If the lengths of the sides of triangle are $3,5,7$, then the largest angle of the triangle is
If the lengths of the sides of triangle are $3,5,7$, then the largest angle of the triangle is
- $\frac{\pi}{2}$
- $\frac{5 \pi}{6}$
- $\frac{2 \pi}{3}$
- $\frac{3 \pi}{4}$
Solution
$\begin{array}{ll} & \text { Let } \mathrm{a}=3, \mathrm{~b}=5, \mathrm{c}=7 \\ & \cos \mathrm{C}=\frac{\mathrm{a}^2+\mathrm{b}^2-\mathrm{c}^2}{2 \mathrm{ab}}=\frac{9+25-49}{2 \times 3 \times 5}=\frac{-15}{30}=-\frac{1}{2} \\ \therefore \quad & \angle C=\frac{2 \pi}{3}\end{array}$
Asked in: MHT CET 2024 (02 May Shift 2)
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