Circles Question
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Grade 11 Circles CCSS: HSG-GMD.A.1
Let triangle ABC be one of the triangles of the inscribed polygon referred to in the previous questions. Here, A would be the center of the circle, B and C would lie on the circle, and ¯BA and ¯AC have lengths determined in the previous question. To determine the length of ¯BC, which will be called a, one can use the formula a=2rsin(360°2n), where r is the radius of the circle and n is the number of sides of the inscribed polygon. Which of the following gives the best reasoning for why the argument of the sine function, which is half of m∠BAC, is 360°2n?
- Since a triangle has 180°, to divide by 2n you must multiply the numerator by 2 as well.
- To account for the conversion from radians to degrees, 1/n is multiplied by 360°/2.
- It depends on the size of the circle, but in this case, where r = 1/2, 360°/n is multiplied by 1/2.
- The inscribed polygon is divided into n triangles, and then half of that would mean dividing a full rotation by 2n.