Points A, B, and C lie on a circle of radius 1. What is the area of triangle ABC?
(1) \(AB^2 = BC^2 + AC^2\)
(2) \(\angle CAB\) equals 30 degrees.
The previous answers in this forum tended for C as the correct answer. I've marked B not C and let me explain why
statement (1) suggests that there's a right triangle, BUT the angle sides might be different and the area of triangle might vary with these angle mesaures. E.g. when angles follow 45-45-90 the area of triangle would be 1, while with 30-60-90 the area of triangle is Sqrt(3)/2 Not Sufficient;
statement (2) Very interesting statement offering the inscribed angle measurement. If we find the angle CAB intercepted at the center, we get (30`)*2 OR 60`. Additionally, with the centrally intercepted angle we have the isosceles triangle with the base angles 60` which convert into the equilateral triangle, since all angles are 60` (BC=OC=OB). SO, side BC is equal to radius 1.
If we continue the line BO from the point O up-to the point D we receive height DC for the side BC of triangle ABC. Now we need to calculate the height which is easy by knowing triangle BCD is a right triangle and angle CBD=60`. So, DC is Sqrt(3). The area of triangle ABC using all these properties ---> base (BC)*height (CD)/2 = 1*Sqrt(3)/2, Sufficient as we can answer the questions area of triangle ABC=Sqrt(3)/2 therefore answer B.
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