Why can't we just use the Sine Rule?

The Sine Rule relates sides to $\sin A$, but our equation is built on $\cos A$. The Cosine Rule is the direct path to converting angles into side lengths, which is exactly what we need to solve for the geometry of the triangle.

Is it possible for the triangle to be obtuse?

The algebra leads us to the identity $b^2 + c^2 = a^2$. This is the definitive signature of a right-angled triangle. There is no ambiguity; the math forces the triangle to be right-angled at vertex $A$.

Do I really need to expand all those fractions?

Yes, the expansion is the crucible where the complexity melts away. By finding a common denominator of $2abc$, you allow the terms to interact and cancel out, revealing the simple Pythagorean truth hidden underneath.

JEE Advanced 1993

LEVELJEE Main

Animated Solution for Mathematics - Trigonometry: If in a triangle $A B C$ , $\frac{2 c o s A}{a} + \frac{c o s B}{b} + \frac{2 c o s C}{c} = \frac{a}{b c} + \frac{b}{c a}$ , then the value of the angle $A$ is .................. degrees.

Enter Numerical Value:

Visualized Solution

Understanding the Triangle and Given Relation

Consider a triangle $A B C$ with sides $a$ , $b$ , and $c$ opposite to angles $A$ , $B$ , and $C$ respectively.
We are given the trigonometric relation: $\frac{2 c o s A}{a} + \frac{c o s B}{b} + \frac{2 c o s C}{c} = \frac{a}{b c} + \frac{b}{c a}$
Our goal is to find the measure of angle $A$ in degrees.

Recalling the Cosine Rule

To simplify the given relation, we can express the cosine terms in terms of the side lengths.
Recall the Cosine Rule for any triangle $A B C$ :
$cos A = \frac{b ^{2} + c ^{2} - a ^{2}}{2 b c}$
$cos B = \frac{a ^{2} + c ^{2} - b ^{2}}{2 a c}$
$cos C = \frac{a ^{2} + b ^{2} - c ^{2}}{2 ab}$

Substituting Cosine Expressions

Let's substitute these cosine formulas directly into our original equation.
The equation becomes: $\frac{2}{a} (\frac{b ^{2} + c ^{2} - a ^{2}}{2 b c}) + \frac{1}{b} (\frac{a ^{2} + c ^{2} - b ^{2}}{2 a c}) + \frac{2}{c} (\frac{a ^{2} + b ^{2} - c ^{2}}{2 ab}) = \frac{a}{b c} + \frac{b}{c a}$
Notice how each term now has a product of sides in the denominator.

Simplifying the Denominators

Let's simplify the denominators of each term on the Left Hand Side (LHS).
First term: $\frac{2}{a} \cdot \frac{b ^{2} + c ^{2} - a ^{2}}{2 b c} = \frac{b ^{2} + c ^{2} - a ^{2}}{ab c}$
Second term: $\frac{1}{b} \cdot \frac{a ^{2} + c ^{2} - b ^{2}}{2 a c} = \frac{a ^{2} + c ^{2} - b ^{2}}{2 ab c}$
Third term: $\frac{2}{c} \cdot \frac{a ^{2} + b ^{2} - c ^{2}}{2 ab} = \frac{a ^{2} + b ^{2} - c ^{2}}{ab c}$

Combining the LHS Fractions

Now, let's write the LHS with a common denominator.
The common denominator for all three terms is $2 ab c$ .
Let's rewrite the first and third terms to have $2 ab c$ in the denominator:
LHS $= \frac{2 ( b ^{2} + c ^{2} - a ^{2} )}{2 ab c} + \frac{a ^{2} + c ^{2} - b ^{2}}{2 ab c} + \frac{2 ( a ^{2} + b ^{2} - c ^{2} )}{2 ab c}$

Aligning the Right Hand Side

Let's look at the Right Hand Side (RHS): $\frac{a}{b c} + \frac{b}{c a}$
To combine these, we find a common denominator, which is $ab c$ :
RHS $= \frac{a ^{2}}{ab c} + \frac{b ^{2}}{ab c} = \frac{a ^{2} + b ^{2}}{ab c}$
To match our LHS denominator of $2 ab c$ , we multiply the numerator and denominator by $2$ :
RHS $= \frac{2 ( a ^{2} + b ^{2} )}{2 ab c}$

Equating Numerators

Now we equate the LHS and RHS:
$\frac{2 ( b ^{2} + c ^{2} - a ^{2} ) + ( a ^{2} + c ^{2} - b ^{2} ) + 2 ( a ^{2} + b ^{2} - c ^{2} )}{2 ab c} = \frac{2 ( a ^{2} + b ^{2} )}{2 ab c}$
Since the denominators are identical and non-zero, we can equate the numerators directly:
$2 (b^{2} + c^{2} - a^{2}) + (a^{2} + c^{2} - b^{2}) + 2 (a^{2} + b^{2} - c^{2}) = 2 (a^{2} + b^{2})$

Expanding the Algebraic Terms

Let's expand the brackets on the LHS:
$2 b^{2} + 2 c^{2} - 2 a^{2} + a^{2} + c^{2} - b^{2} + 2 a^{2} + 2 b^{2} - 2 c^{2} = 2 a^{2} + 2 b^{2}$
Let's group the like terms on the LHS:
For $a^{2}$ : $- 2 a^{2} + a^{2} + 2 a^{2} = a^{2}$
For $b^{2}$ : $2 b^{2} - b^{2} + 2 b^{2} = 3 b^{2}$
For $c^{2}$ : $2 c^{2} + c^{2} - 2 c^{2} = c^{2}$
So, the equation simplifies to: $a^{2} + 3 b^{2} + c^{2} = 2 a^{2} + 2 b^{2}$

Isolating the Variables

Let's rearrange the terms to simplify further:
$a^{2} + 3 b^{2} + c^{2} = 2 a^{2} + 2 b^{2}$
Subtract $a^{2}$ and $2 b^{2}$ from both sides:
$3 b^{2} - 2 b^{2} + c^{2} = 2 a^{2} - a^{2}$
This simplifies beautifully to: $b^{2} + c^{2} = a^{2}$

Applying the Converse of Pythagoras Theorem

Look at the relation we obtained: $b^{2} + c^{2} = a^{2}$
This is the famous Pythagorean relation!
By the Converse of Pythagoras Theorem, the triangle $A B C$ must be a right-angled triangle.
Since $a$ is the hypotenuse (the side opposite to angle $A$ ), the right angle must be at vertex $A$ .
Therefore, the measure of angle $A$ is $9 0^{\circ}$ .

00:00 / 00:00

The Sigma Insight: Properties of Triangles

Solution Diagram

Analyzing the Setup

Imagine you are standing in the middle of a vast, abstract landscape of geometry. You are handed a triangle

A B C

defined by the cryptic equation:

\frac{2 cos A}{a} + \frac{cos B}{b} + \frac{2 cos C}{c} = \frac{a}{b c} + \frac{b}{c a}

At first glance, this looks like a chaotic mess of trigonometry and algebra. However, every complex problem is just a simple truth waiting to be uncovered.

The Rosetta Stone

The first step is to find a common language. We use the Cosine Rule as our translator to bridge the world of angles and the world of sides.

We recall the standard identities:

cos A = \frac{b ^{2} + c ^{2} - a ^{2}}{2 b c}, cos B = \frac{a ^{2} + c ^{2} - b ^{2}}{2 a c}, cos C = \frac{a ^{2} + b ^{2} - c ^{2}}{2 ab}

By substituting these into our equation, we strip away the trigonometric veil to reveal the underlying algebraic skeleton.

The Algebraic Grind

Substituting the expressions yields:

\frac{2}{a} (\frac{b ^{2} + c ^{2} - a ^{2}}{2 b c}) + \frac{1}{b} (\frac{a ^{2} + c ^{2} - b ^{2}}{2 a c}) + \frac{2}{c} (\frac{a ^{2} + b ^{2} - c ^{2}}{2 ab}) = \frac{a ^{2} + b ^{2}}{ab c}

Observe the denominators; they are all products of

a

b

, and

c

. When we multiply the terms, the denominators become

ab c

2 ab c

The equation simplifies to:

\frac{b ^{2} + c ^{2} - a ^{2}}{ab c} + \frac{a ^{2} + c ^{2} - b ^{2}}{2 ab c} + \frac{a ^{2} + b ^{2} - c ^{2}}{ab c} = \frac{a ^{2} + b ^{2}}{ab c}

The Elegant Collapse

We bring everything to a common denominator of

2 ab c

. This is the climax of our derivation:

\frac{2 ( b ^{2} + c ^{2} - a ^{2} ) + ( a ^{2} + c ^{2} - b ^{2} ) + 2 ( a ^{2} + b ^{2} - c ^{2} )}{2 ab c} = \frac{2 ( a ^{2} + b ^{2} )}{2 ab c}

Since the denominator

2 ab c

is non-zero, we equate the numerators and expand the terms:

2 b^{2} + 2 c^{2} - 2 a^{2} + a^{2} + c^{2} - b^{2} + 2 a^{2} + 2 b^{2} - 2 c^{2} = 2 a^{2} + 2 b^{2}

Grouping the like terms, we observe the variables simplify significantly:

a^{2} + 3 b^{2} + c^{2} = 2 a^{2} + 2 b^{2}

Subtracting

a^{2}

and

2 b^{2}

from both sides leaves us with the beautiful, crystalline result:

b^{2} + c^{2} = a^{2}

The Revelation

We have arrived at the heartbeat of geometry. The equation $b^{2} + c^{2} = a^{2}$ is the Pythagorean theorem.

It tells us, with absolute certainty, that our triangle is right-angled. Because

a

is the hypotenuse, the right angle must be at vertex

A

Therefore, angle $A = 9 0^{\circ}$ . You have successfully decoded the geometry of the triangle.

Animated Solution for Mathematics - Trigonometry: If in a triangle $A B C$ , $\frac{2 c o s A}{a} + \frac{c o s B}{b} + \frac{2 c o s C}{c} = \frac{a}{b c} + \frac{b}{c a}$ , then the value of the angle $A$ is .................. degrees.

Enter Numerical Value:

Visualized Solution

Understanding the Triangle and Given Relation

Recalling the Cosine Rule

Substituting Cosine Expressions

Simplifying the Denominators

Combining the LHS Fractions

Aligning the Right Hand Side

Equating Numerators

Expanding the Algebraic Terms

Isolating the Variables

Applying the Converse of Pythagoras Theorem

The Sigma Insight: Properties of Triangles

Analyzing the Setup

The Rosetta Stone

The Algebraic Grind

The Elegant Collapse

The Revelation

Similar Questions

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Enter Numerical Value:

Visualized Solution

Understanding the Triangle and Given Relation

Recalling the Cosine Rule

Substituting Cosine Expressions

Simplifying the Denominators

Combining the LHS Fractions

Aligning the Right Hand Side

Equating Numerators

Expanding the Algebraic Terms

Isolating the Variables

Applying the Converse of Pythagoras Theorem

The Sigma Insight: Properties of Triangles

Analyzing the Setup

The Rosetta Stone

The Algebraic Grind

The Elegant Collapse

The Revelation

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