Q66 / 160 · TS EAMCET 2020 (14 Sep Shift 1) · Mathematics

If the sides of a triangle are in the ratio $\sqrt{3}: \sqrt{5}: \sqrt{8+\sqrt{15}}$, then the largest angle in that triangle is

  • Correct answer
    A
    $\frac{2 \pi}{3}$
  • B
    $\frac{\pi}{2}$
  • C
    $\frac{\pi}{3}$
  • D
    $\frac{5 \pi}{6}$

Solution

Ratio of sides of triangle are $\sqrt{3}: \sqrt{5}: \sqrt{8+\sqrt{15}}$ Let sides of triangle are $\sqrt{3} k, \sqrt{5} k, \sqrt{8+\sqrt{15}} k$ Longest sides is $\sqrt{8+\sqrt{15} k}$ Let $\theta$ is longest angle $\begin{aligned} \therefore \quad \cos \theta & =\frac{3 k^2+5 k^2-(8+\sqrt{15}) k^2}{2(\sqrt{3})(\sqrt{5}) k^2} \\ \cos \theta & =\frac{-\sqrt{15}}{2 \times \sqrt{15}}=\frac{-1}{2} \Rightarrow \theta=\frac{2 \pi}{3} \end{aligned}$

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

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