta\) is the area of the triangle.

• The circumcenter, the center of the circumcircle, is equidistant from all three vertices of the triangle.
• The circumradius is especially useful when analyzing the triangle's geometric properties.

In the given solution, the circumference is pivotal as it connects to the triangle's angles with the relation \(a = 2R\sin{A}\). This makes understanding the circumradius vital in solving complex problems potentially involving trigonometric identities and properties such as the sine rule.

Incircle Radius

The incircle of a triangle is another critical concept, linked directly to the idea of the incircle radius or inradius. The incircle is the largest circle that fits within the triangle and touches all its sides. The radius of this circle, denoted \(r\), is called the inradius. It can be calculated by the formula: \(r = \frac{\Delta}{s}\), where \(\Delta\) is the area of the triangle and \(s\) is the semi-perimeter \(\frac{a+b+c}{2}\).

• This radius provides insight into the triangle's internal properties.
• The inradius is crucial when forming relationships between different parts of the triangle, like its area and semi-perimeter.

In our problem solution, the inradius plays a vital role through the equation \(\Delta = rs\), which links the inradius to the circumradius and ultimately helps determine the value of \(\frac{R}{r}\).

Trigonometric Identities

Trigonometric identities are mathematical equations that involve trigonometric functions and are true for all values of the included variables. These identities can simplify the process of solving triangles, proving other identities, or computing values. Some commonly used trigonometric identities are:

• Pythagorean Identities, such as \(\sin^2 \theta + \cos^2 \theta = 1\)
• Angle Sum and Difference Identities, like \(\sin(A \pm B) = \sin A \cos B \pm \cos A \sin B\)
• Double Angle Identity, specifically \(\sin(2\theta) = 2\sin \theta \cos \theta\)

For our exercise, the identity \(\sin{2\theta} = 2\sin \theta \cos \theta\) is used to simplify expressions involving products of sine and cosine. Trigonometric identities offer powerful tools for breaking down complex trigonometric equations into manageable parts, enabling us to find the general solution more straightforwardly.