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In a $\Delta A B C$, if $\frac{\cos A}{a}=\frac{\cos B}{b}=\frac{\cos C}{c}$ and $a=2$, then its area is
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Verified Answer
The correct answer is:
$\sqrt{3}$
Given, in $\triangle A B C$,
$\frac{\cos A}{a}=\frac{\cos B}{b}=\frac{\cos C}{c}...(i)$
From sine rule,
$\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}...(ii)$
On dividing Eq. (ii) by Eq. (i), we get
$\tan A=\tan B=\tan C$
$\Rightarrow \quad A=B=C$
So, $\triangle A B C$ is an equilateral triangle
$\left(\begin{array}{rrr}
\because \text { in } \triangle A B C, & & A+B+C=180^{\circ} \\
& \Rightarrow & A+A+A=180^{\circ} \\
& \Rightarrow & A=60^{\circ}=B=C
\end{array}\right)$
$\therefore$ Area of equilateral $\triangle A B C$
$\begin{aligned}
&=\frac{\sqrt{3}}{4} a^{2}=\frac{\sqrt{3}}{4}(2)^{2} \quad(\because a=2, \text { given }) \\
&=\frac{\sqrt{3}}{4} \times 4=\sqrt{3}
\end{aligned}$
$\frac{\cos A}{a}=\frac{\cos B}{b}=\frac{\cos C}{c}...(i)$
From sine rule,
$\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}...(ii)$
On dividing Eq. (ii) by Eq. (i), we get
$\tan A=\tan B=\tan C$
$\Rightarrow \quad A=B=C$
So, $\triangle A B C$ is an equilateral triangle
$\left(\begin{array}{rrr}
\because \text { in } \triangle A B C, & & A+B+C=180^{\circ} \\
& \Rightarrow & A+A+A=180^{\circ} \\
& \Rightarrow & A=60^{\circ}=B=C
\end{array}\right)$
$\therefore$ Area of equilateral $\triangle A B C$
$\begin{aligned}
&=\frac{\sqrt{3}}{4} a^{2}=\frac{\sqrt{3}}{4}(2)^{2} \quad(\because a=2, \text { given }) \\
&=\frac{\sqrt{3}}{4} \times 4=\sqrt{3}
\end{aligned}$
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