Given sin A=1/2,sinB=1 and BC=4, calculate the area of triangle ABC
We notice that sin B = 1 => B = 90 degrees.
We also notice that sin A = 1/2 => A = 30 degrees.
Since the triangle ABC is right angled, B = 90 degrees, we'll get the measure of the angle C = 90 - 30 = 60 degrees.
Since we know the length of the side BC, that represents on of the legs of triangle ABC, we'll apply the law of sines to determine the other leg.
sin A/BC = sin C/AB
AB = BC*sin C/sin A
The area of triangle can be calculated in this way:
S = leg1*leg2*sin(angle included)/2
S = AB*BC*sin B/2
S = [(BC*sin C/sin A)*BC*1]/2
S = BC^2*sin C/2sin A
S = 16*sqrt3/4*(1/2)
S = 8sqrt3
The requested area of the triangle is S = 8sqrt3 square units.