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.