For the illustration below AC =16
C=30 degrees
B=90 degrees
A=60 degrees
Solve for BA rounded to the nearest tenth
Solve for BC rounded to the nearest tenth
In triangle ABC, we can use the Law of Sines to solve for side BA and side BC. The Law of Sines states that for any triangle with sides a, b, c and opposite angles A, B, C respectively:
a/sin(A) = b/sin(B) = c/sin(C)
Let's solve for BA first:
BA/sin(C) = AC/sin(A)
BA/sin(30) = 16/sin(60)
Now we can substitute the values and solve for BA:
BA/sqrt(3)/2 = 16/sqrt(3)
Multiply both sides of the equation by sqrt(3)/2 to isolate BA:
BA = (16/sqrt(3)) * (sqrt(3)/2)
BA = 8
Rounded to the nearest tenth, BA = 8.
Now let's solve for BC:
BC/sin(A) = AC/sin(B)
BC/sin(60) = 16/sin(90)
Now we can substitute the values and solve for BC:
BC/1/2 = 16/1
Multiply both sides of the equation by 1/2 to isolate BC:
BC = (16/1) * (1/2)
BC = 8
Rounded to the nearest tenth, BC = 8.
Therefore, BA = 8 and BC = 8.