Question

In a triangle \( ABC \) with usual notations, if \( \frac{\cos A}{a} = \frac{\cos B}{b} = \frac{\cos C}{c} \), then the area of triangle \( ABC \) with \( a = \sqrt{6} \) is

• A. \( \frac{\sqrt{3}}{2} \) sq. units
• B. \( \frac{3\sqrt{3}}{2} \) sq. units
• C. \( \frac{2}{\sqrt{3}} \) sq. units
• D. \( \frac{5\sqrt{3}}{2} \) sq. units

Hint:

When given proportional relationships between the angles and sides of a triangle, use them to derive the area using known trigonometric formulas.

Answer 1

The Correct Option is B

Step 1: Understanding the given condition.
The given condition \( \frac{\cos A}{a} = \frac{\cos B}{b} = \frac{\cos C}{c} \) implies a relationship between the angles and sides of the triangle. We can use this to determine the area of the triangle.

Step 2: Using the area formula for triangles.
The area of a triangle can be expressed as: \[ \text{Area} = \frac{1}{2}ab \sin C \] Using the given relation, we can substitute appropriate values for the sides and angles to find the area. The calculated value for the area turns out to be \( \frac{3\sqrt{3}}{2} \) sq. units.

Step 3: Conclusion.
Thus, the area of the triangle is \( \frac{3\sqrt{3}}{2} \) sq. units, which makes option (B) the correct answer.