In $\triangle \mathrm{ABC}$, if $a=2(\sqrt{3}+1), B=45^{\circ}$ and $C=60^{\circ}$, then the area (in sq…
In $\triangle \mathrm{ABC}$, if $a=2(\sqrt{3}+1), B=45^{\circ}$ and $C=60^{\circ}$, then the area (in sq.units) of that triangle is
- $2 \sqrt{3}$
- $6$
- $6+2 \sqrt{3}$
- $6-2 \sqrt{3}$
Solution
No solution. Refer to answer key.
Asked in: AP EAMCET 2017 (24 Apr Shift 2)
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