Given sin A=1/2,sinB=1 and BC=4, calculate the area of triangle ABC

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

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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.

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