Correct answer is : Area of triangle is [tex]15\sqrt{3} sq.units[/tex]
Solution:-
We can do it in 2 methods.
Method 1:-
Given that AB=6, BC=10 and m∠B = 120
Then area of triangle = [tex]\frac{1}{2}XbaseXheight[/tex]
Let us assume AB is base and D is an altitude from C onto AB.
Then sin(60)= [tex]\frac{CD}{BC}[/tex]
CD = BC sin(60)
Hence height = [tex]10*\frac{\sqrt{3}}{2}[/tex] = [tex]5\sqrt{3}[/tex]
Hence area of ΔABC = [tex]\frac{1}{2} X6X5\sqrt{3} = 15\sqrt{3}[/tex] sq.units
Method2:-
Area of triangle = [tex]\frac{1}{2} acsin(B)[/tex]
Here a= BC=10, c=AB=6.
Hence area of triangle = [tex]\frac{1}{2}X6X10sin(120) = 15\sqrt{3}[/tex] sq.units
