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In triangle $A B C$ with usual notations $b=\sqrt{3}$, $\mathrm{c}=1, \mathrm{~m} \angle \mathrm{A}=30^{\circ}$, then the largest angle of the triangle is
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Verified Answer
The correct answer is:
$120^{\circ}$
By cosine rule, we get
$\begin{aligned}
\mathrm{a}^2 & =b^2+c^2-2 b c \cos A \\
& =(\sqrt{3})^2+(1)^2-2(\sqrt{3})(1) \cos \left(30^{\circ}\right) \\
& =3+1-2 \sqrt{3}\left(\frac{\sqrt{3}}{2}\right) \\
& =4-3 \\
& =1
\end{aligned}$
$\therefore \quad a=1$
$\therefore \quad$ Largest angle is angle $B$
$\begin{array}{ll}
& \cos \mathrm{B}=\frac{\mathrm{c}^2+\mathrm{a}^2-\mathrm{b}^2}{2 \mathrm{ca}}=\frac{1+1-3}{2 \times 1 \times 1}=\frac{-1}{2} \\
\therefore \quad & \mathrm{B}=120^{\circ}
\end{array}$
$\begin{aligned}
\mathrm{a}^2 & =b^2+c^2-2 b c \cos A \\
& =(\sqrt{3})^2+(1)^2-2(\sqrt{3})(1) \cos \left(30^{\circ}\right) \\
& =3+1-2 \sqrt{3}\left(\frac{\sqrt{3}}{2}\right) \\
& =4-3 \\
& =1
\end{aligned}$
$\therefore \quad a=1$
$\therefore \quad$ Largest angle is angle $B$
$\begin{array}{ll}
& \cos \mathrm{B}=\frac{\mathrm{c}^2+\mathrm{a}^2-\mathrm{b}^2}{2 \mathrm{ca}}=\frac{1+1-3}{2 \times 1 \times 1}=\frac{-1}{2} \\
\therefore \quad & \mathrm{B}=120^{\circ}
\end{array}$
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