Program to find area of a Circular Segment

In a circle, if a chord is drawn then that chord divides the whole circle into two parts. These two parts of the circle are called segments of the circle. The smaller area is known as the Minor segment and the larger area is called as the Major segment.
In the figure below, the chord AB divides the circle into minor and major segments. 
 

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We are given radius of circle and angle that forms minor segment. We need to find areas of two segments.
Examples : 
 

Input : 
radius = 21.0
angle = 120.0
Output :
Area of minor segment 270.855
Area of major segment 1114.59

Input :
radius = 10.0
angle = 90.0
Output : 
Area of minor segment 28.5397
Area of major segment 285.619

 

 

Area of the segment :

For that, we join the end points of the chord with the center of the circle resulting in a sector which subtends some 'angle' at the center. And a perpendicular is drawn from the center of the circle on the chord AB. By congruence of triangles, we obtain that the ? AOP = ? BOP = 1/2(angle). 
 

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Formula for Area of Segment : 
 

Area of Segment = Area of sector - Area of Triangle OAB 
 = pi * r2 * (angle/360) -
 Area of Triangle OAB

For detailed information about formula of Area of Sector, refer https://www.geeksforgeeks.org/dsa/area-of-a-sector/. 
 

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In the figure above, assume angle made by sector = X,
so ? AOP = ? BOP = X/2

Area of Triangle AOB = 1/2 * base * height
 = 1/2 * AB * OP

Now in Triangle AOP, By trigonometry
Cos(X/2) = OP/AO i.e. OP = AO * Cos(X/2) 
 OP = r * Cos(X/2)
Sin(X/2) = AP/AO i.e. AP = AO * Sin(X/2) 
 AP = r * Sin(X/2)

So,
Base = AB = AP + PB
 = 2 * AP
 = 2 * r * Sin(X/2)
 
Height = OP = r * Cos(X/2)

Area of triangle = 1/2 * (2 * r * Sin(X/2)) * (r * Cos(X/2))
 = 1/2 * r2 * Sin(X) 
 [Using identity 2 * Sin(A) * Cos(A)]
 = Sin(2 * A))

Hence Area of Segment = pi * r2 * (angle/360) - 1/2 * r2 * Sin(angle)

 

Output : 
 

Area of minor segment = 28.5397
Area of major segment = 285.619

Time Complexity: O(1) 
Auxiliary Space: O(1)

Please suggest if someone has a better solution which is more efficient in terms of space and time.