In a triangle ABC \frac{a}{\cos A}=\frac{b}{\cos B}=\frac{c}{\cos C} What is the area of the triangle if a=6 cm?
- A. 9\sqrt{3} square cm ✓
- B. 12 square cm
- C. 18\sqrt{3} square cm
- D. 24 square cm
Correct Answer: A. 9\sqrt{3} square cm
Explanation
From the Sine Rule, we have \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R. Dividing the given relation by the Sine Rule relation gives \tan A = \tan B = \tan C, which implies A = B = C = 60^{\circ}. Thus, the triangle is equilateral. The area of an equilateral triangle is \frac{\sqrt{3}}{4}a^2 = \frac{\sqrt{3}}{4}(6^2) = 9\sqrt{3} square cm.