Given triangle ABC, A = 120, a = 8, b = 3; determine B to the nearest whole degree
Try the Sine Law
sinB/3 = sin120°/8
sinB = .3247...
B = 18.95°
or
B = 19°
Given triangle ABC, A = 120°, a = 8, b = 3; determine B to the nearest whole degree.
To determine angle B in triangle ABC, we can use the Law of Cosines. The Law of Cosines states that in a triangle with sides a, b, and c, and angle C opposite side c, the following equation holds:
c^2 = a^2 + b^2 - 2ab * cos(C)
In this case, we know side lengths a, b, and angle A. We are trying to find angle B.
Let's substitute the given values into the equation:
c^2 = 8^2 + 3^2 - 2 * 8 * 3 * cos(120)
Simplifying:
c^2 = 64 + 9 - 48cos(120)
To find c, we can take the square root of both sides of the equation:
c = sqrt(64 + 9 - 48cos(120))
c = sqrt(73 + 48cos(120))
Now that we know side c, we can use the Law of Sines to find angle B. The Law of Sines states that in a triangle with sides a, b, and c, and angles A, B, and C respectively, the following equation holds:
sin(A)/a = sin(B)/b = sin(C)/c
In this case, we know side lengths a, b and c, and angle A. We are trying to find angle B.
Let's substitute the known values into the equation:
sin(120)/8 = sin(B)/3 = sin(C)/c
We can simplify this equation as follows:
sin(120)/8 = sin(B)/3
Rearranging the equation to solve for sin(B):
sin(B) = (sin(120)/8) * 3
sin(B) = (sqrt(3)/2) * 3 / 8
sin(B) = sqrt(3)/16
Now we can solve for angle B using the inverse sine function:
B = sin^(-1)(sqrt(3)/16)
Using a scientific calculator, we find B ≈ 6.6 degrees.
Therefore, angle B in triangle ABC is approximately 7 degrees to the nearest whole degree.