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.)
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
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
To find all the solutions to the given triangle, we can use the Law of Sines and the concept of the ambiguous case for the sine function.
First, let's find the values for the first triangle (assuming B ≤ 90°):
1. Use the Law of Sines:
sin(A) / a = sin(B) / b
Substitute the given values:
sin(A) / a = sin(23° 20') / 428
Cross-multiply:
a * sin(23° 20') = 428 * sin(A)
a ≈ 428 * sin(A) / sin(23° 20')
Solve for A:
A ≈ arcsin(a * sin(23° 20') / 428)
2. Use the Law of Sines again:
sin(C) / c = sin(A) / a
Substitute the values we have:
sin(23° 20') / 359 = sin(A) / a
Cross-multiply:
a * sin(23° 20') = 359 * sin(A)
a ≈ 359 * sin(A) / sin(23° 20')
Solve for A:
A ≈ arcsin(a * sin(23° 20') / 359)
Now let's find the values for the second triangle (assuming B' > 90°):
1. Use the Law of Sines again:
sin(A') / a' = sin(B') / b
Substitute the given values:
sin(A') / a' = sin(23° 20') / 428
Cross-multiply:
a' * sin(23° 20') = 428 * sin(A')
a' ≈ 428 * sin(A') / sin(23° 20')
Solve for A':
A' ≈ arcsin(a' * sin(23° 20') / 428)
2. Use the Law of Sines:
sin(C) / c = sin(A') / a'
Substitute the values we have:
sin(23° 20') / 359 = sin(A') / a'
Cross-multiply:
a' * sin(23° 20') = 359 * sin(A')
a' ≈ 359 * sin(A') / sin(23° 20')
Solve for A':
A' ≈ arcsin(a' * sin(23° 20') / 359)
After calculating these values, round them as instructed to the nearest minute for angles A, B, A', and B', and to the nearest whole number for sides a and a'. If any triangle is not possible, enter NONE in each corresponding answer blank.