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In any triangle $A B C$, the sides are $6 \mathrm{~cm}, 10 \mathrm{~cm}$ and $14 \mathrm{~cm}$. Then the triangle is obtuse angled with the obtuse angle equal to
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The correct answer is:
$120^{\circ}$
Since, $\mathrm{c}=14$ is the largest side $\therefore$ Angle $\mathrm{C}$ will be obtuse
$\cos C=\frac{a^{2}+b^{2}-c^{2}}{2 a b}=\frac{(6)^{2}+(10)^{2}-(14)^{2}}{2(6)(10)}$
$=\frac{36+100-196}{2 \times 6 \times 10}=\frac{-1}{2}$
$\Rightarrow C=\cos ^{-1}\left(\frac{-1}{2}\right)=\pi-\frac{\pi}{3}=\frac{2 \pi}{3}=\frac{2 \times 180}{3}=120$
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