Given that AB is a diameter, points A, O, and B are collinear. In this case, A, O, and B do not form a non-degenerate triangle. However, if we consider the lengths of the segments from the center O to
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Here is the solution for question 14:
a) State the special name of the triangle AOB.
Given that AB is a diameter, points A, O, and B are collinear. In this case, A, O, and B do not form a non-degenerate triangle. However, if we consider the lengths of the segments from the center O to the points A and B, we have OA = OB (both are radii of the circle). A figure with two equal sides is generally referred to as isosceles. Therefore, if we consider the segments OA and OB, the "triangle" AOB would be considered a degenerate isosceles triangle.
b) Find the value of angle AOB.
Since
