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KCET · Maths · Properties of Triangles

This sides of a triangle are \(6+2 \sqrt{3}, 4 \sqrt{3}\) and \(\sqrt{24}\). The tangent of the smallest angle of the triangle is

  1. A \(\frac{1}{\sqrt{3}}\)
  2. B \(\sqrt{2}-1\)
  3. C \(\sqrt{3}\)
  4. D 1
Verified Solution

Answer & Solution

Correct Answer

(A) \(\frac{1}{\sqrt{3}}\)

Step-by-step Solution

Detailed explanation

Given that, side of triangles are
\[
\mathrm{a}=6+2 \sqrt{3}, \mathrm{~b}=4 \sqrt{3} \text { and } \mathrm{c}=\sqrt{24}
\]
Here, we observe that the side ' \(c\) ' is small, hence the angle \(C\) is also small.
\[
\text { Then, } \begin{aligned}
& \cos C=\frac{a^{2}+b^{2}-c^{2}}{2 a b} \\
=& \frac{(6+2 \sqrt{3})^{2}+(4 \sqrt{3})^{2}-(\sqrt{24})^{2}}{2(6+2 \sqrt{3})(4 \sqrt{3})}
\end{aligned}
\]
\[
\begin{aligned}
&\Rightarrow \quad \cos C=\frac{36+12+48-24+24 \sqrt{3}}{16 \sqrt{3}(3+\sqrt{3})} \\
&\Rightarrow \quad \cos C=\frac{72+24 \sqrt{3}}{16 \sqrt{3}(3+\sqrt{3})}=\frac{24(3+\sqrt{3})}{16 \sqrt{3}(3+\sqrt{3})} \\
&\Rightarrow \quad \cos C=\frac{3}{2 \sqrt{3}}=\frac{\sqrt{3}}{2} \\
&\Rightarrow \quad \cos C=\cos 30^{\circ} \Rightarrow \angle C=30^{\circ} \\
&\text { The smallest angle } C=30^{\circ} \\
&\text { Hence, the tangent of smallest angle is } \\
&\text { tan } C=\tan 30^{\circ} \\
&=\frac{1}{\sqrt{3}}
\end{aligned}
\]