Question

In any triangle \(ABC\), \[ \frac{\cos A}{a}+\frac{\cos B}{b}+\frac{\cos C}{c} = \]

• A. \(a^2+b^2+c^2\)
• B. \(\dfrac{a^2+b^2+c^2}{2abc}\)
• C. \(\dfrac{2abc}{a^2+b^2+c^2}\)
• D. \(a+b+c\)

Hint:

In triangle problems involving \(\cos A,\cos B,\cos C\), use the cosine rule: \[ \cos A=\frac{b^2+c^2-a^2}{2bc}. \] Then simplify after making a common denominator.

Answer 1

The Correct Option is B

Step 1: Use cosine rule.
In triangle \(ABC\), \[ \cos A=\frac{b^2+c^2-a^2}{2bc} \] \[ \cos B=\frac{c^2+a^2-b^2}{2ca} \] \[ \cos C=\frac{a^2+b^2-c^2}{2ab} \]

Step 2: Divide each by corresponding side.
\[ \frac{\cos A}{a}=\frac{b^2+c^2-a^2}{2abc} \] \[ \frac{\cos B}{b}=\frac{c^2+a^2-b^2}{2abc} \] \[ \frac{\cos C}{c}=\frac{a^2+b^2-c^2}{2abc} \]

Step 3: Add all three expressions.
\[ \frac{\cos A}{a}+\frac{\cos B}{b}+\frac{\cos C}{c} \] \[ = \frac{b^2+c^2-a^2+c^2+a^2-b^2+a^2+b^2-c^2}{2abc} \]

Step 4: Simplify the numerator.
Combining like terms, \[ b^2+c^2-a^2+c^2+a^2-b^2+a^2+b^2-c^2 = a^2+b^2+c^2 \] Thus, \[ \frac{\cos A}{a}+\frac{\cos B}{b}+\frac{\cos C}{c} = \frac{a^2+b^2+c^2}{2abc} \]

Step 5: Final conclusion.
Therefore, \[ \boxed{\frac{a^2+b^2+c^2}{2abc}} \]