Ambiguous Triangles - Trigonometry

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Question

If $\small B$ = $\small 90^o$ , $\small a=7$ , and $\small b=8$ find $\small \small \angle A$ to the nearest degree.

Answer

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 $\frac{\sin A}{a}=\frac{\sin B}{b}$ , we get $b\cdot \frac{\sin A}{a}=\sin B$ . In this equation, if $\sin B > 1$ , no angle A that satisfies the triangle can be found. If $\sin B = 1$ , $B=90^\circ$ and there is a right triangle determined. Finally, if $\sin B < 1$ , two measures of angle B can be calculated: an acute angle B and an obtuse angle $B'=180^\circ-B$ . In this case, there may be one or two triangles determined. If $B'+A\geq 180^\circ$ , then the angle B' is not a solution.

In this problem, $\sin B = 1$ , $B=90^\circ$ and there is one right triangle determined. 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:

$\small \frac{b}{\sin\angle B} = \frac{a}{\sin \angle A}$

Inputting the lengths of the triangle into this equation

$\small \frac{8}{\sin(90^o)} = \frac{7}{\sin\angle A}$

Isolating $\small \angle A$

$\small \angle A = \sin^{-1}(\frac{7\sin(90^o)}{8})$

$\small \angle A = 61^o$

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Question

If $\small c=10.3$ , $\small a=7.4$ , and $\small \angle A$ = $\small 34^o$ find $\small \angle C$ to the nearest degree.

Answer

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, $\sin C =.559<1$ , 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

$\small \frac{10.3}{\sin\angle C} = \frac{7.4}{\sin\angle 34^o}$

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

$\small \angle C = \sin^{-1}(\frac{10.3\sin\angle34^o}{7.4})$

$\small \angle C =$ $\small 51^o$

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 less than the other given side . Therefore, we have a second solution. Find it by following the below steps:

$\angle C'=180^\circ-C=180^\circ-51^\circ=129^\circ$ .

$C'+A=129^\circ+34^\circ=163^\circ\leq 180^\circ$ , so $\angle C'$ is a solution.

Therefore there are two values for an angle, $\angle C=51^\circ$ and $\angle C'=129^\circ$

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Question

If $A=5^\circ$ , $\small b=10$ , $\small a=4$ , find $\small \angle B$ to the nearest tenth of a degree.

Answer

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 $\frac{\sin A}{a}=\frac{\sin B}{b}$ , we get $b\cdot \frac{\sin A}{a}=\sin B$ . In this equation, if $\sin B > 1$ , no $\angle A$ that satisfies the triangle can be found. If $\sin B = 1$ , $B=90^\circ$ and there is a right triangle determined. Finally, if $\sin B < 1$ , two measures of $\angle B$ can be calculated: an acute $\angle B$ and an obtuse $\angle$ $B'=180^\circ-B$ . In this case, there may be one or two triangles determined. If $B'+A\geq 180^\circ$ , then the $\angle B'$ is not a solution.

In this problem, $\sin\angle A=.087<1$ , 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:

$\small \frac{b}{\sin\angle B} = \frac{a}{\sin\angle A}$

Inputting the values from the problem

$\small \frac{10}{\sin\angle B} = \frac{4}{\sin\angle 5^o}$

$\small \angle B = \sin^{-1}(\frac{10\sin\angle 5^o}{4})$

$\small \small \angle B = 12.6^o$

When the original given angle ( $\angle A=5^\circ$ ) 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 less than the other given side . Therefore, we have a second solution. Find it by following the below steps:

$\angle B'=180^\circ-B=180^\circ-12.6^\circ=167.4^\circ$

$\angle B'+\angle A=167.4^\circ+5^\circ=172.4^\circ$ $\leq180^\circ$ , so $\angle B'$ is a solution.

Therefore there are two values for an angle, $\angle B=12.6^\circ$ and $\angle B'=167.4^\circ$ .

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Question

Solve for . Image not drawn to scale. There may be more than one answer.

Answer

To solve, use Law of Sines, $\frac{a}{sinA} = \frac{b}{sinB}$ , where A is the angle across from side a, and B is the angle across from side b. In this case, our proportion is set up like this:

$\frac{7}{sin(40)}=\frac{8}{sin(x)}$ cross-multiply

evaluate the right side using a calculator

divide both sides by 7

solve for x by evaluating $sin^{-1}(0.734)$ in a calculator

There is another solution as well. If has a sine of 0.734, so will its supplementary angle, .

Since is still less than , is a possible value for x.

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Question

Solve for . Image not drawn to scale; there may be more than one solution.

Answer

To solve, use Law of Sines, $\frac{a}{sinA} = \frac{b}{sinB}$ , where A is the angle across from side a, and B is the angle across from side b. In this case, our proportion is set up like this:

$\frac{4}{sin(25)}=\frac{9}{sin(x)}$ Cross-multiply.

Evaluate the right side using a calculator.

Divide both sides by 4.

Solve for x by evaluating $sin^{-1}(0.951)$ in a calculator.

There is another solution as well. If has a sine of 0.951, so will its supplementary angle, .

Since is still less than , is a possible value for x.

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Question

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

Answer

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.

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Question

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

Answer

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, $\angle C = 26.3^\circ$ .

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Question

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

Answer

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, $\angle C=54.72^\circ$

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Question

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

Answer

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.

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Question

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

Answer

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.

Compare your answer with the correct one above

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