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Trigonometry : Ambiguous Triangles

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Example Questions

Example Question #1 : Ambiguous Triangles

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

Ambiguous triangle 1

Possible Answers:

47.25^o or 132.75^o

42.75^o or 137.25^o

34.22^o

47.25^o

Correct answer:

47.25^o or 132.75^o

Explanation:

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

7sin(x)=8sin(40) evaluate the right side using a calculator

7sin(x)= 5.14 divide both sides by 7

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

x = 47.25^o

There is another solution as well. If 47.25^o has a sine of 0.734, so will its supplementary angle, 132.75^o .

Since 132.75^o + 40^o is still less than 180^o , is a possible value for x.

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Example Question #1 : Ambiguous Triangles

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

Ambiguous triangle 2

Possible Answers:

25^o or 155^o

25^o

71.97^o

71.97^o or 108.03^o

Correct answer:

71.97^o or 108.03^o

Explanation:

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.

4sin(x)=9sin(25) Evaluate the right side using a calculator.

4sin(x)= 3.804 Divide both sides by 4.

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

x = 71.97^o

There is another solution as well. If 71.97^o has a sine of 0.951, so will its supplementary angle, 108.03^o .

Since 108.03^o + 25^o is still less than 180^o , is a possible value for x.

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Example Question #1 : Ambiguous Triangles

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

Possible Answers:

$\small 52^o$

$\small 74^o$

$\small 61^o$

$\small 80^o$

Correct answer:

$\small 61^o$

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 $\frac{\sin A}{a}=\frac{\sin B}{b}$ , 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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Example Question #1 : Ambiguous Triangles

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

Possible Answers:

$\angle B= 172.4^\circ$

$\angle B= 167.4^\circ$

$\small 12.6^o$

$\angle B=12.6^\circ$ and $\angle B'=167.4^\circ$

$\angle B=12.6^\circ$ and $\angle B'=172.4^\circ$

Correct answer:

$\angle B=12.6^\circ$ and $\angle B'=167.4^\circ$

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 $\frac{\sin A}{a}=\frac{\sin B}{b}$ , 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 a=4 , which is less than the other given side b=10 . 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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Example Question #1 : Ambiguous Triangles

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

Possible Answers:

$\angle C = 78^\circ$

$\angle C = 51^\circ$ and $\angle C'=78^\circ$

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

$\angle C = 51^\circ$

$\angle C = 129^\circ$

Correct answer:

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

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, $\sin C =.559<1$ , 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 a=7.4 , which is less than the other given side c=10.3 . 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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Example Question #6 : Ambiguous Triangles

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

Possible Answers:

$\angle C = 51^\circ$

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

No solution

$\angle C = 129^\circ$

Correct answer:

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.

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Example Question #7 : Ambiguous Triangles

If $\small c=10.3$ , a=13 , and $\small \angle A$ = $\small 34^o$ find $\small \angle C$ to the nearest degree.

Possible Answers:

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

No solution

$\angle C = 26.3^\circ$

$\angle C = 206.3^\circ$

Correct answer:

$\angle C = 26.3^\circ$

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 a=13 , which is greater than the other given side c=10.3 . Therefore, we have only one solution, $\angle C = 26.3^\circ$ .

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Example Question #8 : Ambiguous Triangles

If $\small c=10.3$ , a=12 , and $\angle A =108^\circ$ find $\small \angle C$ to the nearest degree.

Possible Answers:

$35.28^\circ$

No solution

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

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

$54.72^\circ$

Correct answer:

$54.72^\circ$

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 a=12 , which is greater than the other given side c=10.3 . Therefore this problem has one and only one solution, $\angle C=54.72^\circ$

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Example Question #9 : Ambiguous Triangles

If c=18 , a=5 , and $\small \angle A$ = $\small 34^o$ find $\small \angle C$ to the nearest degree.

Possible Answers:

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

$27.19^\circ$

No solution

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

$8.94^\circ$

Correct answer:

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.

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Example Question #61 : Law Of Cosines And Law Of Sines

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

Possible Answers:

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

$8.11^\circ$

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

$49.89^\circ$

no solution

Correct answer:

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.

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Trigonometry : Ambiguous Triangles

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Example Question #1 : Ambiguous Triangles

Example Question #1 : Ambiguous Triangles

Example Question #1 : Ambiguous Triangles

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Example Question #61 : Law Of Cosines And Law Of Sines

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