Given a triangle A B C with b = 40 , and c = 40 Find angle A

Answer 1

b=c=40

use the Law of Cosines

a^2=40^2+40^2-2(40)(40)CosA

CosA = (a^2-3200)/(-3200)

A = Cos^-1[1 -(a^2)/3200] degrees

0 < A < 180

0 < a < 80

if a=40, A= 60 degrees

if a = close to 80, A= close to 180 degrees

if a= close to 0, A = close to 0 degrees

if a= 40sqr2 = about 56.56, A= 90 degrees

by convention, sides are small letters

angles are capital letters

a is the side opposite angle A

b is the side opposite angle B

c is the side opposite angle C