Finding the angle in a right-angle triangle with python 3 - python-3.x

Question

I manage to solve this question for only 1 case out of the 5 in the system. I'm certain that my method is correct, but for some reason it doesn't work out for other cases.
Below is my code
import math
AB = int(input("Enter a value for side AB: "))
while(AB>100 and AB<=0):
AB = input("Enter a value for side AB: ")
BC = int(input("Enter a value for side BC: "))
while(BC>100 and BC<=0):
BC = input("Enter a value for side BC: ")
hyp = math.sqrt(math.pow(AB,2) + math.pow(BC,2)) #find side AC
mhyp = hyp/2 #find side MC
sind = (mhyp/BC) #sin(deg)=opp/hypotenuse
degree = round(((math.asin(sind))/((math.pi)*2))*360,0) #find the angle
print("%d" %(degree) + "°")
For the case when AC and BC are 10, it did yield 45 degrees as the angle. But when AC=1 and BC=100, it produces an error since arcsine cannot accept value beyond 1.7 radians. Same goes for AC=20 and BC=10 and AC=100 and BC=1..
Is this question solvable?

Answer 1

It is a matter of primarily your angle coming in radians and then you converting it into degrees. Rest everything then will fall in place :)
Also, We can see that BM is equal to MC basis the median to hypotenuse property making the triangle MBC as isosceles and hence Angle MBC = Angle MCB.
import math
if __name__ == '__main__':
AB = input()
assert 0<int(AB)<=100
BC = input()
assert 0<int(BC)<=100
assert (int(AB) >= 0 and float(AB).is_integer() == True) and (int(BC) >= 0 and float(BC).is_integer() == True)
AC = math.sqrt((int(AB)**2) + (int(BC)**2))
tan_acb_rad = int(AB)/int(BC)
acb_rad = math.atan(tan_acb_rad) #the angle comes in radians
acb_deg = math.degrees(acb_rad) #you have to convert it into degrees
ang_mbc = acb_deg
print(str(int(round(ang_mbc,0)))+u"\N{DEGREE SIGN}") #adding the degree symbol at end

Answer 2

import math
AB =int(input())
BC =int(input())
degree=u'\N{DEGREE SIGN}'
print(str(int(round(math.degrees(math.atan(AB/BC)))))+degree)

Answer 3

I know this isn't exactly your case but it still might explain your issue
Inverse Sine of a Value Greater than One
Think about what the sine wave or curve looks like. As the angles
change, the sine of the angle goes up and down, but it never goes
above 1 or below -1. In other words, there is no angle with a sine
that is greater than 1.
When you use the sine key, you put in an angle and get out the sine of
that angle. For instance, when you do sin(30) you are finding the
sine of 30 degrees and the calculator says it's 0.5. When you use the
inverse sine (shift-sine) you put in the value of the sine and the
calculator tells you the angle. So, the inverse sine of 0.5 is 30
since a 30 degree angle has a sine of 0.5.
So when you ask the calculator to do the inverse sine of 1.732, you
are asking it what angle has a sine of 1.732. But as we said above,
there is no angle that has a sine greater than 1. The inverse sine or
arcsin of 1.732 does not exist. That's what the calculator is saying.
It sounds like your problem is asking you to try to find angle B in a
triangle with a = 40, b = 80, and A = 60 degrees. Try constructing
such a triangle, and see what happens. We make angle A and mark 80
units on one of its rays for side b, then swing an arc around the
resulting point C with a radius or length of 40, so that its
intersection with the other ray will give point B. What happens?
B
/
/
/
/
/
c / ooooooooo
/ oooo oooo
/ ooo \ ooo
/ oo \
/ o \ a=40
/ o \
/ o \
/ o \
/60 o \
A----------------o----------------C-----------
b=80
So the calculator is correct: there is no such triangle! Obviously we
can't measure angle B at the top if side 'a' is not long enough to
complete the triangle and form that angle.

Answer 4

You made mistake in geometry - BM is median, not a height to AC (they accidentally coincide for isoseles right-angle triangle, but differ in general case).
So angle BMC is not right in general case, and you cannot get sin(theta) as ratio of MC and BC
But there is well-known property of right triangle - center of circumcircle lies in the middle of hypotenuse, so M point in this case is circumcenter.
This fact immediately tells us that BM=MC (two radii), BMC triangle is isosceles and theta = ACB angle.
Solution is pretty simple (note atan using):
import math
AB = 1
BC = 1
theta = math.atan(AB / BC)
degrees = round(theta * 180 / math.pi)
print(degrees)
>> 45
AB = 3
BC = 4
...
>> 37