A) \[{{135}^{o}}\]
B) \[{{90}^{o}}\]
C) \[{{60}^{o}}\]
D) \[{{120}^{o}}\]
Correct Answer: D
Solution :
We have, \[b=\sqrt{3},\,c=1,\,\angle A={{30}^{o}}\] \[\cos A=\frac{{{b}^{2}}+{{c}^{2}}-{{a}^{2}}}{2bc}\Rightarrow \frac{\sqrt{3}}{2}=\frac{{{(\sqrt{3})}^{2}}+{{1}^{2}}-{{a}^{2}}}{2.\sqrt{3}.1}\] \[\therefore \] \[a=1,\,\,b=\sqrt{3},\,\,c=1\]. b is the largest side. Therefore, the largest angle B is given by \[\cos B=\frac{{{a}^{2}}+{{c}^{2}}-{{b}^{2}}}{2ac}=\frac{1+1-3}{2.1.1}=-\frac{1}{2}\] \ \[B=120{}^\circ \].
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