To find the area of the triangle ABC where AB = 10, AC = 20.
sinA = sqrt3/2.
We know that the area of a triangle when AB and BC are known and the angle A is given by:
Area = (1/2)AB*AC sin A....(1)
Therefore substitute AB = 10 and AC = 20 and sinA = (sqrt3)/2 in the formula (1) and we get:
Area = (1/2) 10*20*(sqrt3)/2 = (200*sqrt3)/4 = 50 sqrt3.
Therefore the area of triangle = 50 sq units.
We'll have many formulas to compute the area of a triangle.
The one which is very oftenly used is:
S = base*height/2
Another one, less applied, is:
S = a*b*sin(a,b)/2
Knowing the values of the lengths of the sides AB and AC, also the measure of an angle of the triangle, we'll use the formula of the area:
S = AB*AC*sin A/2
S = 10*20*sinA/2 = 100*sin A
But sin A = sqrt3/2
S = 100*sqrt3/2
We'll divide by 2 both denominator and numerator:
S = 50*sqrt3
S = 86.60 square units