Q90 / 150 · MHT CET 2024 (11 May Shift 1) · Mathematics

With usual notations, if the lengths of the sides of a triangle are $7 \mathrm{~cm}, 4 \sqrt{3} \mathrm{~cm}$ and $\sqrt{13} \mathrm{~cm}$, then the measures of the smallest angle is

  • A
    $\frac{\pi}{2}$
  • Correct answer
    B
    $\frac{\pi}{6}$
  • C
    $\frac{\pi}{3}$
  • D
    $\frac{\pi}{4}$

Solution

$\begin{aligned} & \text { Let } a=7, b=4 \sqrt{3}, c=\sqrt{13} \\ \therefore \quad & \cos C=\frac{a^2+b^2-c^2}{2 a b}=\frac{49+48-13}{2 \times 7 \times 4 \sqrt{3}}=\frac{\sqrt{3}}{2} \\ \therefore \quad & \angle C=\frac{\pi}{6}\end{aligned}$

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

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