in triangle ABC if 2LA=3L B=6Lc determine Angle A,b,c
clearly
c = 2b
b = 2/3 a
a+b+c = 180
a + 2/3 a + 2(2/3 a) = 180
a + 2/3 a + 4/3 a = 180
3a = 180
a = 60
so, the angles a,b,c are
60,40,80
To find the measures of angles A, B, and C in triangle ABC, we can use the Law of Cosines.
First, let's label the side lengths of the triangle:
AB = c
BC = a
AC = b
According to the given information, we have:
2A = 3B (or A = (3/2)B) -- (1)
B = 6C -- (2)
Using the Law of Cosines, we have:
a^2 = b^2 + c^2 - 2bc*cos(A) -- (3)
b^2 = c^2 + a^2 - 2ac*cos(B) -- (4)
c^2 = a^2 + b^2 - 2ab*cos(C) -- (5)
Since we have the relationship between A and B in equation (1), we can express A in terms of B:
A = (3/2)B
Substituting this into equation (3), we get:
a^2 = b^2 + c^2 - 2bc*cos((3/2)B)
Similarly, substituting B in terms of C from equation (2) into equation (4), we have:
b^2 = c^2 + a^2 - 2ac*cos(6C)
Now, let's solve these equations to find the measures of angles A, B, and C.