In $\triangle \mathrm{ABC}$, if $b=2, c=\sqrt{3},\left\lfloor\mathrm{~A}=30^{\circ}\right.$, then its…
- $\sqrt{3}-1$
- $\sqrt{3}+1$
- $\frac{\sqrt{3}+1}{2}$
- $\frac{\sqrt{3}-1}{2}$
Solution
Asked in: AP EAMCET 2017 (24 Apr Shift 2)
Asked in: AP EAMCET 2017 (24 Apr Shift 2)