# Calculate the area of the triangle ABC if AB=10, AC=20, sinA=sqrt3/2.

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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**