In a triangle ABC \(\rm \frac{a}{\cos A}=\frac{b}{\cos B}=\frac{c}{\cos C}\) What is the area of the triangle if a = 6 cm?

Explanation:

Given,

\(\rm \frac{a}{\cos A}=\frac{b}{\cos B}=\frac{c}{\cos C}\)

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

⇒b²+ c ² – a ² = c ² + a ² – b ² = a ² + b ² – c ²

⇒ a ² = b ² = c ²  

⇒ a = b = c

Hence, ABC is an equilateral triangle

Side of the equilateral triangle, a = 6 cm

The area of the equilateral triangle is,

\( \text{Area} = \frac{\sqrt{3}}{4} a^2 \)

⇒ \( \text{Area} = \frac{\sqrt{3}}{4} \times 6^2 \)

⇒ \( \text{Area} = \frac{\sqrt{3}}{4} \times 36 \)

⇒ \( \text{Area} = 9\sqrt{3} \, \text{cm}^2 \)

∴ The area of the triangle is 9 × √3 square cm.