Find all solutions to the following triangle. (Round your answers for angles A, B, A', and B' to the nearest minute. Round your answers for sides a and a' to the nearest whole number. If either triangle is not possible, enter NONE in each corresponding answer blank.)

Question

C = 23¡ã 20', c = 359 m, b = 428 m
First triangle (assume B ¡Ü 90¡ã):


A

=

¡ã '

B

=

¡ã '

a

=

m


Second triangle (assume B' > 90¡ã):


A'

=

¡ã '

B'

=

¡ã '

a'

=

m

Answer 1

The sine rule states sinA/a = sinB/b = sinC/c. Since we know C, b, and c, we can find sinB = sinC * (b/c), which gives us sinB = sin(23° 20') * (428/359). Now, we use a calculator to find the approximate value of sinB:

sinB ≈ 0.714236

We can find angle B using the inverse sine function:

B ≈ arcsin(0.714236)

B ≈ 45° 54'

Since B ≤ 90°, this value for B is valid for the first triangle.

Now, we can find angle A:

A = 180° - C - B

A ≈ 180° - 23° 20' - 45° 54'

A ≈ 111° 46'

Next, we can find side a using the sine rule:

sinA/a = sinB/b

sin(111° 46')/a = sin(45° 54')/428

a ≈ 428 * (sin(111° 46')/sin(45° 54'))

a ≈ 466

Now, let's try to find the second triangle, assuming B' > 90°. If B' > 90°, then we can find a supplementary angle, B'':

B'' = 180° - B ≈ 180° - 45° 54'

B'' ≈ 134° 6'

Since B'' > 90°, it's possible to have a triangle with B' > 90°. We can find A':

A' = 180° - C - B''

A' ≈ 180° - 23° 20' - 134° 6'

A' ≈ 22° 34'

Now, we can find side a' using the sine rule:

sinA'/a' = sinB''/b

sin(22° 34')/a' = sin(134° 6')/428

a' ≈ 428 * (sin(22° 34')/sin(134° 6'))

a' ≈ 167

So, we have two triangles:

First triangle (assume B ≤ 90°):
A ≈ 111° 46'
B ≈ 45° 54'
a ≈ 466 m

Second triangle (assume B' > 90°):
A' ≈ 22° 34'
B' ≈ 134° 6'
a' ≈ 167 m