For problems 1 and 2, determine how many solutions there are for each triangle. You do not have to solve the triangle.
1. A = 29°, a = 13, c = 27
2. A = 100.1°, a = 20, b = 11
For problems 3-6, solve each triangle using the Law of Sines. If there is no solution, write “no solution.” Round each answer to the nearest tenth.
3. A = 41°, B =61°, c = 19
4. A = 125.4°, a = 33, b = 41
5. A = 76°, a = 15, b = 5
6. A = 97.5°, a = 13, b = 9
Your exercise probably deals with the
"ambiguous case", that is, the information usually consists of two sides and a non-contained angle.
make sketches marking the given information, even though in reality the triangle may not be possible.
e.g. #1
by the sine law:
Sin C/27 = sin 29°/13
sinC = 1.0069
this is not possible since the sine of any angle must be between -1 and +1.
So, no solution is possible
#2,
sinB/11 = sin100.1/20
sinB = .54147
angle B = 32.784
but the sine is + in quadrants I or II
so angle B could also have been 180-32.784 = 147.726°.
However, a triangle cannot have two obtuse angles, so the second case is not possible.
Find the third side and angle in the usual way.
#5
sinB/5 = sin76/15
sinB = .3234
angle B = 18.9° or 161.1°
In this case the sum of 161.1 + 76 is already > 180, so we have to go with the solution of
angle B = 18.9°.
The rest of the question is routine.
1. To determine the number of solutions for triangle 1, we can use the Law of Sines. The Law of Sines states that the ratio of a side length to the sine of the opposite angle is constant for all sides and angles in a triangle.
We are given angle A (29°), side a (13), and side c (27). To check for a unique solution, we can set up the following proportion:
sin(A)/a = sin(C)/c
sin(29°)/13 = sin(C)/27
Now we can solve for sin(C):
sin(C) = (sin(29°)/13) * 27
sin(C) ≈ 0.614
Since sine values are limited to the range of -1 to 1, we can determine that there is only one solution for angle C. Therefore, there is only one solution for triangle 1.
2. Similarly, for triangle 2, we can use the Law of Sines to determine the number of solutions. We are given angle A (100.1°), side a (20), and side b (11). Setting up the proportion:
sin(A)/a = sin(B)/b
sin(100.1°)/20 = sin(B)/11
Now we can solve for sin(B):
sin(B) = (sin(100.1°)/20) * 11
sin(B) ≈ 0.410
Again, since sine values are limited to the range of -1 to 1, we can determine that there is only one solution for angle B. Therefore, there is only one solution for triangle 2.
3. To solve triangle 3 using the Law of Sines, we are given angle A (41°), angle B (61°), and side c (19). We can set up the proportion:
a/sin(A) = c/sin(C)
a/sin(41°) = 19/sin(C)
Now we can solve for a:
a = (19*sin(41°))/sin(C)
Using a calculator, we find that a ≈ 12.2.
Now, to find angle C:
sin(C) = (19*sin(61°))/a
Using a calculator, we find that sin(C) ≈ 0.727.
Since sine values are limited to the range of -1 to 1, we cannot find a valid value for angle C. Therefore, there is no solution for triangle 3.
4. For triangle 4, we are given angle A (125.4°), side a (33), and side b (41). Setting up the proportion:
a/sin(A) = b/sin(B)
33/sin(125.4°) = 41/sin(B)
Now we can solve for sin(B):
sin(B) = (41*sin(125.4°))/33
Using a calculator, we find that sin(B) ≈ 0.967.
To find angle B, we take the inverse sine:
B ≈ sin^(-1)(0.967)
Using a calculator, we find that B ≈ 75.3°.
Now, to find angle C:
C = 180° - A - B
C = 180° - 125.4° - 75.3°
C ≈ 79.3°
So, for triangle 4, angles B ≈ 75.3° and C ≈ 79.3°.
5. For triangle 5, we are given angle A (76°), side a (15), and side b (5). Setting up the proportion:
a/sin(A) = b/sin(B)
15/sin(76°) = 5/sin(B)
Now we can solve for sin(B):
sin(B) = (5*sin(76°))/15
Using a calculator, we find that sin(B) ≈ 0.177.
To find angle B, we take the inverse sine:
B ≈ sin^(-1)(0.177)
Using a calculator, we find that B ≈ 11.8°.
Now, to find angle C:
C = 180° - A - B
C = 180° - 76° - 11.8°
C ≈ 92.2°
So, for triangle 5, angles B ≈ 11.8° and C ≈ 92.2°.
6. For triangle 6, we are given angle A (97.5°), side a (13), and side b (9). Setting up the proportion:
a/sin(A) = b/sin(B)
13/sin(97.5°) = 9/sin(B)
Now we can solve for sin(B):
sin(B) = (9*sin(97.5°))/13
Using a calculator, we find that sin(B) ≈ 0.862.
To find angle B, we take the inverse sine:
B ≈ sin^(-1)(0.862)
Using a calculator, we find that B ≈ 59.4°.
Now, to find angle C:
C = 180° - A - B
C = 180° - 97.5° - 59.4°
C ≈ 23.1°
So, for triangle 6, angles B ≈ 59.4° and C ≈ 23.1°.
To determine the solution or number of solutions for each triangle in problems 1 and 2, we need to apply the Law of Sines. The Law of Sines states that the ratio of the length of a side of a triangle to the sine of its opposite angle is constant.
For problem 1:
Given A = 29°, a = 13, and c = 27, we can determine the ratio using the Law of Sines:
sin(A)/a = sin(C)/c
sin(29°)/13 = sin(C)/27
Next, we can solve for sin(C) by cross-multiplying:
sin(C) = (sin(29°) / 13) * 27
By taking the arcsine (sin^-1) of both sides, we can find the value of angle C:
C = sin^-1[(sin(29°) / 13) * 27]
Now, to determine the number of solutions, we need to consider the value of angle C. If C is acute (less than 90°), there will be one solution. If C is 90°, the triangle will be a right triangle and there will be two solutions. If C is greater than 90°, there will be no solution.
For problem 2:
Given A = 100.1°, a = 20, and b = 11, we can apply the Law of Sines in the same way as problem 1:
sin(A)/a = sin(B)/b
sin(100.1°)/20 = sin(B)/11
Solving for sin(B) by cross-multiplying:
sin(B) = (sin(100.1°) / 20) * 11
Finding angle B by taking the arcsine:
B = sin^-1[(sin(100.1°) / 20) * 11]
Similarly, we need to determine the number of solutions based on the value of angle B. If B is acute, there will be one solution. If B is 90°, there will be two solutions. If B is greater than 90°, there will be no solution.
Now, let's move on to problems 3-6 and solve each triangle using the Law of Sines. Remember to round each answer to the nearest tenth.
3. A = 41°, B = 61°, c = 19:
To solve this triangle, we can use the Law of Sines:
a/sin(A) = c/sin(C)
a/sin(41°) = 19/sin(C)
Solving for sin(C) by cross-multiplying:
sin(C) = (19 / sin(41°)) * sin(C)
Taking the arcsine:
C = sin^-1[(19 / sin(41°)) * sin(C)]
Similarly, solve for angle B using:
b/sin(B) = c/sin(C)
sin(B) = (19 / sin(61°)) * sin(C)
B = sin^-1[(19 / sin(61°)) * sin(C)]
If all three angles (A, B, and C) are less than 180°, there is one solution. If any of the angles are greater than or equal to 180°, there is no solution.
4. A = 125.4°, a = 33, b = 41:
Using the Law of Sines:
a/sin(A) = b/sin(B)
33/sin(125.4°) = 41/sin(B)
Solving for sin(B) by cross-multiplying:
sin(B) = (41 / sin(125.4°)) * sin(B)
Taking the arcsine:
B = sin^-1[(41 / sin(125.4°)) * sin(B)]
If angle B is acute (< 90°), there will be two solutions. If B is 90°, there will be one solution. If B is greater than 90°, there will be no solution.
5. A = 76°, a = 15, b = 5:
Using the Law of Sines:
a/sin(A) = b/sin(B)
15/sin(76°) = 5/sin(B)
Solving for sin(B) by cross-multiplying:
sin(B) = (5 / sin(76°)) * sin(B)
Taking the arcsine:
B = sin^-1[(5 / sin(76°)) * sin(B)]
If angle B is acute, there will be two solutions. If B is 90°, there will be one solution. If B is greater than 90°, there will be no solution.
6. A = 97.5°, a = 13, b = 9:
Using the Law of Sines:
a/sin(A) = b/sin(B)
13/sin(97.5°) = 9/sin(B)
Solving for sin(B) by cross-multiplying:
sin(B) = (9 / sin(97.5°)) * sin(B)
Taking the arcsine:
B = sin^-1[(9 / sin(97.5°)) * sin(B)]
If angle B is acute, there will be two solutions. If B is 90°, there will be one solution. If B is greater than 90°, there will be no solution.