Find all solutions to the following triangle. (Round your answers for angles A, C, A', and C' to the nearest whole number. Round your answers for sides c and c' to two decimal places. If either triangle is not possible, enter NONE in each corresponding answer blank.)
B = 113¡ã, b = 0.62 cm, a = 0.98 cm
First triangle (assume A ¡Ü 90¡ã):
A
=
¡ã
C
=
¡ã
c
=
cm
Second triangle (assume A' > 90¡ã):
A'
=
¡ã
C'
=
¡ã
c'
=
cm
To find the solutions to the given triangle, we'll use the Law of Sines and the Law of Cosines. Here's how we can proceed:
First, let's assume that triangle ABC is the first triangle, where angle A is less than or equal to 90 degrees.
1. Use the Law of Sines to find angle A:
sin(A) / a = sin(B) / b
sin(A) / 0.98 = sin(113) / 0.62
sin(A) = (0.62 * sin(113)) / 0.98
A = sin^(-1)((0.62 * sin(113)) / 0.98)
2. Use the Law of Sines to find angle C:
sin(C) / c = sin(B) / b
sin(C) / c = sin(113) / 0.62
sin(C) = (0.62 * sin(113)) / c
C = sin^(-1)((0.62 * sin(113)) / c)
3. Use the Law of Cosines to find side c:
c^2 = a^2 + b^2 - 2ab*cos(C)
c^2 = 0.98^2 + 0.62^2 - 2 * 0.98 * 0.62 * cos(C)
Solve for c by taking the square root of both sides: c = sqrt(0.98^2 + 0.62^2 - 2 * 0.98 * 0.62 * cos(C))
Now, let's assume that triangle A'B'C' is the second triangle, where angle A' is greater than 90 degrees.
1. Use the Law of Sines to find angle A':
sin(A') / a = sin(B) / b
sin(A') / 0.98 = sin(113) / 0.62
sin(A') = (0.62 * sin(113)) / 0.98
A' = sin^(-1)((0.62 * sin(113)) / 0.98)
2. Use the Law of Sines to find angle C':
sin(C') / c' = sin(B) / b
sin(C') / c' = sin(113) / 0.62
sin(C') = (0.62 * sin(113)) / c'
C' = sin^(-1)((0.62 * sin(113)) / c')
3. Use the Law of Cosines to find side c':
c'^2 = a^2 + b^2 + 2ab*cos(C')
c'^2 = 0.98^2 + 0.62^2 + 2 * 0.98 * 0.62 * cos(C')
Solve for c' by taking the square root of both sides: c' = sqrt(0.98^2 + 0.62^2 + 2 * 0.98 * 0.62 * cos(C'))
Round the angles A, C, A', C' to the nearest whole number, and round the sides c and c' to two decimal places.
In case any of the triangles are not possible (if the values obtained are not valid for a triangle), enter "NONE" for each corresponding answer blank.
To find the solutions to the triangle, we can use the Law of Sines, which states that:
a / sin(A) = b / sin(B) = c / sin(C)
First, let's find the angles A and C for the first triangle:
sin(A) = (a * sin(B)) / b
sin(A) = (0.98 cm * sin(113°)) / 0.62 cm
sin(A) = 1.258 cm / 0.62 cm
sin(A) ≈ 2.03
Since the value is greater than 1, there is no solution for angle A. Therefore, the first triangle is not possible.
Next, let's find the angles A' and C' for the second triangle:
sin(A') = (a * sin(B)) / b
sin(A') = (0.98 cm * sin(113°)) / 0.62 cm
sin(A') = 1.258 cm / 0.62 cm
sin(A') ≈ 2.03
Similarly, the value is greater than 1, so there is no solution for angle A'. Therefore, the second triangle is also not possible.
Therefore, the solutions to the triangle are NONE for both triangles.