Given the angle subtended by an arc at the circle circumference X, the task is to find the angle subtended by an arc at the centre of a circle.
Input: X = 30
Input: X = 90
- When we draw the radius AD and the chord CB, we get three small triangles.
- The three triangles ABC, ADB and ACD are isosceles as AB, AC and AD are radiuses of the circle.
- So in each of these triangles, the two acute angles (s, t and u) in each are equal.
- From the diagram, we can see
D = t + u (i)
- In triangle ABC,
s + s + A = 180 (angles in triangle) ie, A = 180 - 2s (ii)
- In triangle BCD,
(t + s) + (s + u) + (u + t) = 180 (angles in triangle again) so 2s + 2t + 2u = 180 ie 2t + 2u = 180 - 2s (iii)
A = 2t + 2u = 2D from (i), (ii) and (iii)
- Hence Proved that ‘the angle at the centre is twice the angle at the circumference‘.
Below is the implementation of the above approach:
Time Complexity: O(1)
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