In a triangle ABC, if (cos A)/a=(cos B)/b=(cos C)/c, and the side a=2, then area of the triangle is

Correct answer: (D) √3. (cos A)/a=(cos B)/b=(cos C)/c ⇒ (cos A)/(2 R sin A)=(cos B)/(2 R sin B)=(cos C)/(2 R sin C) ⇒ cot A=cot B=cot C ⇒ A=B=C=60^° ⇒ Δ ABC is equilateral Hence, area of Δ=(√3)/4 a^2=√3 .

Q47 / 125 · BITSAT 2016 · Mathematics

In a $\triangle \mathrm{ABC},$ if $\frac{\cos A}{\mathrm{a}}=\frac{\cos \mathrm{B}}{\mathrm{b}}=\frac{\cos \mathrm{C}}{\mathrm{c}},$ and the side $\mathrm{a}=2,$ then area of the triangle is

A
1
B
2
C
$\frac{\sqrt{3}}{2}$
Correct answer
D
$\sqrt{3}$

Solution

$\frac{\cos A}{a}=\frac{\cos B}{b}=\frac{\cos C}{c}$ $\Rightarrow \frac{\cos \mathrm{A}}{2 \mathrm{R} \sin \mathrm{A}}=\frac{\cos \mathrm{B}}{2 \mathrm{R} \sin \mathrm{B}}=\frac{\cos \mathrm{C}}{2 \mathrm{R} \sin \mathrm{C}}$ $\Rightarrow \cot A=\cot B=\cot C$ $\Rightarrow \mathrm{A}=\mathrm{B}=\mathrm{C}=60^{\circ} \Rightarrow \Delta \mathrm{ABC}$ is equilateral Hence, area of $\Delta=\frac{\sqrt{3}}{4} \mathrm{a}^{2}=\sqrt{3}$.

Hence, the correct answer is option: 'D'.

In a $\triangle \mathrm{ABC},$ if $\frac{\cos A}{\mathrm{a}}=\frac{\cos \mathrm{B}}{\mathrm{b}}=\frac{\cos \mathrm{C}}{\mathrm{c}},$ and the side $\mathrm{a}=2,$ then area of the triangle is

Solution

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