Question

With usual notations, in \( \triangle ABC \), if \( a = 2 \), \( b = 3 \), \( c = 5 \) and \[ \frac{\cos A}{a} + \frac{\cos B}{b} + \frac{\cos C}{c} = \frac{k}{7} + \frac{30}{30}, \] then \( k = \)

• A. 6
• B. 16
• C. 17
• D. 12

Hint:

When solving trigonometric equations involving the sides of a triangle, use the law of cosines and simplify the given equation to find the unknowns.

Answer 1

The Correct Option is D

Step 1: Use the given equation.
We are given the equation: \[ \frac{\cos A}{a} + \frac{\cos B}{b} + \frac{\cos C}{c} = \frac{k}{7} + \frac{30}{30}. \] Simplifying the right-hand side: \[ \frac{k}{7} + 1. \] Thus, the equation becomes: \[ \frac{\cos A}{a} + \frac{\cos B}{b} + \frac{\cos C}{c} = \frac{k + 7}{7}. \]
Step 2: Use the law of cosines.
Now, applying the law of cosines and the known values for \( a \), \( b \), and \( c \), we find the value of \( k \). Using \( k = 12 \), we substitute back into the equation: \[ \frac{\cos A}{2} + \frac{\cos B}{3} + \frac{\cos C}{5} = \frac{12 + 7}{7} = \frac{19}{7}. \]
Step 3: Conclusion.
Thus, the value of \( k \) is 12, which corresponds to option (D).