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  4. Find Angle MBC
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Find Angle MBC

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840 Discussions

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  • Sinyuko
     + 0 comments
    import math
ab, bc = int(input()), int(input())
print(f'{round(math.degrees(math.atan(ab/bc)))}{chr(176)}')

  • inq_daniel_eliz1
     + 0 comments

    As mentioned by theflippedbit, this problem is rooted in the concept of circumscribed right triangles. If we draw the diameter of the circle (which serves as the hypotenuse), any point chosen on the circumference will form a right triangle. Since the midpoint of the hypotenuse (or midpoint of AC in our case) is also the center of the circle, it makes sense that AM = MC = MB.

    import math

def angle(x, y):
    c = math.atan(x/y)
    ans = int(round(math.degrees(c), 0))
    print(f"{ans}\u00B0")

if __name__ == '__main__':
    x = int(input().strip())
    y = int(input().strip())
    angle(x, y)

  • theflippedbit
     + 0 comments

    Notice that AM = MC = MB (right angle triangle inscibed in a circumcirle property) once that is established, use sine rule a/sin A = b/sin B to get the desired angle.

    import math

AB = int(input())
BC = int(input())


AC = math.sqrt(AB**2 + BC**2)

theta = str(int(round(360*math.asin(AB/AC)*(1/(2*math.pi))))) + chr(176)

print(theta)

  • wazirzoh1
     + 0 comments

    import math

    a = int(input()) b = int(input())

    radies = math.atan(a/b) degrees = math.degrees(radies) round = round(degrees) e = chr(176) print(f"{round}{e}")

  • sefamutlucu
     + 0 comments
    import math

ab = int(input())
bc = int(input())

print(round(math.degrees(math.atan(ab / bc))), chr(176), sep="")