An equilateral triangle is inscribed in a circle. The radius of the circle is 21 in. Determine the side length of the triangle to the nearest 10th of an inch
3 isosceles triangles around circle center with equal legs of 21 in and included angle of 360/3 = 120 degrees
other 2 angles are (180-120)/2 = 30 deg
law of sines
sin 120/base = sin 30/21
.866/base = .5/21
so base, which is a side of our triangle = .866 *21 * 2 = 36.4 in
Thanks so much!
/ KOS di
To find the side length of the equilateral triangle inscribed in a circle, we can use the properties of an equilateral triangle.
In an equilateral triangle, all sides are equal in length. Let's assume the side length of the equilateral triangle is "s".
When an equilateral triangle is inscribed in a circle, the center of the circle coincides with the centroid of the triangle. The centroid is the point where the medians of the triangle intersect.
In an equilateral triangle, the median is a line segment drawn from a vertex to the midpoint of the opposite side. The centroid is the point where the three medians intersect, and it divides each median in a ratio of 2:1.
So, in this case, the distance from the center of the circle to one of the vertices of the equilateral triangle is equal to 2/3 times the length of the median.
Given that the radius of the circle is 21 inches, the distance from the center to one of the vertices is (2/3) * s.
We can set up the equation:
(2/3) * s = 21
To solve for s, we can multiply both sides of the equation by 3/2:
s = (21) * (3/2)
s = 31.5 inches
Therefore, the side length of the equilateral triangle is 31.5 inches to the nearest 10th of an inch.