Question

If in a $\Delta ABC$, $a=\sqrt{3},c=3$ and $A={30}^{\circ }$

then the angle $B$ has the measure?

• A. ${90}^{\circ }$
• B. ${60}^{\circ }$
• C. ${30}^{\circ }$
• D. ${120}^{\circ }$

Answer 1

Answer: (A), (C)

${90}^{\circ }$
${30}^{\circ }$

$cosA=\frac{b^2+c^2-a^2}{2bc}\Rightarrow \frac{\sqrt{3}}{2}=\frac{b^2+9-3}{2\ast 3\ast b}\Rightarrow b^2-3\sqrt{3}b+6=0\Rightarrow (b-\sqrt{3})(b-2\sqrt{3})=0$

Therefore, $b=\sqrt{3},2\sqrt{3}$

Now , $cos{B}_{atb=\sqrt{3}}=\frac{3+9-3}{2\ast 3\ast \sqrt{3}}=\frac{\sqrt{3}}{2}\Rightarrow B={30}^o$

$cos{B}_{atb=2\sqrt{3}}=\frac{3+9-12}{2\ast 3\ast \sqrt{3}}=0\Rightarrow B={90}^o$

Therefore, Correct Options are $A,C$