Challenge question of the month 2

Two sides of a triangle are $\sqrt{6}$ and 4. The angle opposite to the smaller side is 30 degrees. The area of the triangle is

(a) $2\sqrt{3}$
(b)$3\sqrt{2}$
(c)$2\sqrt{3} + \sqrt{2}$
(d)$2\sqrt{3} - \sqrt{2}$

Question

Challenge question of the month 2

Two sides of a triangle are $\sqrt{6}$ and 4. The angle opposite to the smaller side is 30 degrees. The area of the triangle is 

(a) $2\sqrt{3}$
(b)$3\sqrt{2}$
(c)$2\sqrt{3} + \sqrt{2}$
(d)$2\sqrt{3} - \sqrt{2}$

vishweshshrimali5 · Answer

@mukushla 
@mahmit2012 
@mattfeury 
@KingGeorge 
@sami-21 
@satellite73 
@lgbasallote 
@walne

vishweshshrimali5 · Answer

@UnkleRhaukus 
@eliassaab

vishweshshrimali5 · Answer

King george I did find the angle opposite to side $\sqrt {6}$ using the sine rule of a triangle

anonymous · Answer

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