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.