Question

$\mathrm{The}\mathrm{sides}\mathrm{of}a\mathrm{triangle}\mathrm{are}\mathrm{given}\mathrm{as}a=3,b=5,c=7,\mathrm{then}\mathrm{the}\mathrm{smallest}\mathrm{angle}\mathrm{of}\mathrm{triangle}is:$

• A. ${\cos }^{-1}\left(\frac{11}{14}\right)$
• B. ${\cos }^{-1}\left(\frac{13}{14}\right)$
• C. ${\sin }^{-1}\left(\frac{13}{14}\right)$
• D. ${\cos }^{-1}\left(\frac{-1}{2}\right)$

Answer 1

The correct option is B ${\cos }^{-1}\left(\frac{13}{14}\right)$

Three sides of a triangle are given, so we have to use cosine rule.
And we know that the smallest angle will be the one which is opposite to the smallest side, so the smallest angle will be opposite to side 'a'.
$\therefore \cos A=\frac{b^2+c^2-a^2}{2bc}=\frac{5^2+7^2-3^2}{2\times 5\times 7}=\frac{13}{14}\Rightarrow A={\cos }^{-1}\left(\frac{13}{14}\right)$