Trigonometry - Law of Cosines and Law of Sines

1

If c=70, a=50, and $\angle A=122^\circ$ find $\angle C$ to the nearest degree.

$8.11^\circ$

$49.89^\circ$

$8.11^\circ$ and $171.89^\circ$

$49.89^\circ$ and $130.11^\circ$

no solution

Explanation

Notice that the given information is Angle-Side-Side, which is the ambiguous case. Therefore, we should test to see if there are no triangles that satisfy, one triangle that satisfies, or two triangles that satisfy this. From $\frac{\sin A}{a}=\frac{\sin C}{c}$ , we get $c\cdot \frac{\sin A}{a}=\sin C$ . In this equation, if $\sin C >1$ , no $\angle A$ that satisfies the triangle can be found. If $\sin C=1, C=90^\circ$ and there is a right triangle determined. Finally, if $\sin C <1$ , two measures of $\angle C$ can be calculated: an acute $\angle C$ and an obtuse angle $C'=180^\circ-C$ . In this case, there may be one or two triangles determined. If $C'+A \geq 180^\circ$ , then the $\angle C'$ is not a solution.

In this problem, $\sin C = 1.18>1$ , which means that there are no solutions to $\angle C$ that satisfy this triangle. If you got answers for this triangle, check that you set up your Law of Sines equation properly at the start of the problem.

2

If c=70, a=50, and $\angle A=122^\circ$ find $\angle C$ to the nearest degree.

$8.11^\circ$

$49.89^\circ$

$8.11^\circ$ and $171.89^\circ$

$49.89^\circ$ and $130.11^\circ$

no solution

Explanation

Notice that the given information is Angle-Side-Side, which is the ambiguous case. Therefore, we should test to see if there are no triangles that satisfy, one triangle that satisfies, or two triangles that satisfy this. From $\frac{\sin A}{a}=\frac{\sin C}{c}$ , we get $c\cdot \frac{\sin A}{a}=\sin C$ . In this equation, if $\sin C >1$ , no $\angle A$ that satisfies the triangle can be found. If $\sin C=1, C=90^\circ$ and there is a right triangle determined. Finally, if $\sin C <1$ , two measures of $\angle C$ can be calculated: an acute $\angle C$ and an obtuse angle $C'=180^\circ-C$ . In this case, there may be one or two triangles determined. If $C'+A \geq 180^\circ$ , then the $\angle C'$ is not a solution.

In this problem, $\sin C = 1.18>1$ , which means that there are no solutions to $\angle C$ that satisfy this triangle. If you got answers for this triangle, check that you set up your Law of Sines equation properly at the start of the problem.

3

If c=10.3, a=7.4, and $\angle A=122^\circ$ find $\angle C$ to the nearest degree.

$\angle C = 51^\circ$

$\angle C = 129^\circ$

$\angle C=51^\circ$ and $\angle C'=129^\circ$

No solution

Explanation

Notice that the given information is Angle-Side-Side, which is the ambiguous case. Therefore, we should test to see if there are no triangles that satisfy, one triangle that satisfies, or two triangles that satisfy this. From $\frac{\sin A}{a}=\frac{\sin C}{c}$ , we get $c\cdot \frac{\sin A}{a}=\sin C$ . In this equation, if $\sin C >1$ , no $\angle A$ that satisfies the triangle can be found. If $\sin C=1, C=90^\circ$ and there is a right triangle determined. Finally, if $\sin C <1$ , two measures of $\angle C$ can be calculated: an acute $\angle C$ and an obtuse angle $C'=180^\circ-C$ . In this case, there may be one or two triangles determined. If $C'+A \geq 180^\circ$ , then the $\angle C'$ is not a solution.

In this problem, $\sin C = 1.18>1$ , which means that there are no solutions to $\angle C$ that satisfy this triangle. If you got answers for this triangle, check that you set up your Law of Sines equation properly at the start of the problem.

4

If c=10.3, a=7.4, and $\angle A=122^\circ$ find $\angle C$ to the nearest degree.

$\angle C = 51^\circ$

$\angle C = 129^\circ$

$\angle C=51^\circ$ and $\angle C'=129^\circ$

No solution

Explanation

Notice that the given information is Angle-Side-Side, which is the ambiguous case. Therefore, we should test to see if there are no triangles that satisfy, one triangle that satisfies, or two triangles that satisfy this. From $\frac{\sin A}{a}=\frac{\sin C}{c}$ , we get $c\cdot \frac{\sin A}{a}=\sin C$ . In this equation, if $\sin C >1$ , no $\angle A$ that satisfies the triangle can be found. If $\sin C=1, C=90^\circ$ and there is a right triangle determined. Finally, if $\sin C <1$ , two measures of $\angle C$ can be calculated: an acute $\angle C$ and an obtuse angle $C'=180^\circ-C$ . In this case, there may be one or two triangles determined. If $C'+A \geq 180^\circ$ , then the $\angle C'$ is not a solution.

In this problem, $\sin C = 1.18>1$ , which means that there are no solutions to $\angle C$ that satisfy this triangle. If you got answers for this triangle, check that you set up your Law of Sines equation properly at the start of the problem.

5

If $\small c=10.3$ , , and $\small \angle A$ = $\small 34^o$ find $\small \angle C$ to the nearest degree.

$\angle C = 26.3^\circ$

$\angle C = 206.3^\circ$

$\angle C = 26.3^\circ$ and $\angle C' = 206.3^\circ$

No solution

Explanation

Notice that the given information is Angle-Side-Side, which is the ambiguous case. Therefore, we should test to see if there are no triangles that satisfy, one triangle that satisfies, or two triangles that satisfy this. From $\frac{\sin A}{a}=\frac{\sin C}{c}$ , we get $c\cdot \frac{\sin A}{a}=\sin C$ . In this equation, if $\sin C >1$ , no $\angle A$ that satisfies the triangle can be found. If $\sin C=1, C=90^\circ$ and there is a right triangle determined. Finally, if $\sin C <1$ , two measures of $\angle C$ can be calculated: an acute $\angle C$ and an obtuse angle $C'=180^\circ-C$ . In this case, there may be one or two triangles determined. If $C'+A \geq 180^\circ$ , then the $\angle C'$ is not a solution.

In this problem, $\sin C =.44<1$ , so there may be one or two angles that satisfy this triangle. Since we have the length of two sides of the triangle and the corresponding angle of one of the sides, we can use the Law of Sines to find the angle that we are looking for. This goes as follows:

$\small \small \frac{c}{\sin\angle C} = \frac{a}{\sin\angle A}$

Inputting the values of the problem

$\frac{10.3}{\sin C}=\frac{13}{\sin 34^\circ}$

Rearranging the equation to isolate $\small \angle C$

$\angle C=\sin^-^1\left (\frac{10.3\sin34^\circ}{13} \right )$

$\angle C = 26.3^\circ$

When the original given angle ( $\angle A=34^\circ$ ) is acute, there will be:

One solution if the side opposite the given angle is equal to or greater than the other given side
No solution, one solution (right triangle), or two solutions if the side opposite the given angle is less than the other given side

In this problem, the side opposite the given angle is , which is greater than the other given side . Therefore, we have only one solution, $\angle C = 26.3^\circ$ .

6

If $\small c=10.3$ , , and $\small \angle A$ = $\small 34^o$ find $\small \angle C$ to the nearest degree.

$\angle C = 26.3^\circ$

$\angle C = 206.3^\circ$

$\angle C = 26.3^\circ$ and $\angle C' = 206.3^\circ$

No solution

Explanation

Notice that the given information is Angle-Side-Side, which is the ambiguous case. Therefore, we should test to see if there are no triangles that satisfy, one triangle that satisfies, or two triangles that satisfy this. From $\frac{\sin A}{a}=\frac{\sin C}{c}$ , we get $c\cdot \frac{\sin A}{a}=\sin C$ . In this equation, if $\sin C >1$ , no $\angle A$ that satisfies the triangle can be found. If $\sin C=1, C=90^\circ$ and there is a right triangle determined. Finally, if $\sin C <1$ , two measures of $\angle C$ can be calculated: an acute $\angle C$ and an obtuse angle $C'=180^\circ-C$ . In this case, there may be one or two triangles determined. If $C'+A \geq 180^\circ$ , then the $\angle C'$ is not a solution.

In this problem, $\sin C =.44<1$ , so there may be one or two angles that satisfy this triangle. Since we have the length of two sides of the triangle and the corresponding angle of one of the sides, we can use the Law of Sines to find the angle that we are looking for. This goes as follows:

$\small \small \frac{c}{\sin\angle C} = \frac{a}{\sin\angle A}$

Inputting the values of the problem

$\frac{10.3}{\sin C}=\frac{13}{\sin 34^\circ}$

Rearranging the equation to isolate $\small \angle C$

$\angle C=\sin^-^1\left (\frac{10.3\sin34^\circ}{13} \right )$

$\angle C = 26.3^\circ$

When the original given angle ( $\angle A=34^\circ$ ) is acute, there will be:

One solution if the side opposite the given angle is equal to or greater than the other given side
No solution, one solution (right triangle), or two solutions if the side opposite the given angle is less than the other given side

In this problem, the side opposite the given angle is , which is greater than the other given side . Therefore, we have only one solution, $\angle C = 26.3^\circ$ .

7

If $\small c=10.3$ , , and $\angle A =108^\circ$ find $\small \angle C$ to the nearest degree.

$35.28^\circ$

$54.72^\circ$

$35.28^\circ$ and $144.72^\circ$

$54.72^\circ$ and $125.28^\circ$

No solution

Explanation

Notice that the given information is Angle-Side-Side, which is the ambiguous case. Therefore, we should test to see if there are no triangles that satisfy, one triangle that satisfies, or two triangles that satisfy this. From $\frac{\sin A}{a}=\frac{\sin C}{c}$ , we get $c\cdot \frac{\sin A}{a}=\sin C$ . In this equation, if $\sin C >1$ , no $\angle A$ that satisfies the triangle can be found. If $\sin C=1, C=90^\circ$ and there is a right triangle determined. Finally, if $\sin C <1$ , two measures of $\angle C$ can be calculated: an acute $\angle C$ and an obtuse angle $C'=180^\circ-C$ . In this case, there may be one or two triangles determined. If $C'+A \geq 180^\circ$ , then the $\angle C'$ is not a solution.

In this problem, $\sin C =.82<1$ , so there may be one or two angles that satisfy this triangle. Since we have the length of two sides of the triangle and the corresponding angle of one of the sides, we can use the Law of Sines to find the angle that we are looking for. This goes as follows:

$\small \small \frac{c}{\sin\angle C} = \frac{a}{\sin\angle A}$

Inputting the values of the problem

$\frac{12}{\sin 108^\circ}=\frac{10.3}{\sin C}$

Rearranging the equation to isolate $\small \angle C$

$\angle C=\sin^-^1\left (\frac{10.3\sin108^\circ}{12} \right )$

$\angle C=54.72^\circ$

When the original given angle ( $\angle A=108^\circ$ ) is obtuse, there will be:

No solution when the side opposite the given angle is less than or equal to the other given side
One solution if the side opposite the given angle is greater than the other given side

In this problem, the side opposite the given angle is , which is greater than the other given side . Therefore this problem has one and only one solution, $\angle C=54.72^\circ$

8

If $\small c=10.3$ , , and $\angle A =108^\circ$ find $\small \angle C$ to the nearest degree.

$35.28^\circ$

$54.72^\circ$

$35.28^\circ$ and $144.72^\circ$

$54.72^\circ$ and $125.28^\circ$

No solution

Explanation

Notice that the given information is Angle-Side-Side, which is the ambiguous case. Therefore, we should test to see if there are no triangles that satisfy, one triangle that satisfies, or two triangles that satisfy this. From $\frac{\sin A}{a}=\frac{\sin C}{c}$ , we get $c\cdot \frac{\sin A}{a}=\sin C$ . In this equation, if $\sin C >1$ , no $\angle A$ that satisfies the triangle can be found. If $\sin C=1, C=90^\circ$ and there is a right triangle determined. Finally, if $\sin C <1$ , two measures of $\angle C$ can be calculated: an acute $\angle C$ and an obtuse angle $C'=180^\circ-C$ . In this case, there may be one or two triangles determined. If $C'+A \geq 180^\circ$ , then the $\angle C'$ is not a solution.

In this problem, $\sin C =.82<1$ , so there may be one or two angles that satisfy this triangle. Since we have the length of two sides of the triangle and the corresponding angle of one of the sides, we can use the Law of Sines to find the angle that we are looking for. This goes as follows:

$\small \small \frac{c}{\sin\angle C} = \frac{a}{\sin\angle A}$

Inputting the values of the problem

$\frac{12}{\sin 108^\circ}=\frac{10.3}{\sin C}$

Rearranging the equation to isolate $\small \angle C$

$\angle C=\sin^-^1\left (\frac{10.3\sin108^\circ}{12} \right )$

$\angle C=54.72^\circ$

When the original given angle ( $\angle A=108^\circ$ ) is obtuse, there will be:

No solution when the side opposite the given angle is less than or equal to the other given side
One solution if the side opposite the given angle is greater than the other given side

In this problem, the side opposite the given angle is , which is greater than the other given side . Therefore this problem has one and only one solution, $\angle C=54.72^\circ$

9

If , , and $\small \angle A$ = $\small 34^o$ find $\small \angle C$ to the nearest degree.

$8.94^\circ$

$27.19^\circ$

$8.94^\circ$ and $171.06^\circ$

$27.19^\circ$ and $152.81^\circ$

No solution

Explanation

Notice that the given information is Angle-Side-Side, which is the ambiguous case. Therefore, we should test to see if there are no triangles that satisfy, one triangle that satisfies, or two triangles that satisfy this. From $\frac{\sin A}{a}=\frac{\sin C}{c}$ , we get $c\cdot \frac{\sin A}{a}=\sin C$ . In this equation, if $\sin C >1$ , no $\angle A$ that satisfies the triangle can be found. If $\sin C=1, C=90^\circ$ and there is a right triangle determined. Finally, if $\sin C <1$ , two measures of $\angle C$ can be calculated: an acute $\angle C$ and an obtuse angle $C'=180^\circ-C$ . In this case, there may be one or two triangles determined. If $C'+A \geq 180^\circ$ , then the $\angle C'$ is not a solution.

In this problem, $\sin C=2.01>1$ , which means that there are no solutions to $\angle C$ that satisfy this triangle. If you got answers for this triangle, check that you set up your Law of Sines equation properly at the start of the problem.

10

If , , and $\small \angle A$ = $\small 34^o$ find $\small \angle C$ to the nearest degree.

$8.94^\circ$

$27.19^\circ$

$8.94^\circ$ and $171.06^\circ$

$27.19^\circ$ and $152.81^\circ$

No solution

Explanation

Notice that the given information is Angle-Side-Side, which is the ambiguous case. Therefore, we should test to see if there are no triangles that satisfy, one triangle that satisfies, or two triangles that satisfy this. From $\frac{\sin A}{a}=\frac{\sin C}{c}$ , we get $c\cdot \frac{\sin A}{a}=\sin C$ . In this equation, if $\sin C >1$ , no $\angle A$ that satisfies the triangle can be found. If $\sin C=1, C=90^\circ$ and there is a right triangle determined. Finally, if $\sin C <1$ , two measures of $\angle C$ can be calculated: an acute $\angle C$ and an obtuse angle $C'=180^\circ-C$ . In this case, there may be one or two triangles determined. If $C'+A \geq 180^\circ$ , then the $\angle C'$ is not a solution.

In this problem, $\sin C=2.01>1$ , which means that there are no solutions to $\angle C$ that satisfy this triangle. If you got answers for this triangle, check that you set up your Law of Sines equation properly at the start of the problem.

Law of Cosines and Law of Sines

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