Question

In $\Delta$ $ABC$ the sides opposite to angles $A,B,C$ are denoted by $a,b,c$ respectively.

If $\frac{cosA}{a}=\frac{cosB}{b}=\frac{cosC}{c}$ and $a=2$,

then its area is?

• A. $2\sqrt{3}$
• B. $\sqrt{3}$
• C. $\frac{\sqrt{3}}{2}$
• D. $\frac{\sqrt{3}}{4}$

Answer 1

The correct option is B $\sqrt{3}$

Given that $\frac{cosA}{a}=\frac{cosB}{b}=\frac{cosC}{c},$

Since we already know that sine rule tells us $\frac{sinA}{a}=\frac{sinB}{b}=\frac{sinC}{c}$,

Comparing the two equations gives us that $\frac{a}{b}=\frac{cosA}{cosB}=\frac{sinA}{sinB}$

$\Rightarrow tanA=tanB$ or $A=B$

Similarly, we can conclude that $B=C$ and $A=C$

Thus, we have an equilateral triangle with side equal to 2 unit.

Area thus becomes $\frac{\sqrt{3}}{4}\times 2^2=\sqrt{3}$