Measurements in a Triangle

[Samin751]

import math

a,b,c=float(input("a(unit)=")),float(input("b(unit)=")),float(input("c(unit)="))

print("\n")

A=math.acos((b2+c2-a2)/(2bc))
B=math.acos((c2+a2-b2)/(2ca))
C=math.acos((a2+b2-c**2)/(2ab))

s=(a+b+c)/2

Area=math.pow(s*(s-a)(s-b)(s-c),0.5)

Out_r=a/math.sin(A)/2
In_r=(a+b-c)/2*math.tan(C/2)

Ex_r_a=a/(math.tan(B/2)+math.tan(C/2))
Ex_r_b=b/(math.tan(C/2)+math.tan(A/2))
Ex_r_c=c/(math.tan(A/2)+math.tan(B/2))

Med_a=math.pow((2*(cc+bb)-aa)/4,.5)
Med_b=math.pow((2(cc+aa)-bb)/4,.5)
Med_c=math.pow((2(aa+bb)-c*c)/4,.5)

Bisec_A=2Area/(amath.cos((B-C)/2))
Bisec_B=2Area/(bmath.cos((C-A)/2))
Bisec_C=2Area/(cmath.cos((A-B)/2))

print("A=",A180/math.pi,"°")
print("B=",B180/math.pi,"°")
print("C=",C*180/math.pi,"°","\n")

print("Circumference=",s*2," unit")
print("Area=",Area," sq.unit","\n")

print("h(a)=",Area2/a," unit")
print("h(b)=",Area2/b," unit")
print("h(c)=",Area*2/c," unit","\n")

print("Median(a)=",Med_a," unit")
print("Median(b)=",Med_b," unit")
print("Median(c)=",Med_c," unit","\n")

print("Bisector(A)=",Bisec_A," unit")
print("Bisector(B)=",Bisec_B," unit")
print("Bisector(C)=",Bisec_C," unit","\n")

print("Outradius=",Out_r," unit")
print("Inradius=",In_r," unit")

print("Extraradius(a)=",Ex_r_a," unit")
print("Extraradius(b)=",Ex_r_b," unit")
print("Extraradius(c)=",Ex_r_c," unit","\n")

if (a+b<=c or b+c<=a or c+a<b or c+a<=b):
print("This is Not a Triangle")
elif a==b and b==c:
print("Equilateral Triangle")
elif (a==b and b!=c or b==c and c!=a or c==a and a!=b):
print("Isosceles Triangle")
elif a!=b and b!=c:
print("Scalene Triangle","\n")

if (A==math.pi/2 or B==math.pi/2 or C==math.pi/2):
print("Right Triangle")
elif(A>math.pi/2 or B>math.pi/2 or C>math.pi/2):
print("Obtuse Triangle")
else:
print("Acute Triangle")

[Youssef-AMARZOU]

Pretty solid triangle calculator. One thing though - what happens if someone types in numbers that can't actually form a triangle? Like 1, 2, 10? The acos will blow up with a domain error. Might wanna check that a+b>c and all that before running the math. Also the median formula looks off, pretty sure it should be sqrt((2b²+2c²-a²)/4) not what's there. Could be wrong tho, been a while since I did this stuff.