All Trigonometry Resources
Example Questions
Example Question #1 : Ambiguous Triangles
Solve for . Image not drawn to scale. There may be more than one answer.
To solve, use Law of Sines, , 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:
cross-multiply
evaluate the right side using a calculator
divide both sides by 7
solve for x by evaluating 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.
Example Question #1 : Ambiguous Triangles
Solve for . Image not drawn to scale; there may be more than one solution.
To solve, use Law of Sines, , 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:
Cross-multiply.
Evaluate the right side using a calculator.
Divide both sides by 4.
Solve for x by evaluating 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.
Example Question #1 : Ambiguous Triangles
If = , , and find to the nearest degree.
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 , we get . In this equation, if , no angle A that satisfies the triangle can be found. If , and there is a right triangle determined. Finally, if , two measures of angle B can be calculated: an acute angle B and an obtuse angle . In this case, there may be one or two triangles determined. If , then the angle B' is not a solution.
In this problem, , 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:
Inputting the lengths of the triangle into this equation
Isolating
Example Question #1 : Ambiguous Triangles
If , , , find to the nearest tenth of a degree.
and
and
and
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 , we get . In this equation, if , no that satisfies the triangle can be found. If , and there is a right triangle determined. Finally, if , two measures of can be calculated: an acute and an obtuse . In this case, there may be one or two triangles determined. If , then the is not a solution.
In this problem, , 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:
Inputting the values from the problem
When the original given angle () 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:
, so is a solution.
Therefore there are two values for an angle, and .
Example Question #1 : Ambiguous Triangles
If , , and = find to the nearest degree.
and
and
and
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 , we get . In this equation, if , no that satisfies the triangle can be found. If and there is a right triangle determined. Finally, if , two measures of can be calculated: an acute and an obtuse angle . In this case, there may be one or two triangles determined. If , then the is not a solution.
In this problem, , 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:
Inputting the values of the problem
Rearranging the equation to isolate
When the original given angle () 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:
.
, so is a solution.
Therefore there are two values for an angle, and
Example Question #6 : Ambiguous Triangles
If c=10.3, a=7.4, and find to the nearest degree.
and
No solution
No solution
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 , we get . In this equation, if , no that satisfies the triangle can be found. If and there is a right triangle determined. Finally, if , two measures of can be calculated: an acute and an obtuse angle . In this case, there may be one or two triangles determined. If , then the is not a solution.
In this problem, , which means that there are no solutions to 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.
Example Question #7 : Ambiguous Triangles
If , , and = find to the nearest degree.
and
No solution
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 , we get . In this equation, if , no that satisfies the triangle can be found. If and there is a right triangle determined. Finally, if , two measures of can be calculated: an acute and an obtuse angle . In this case, there may be one or two triangles determined. If , then the is not a solution.
In this problem,, 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:
Inputting the values of the problem
Rearranging the equation to isolate
When the original given angle () 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, .
Example Question #8 : Ambiguous Triangles
If , , and find to the nearest degree.
No solution
and
and
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 , we get . In this equation, if , no that satisfies the triangle can be found. If and there is a right triangle determined. Finally, if , two measures of can be calculated: an acute and an obtuse angle . In this case, there may be one or two triangles determined. If , then the is not a solution.
In this problem, , 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:
Inputting the values of the problem
Rearranging the equation to isolate
When the original given angle () 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,
Example Question #9 : Ambiguous Triangles
If , , and = find to the nearest degree.
and
No solution
and
No solution
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 , we get . In this equation, if , no that satisfies the triangle can be found. If and there is a right triangle determined. Finally, if , two measures of can be calculated: an acute and an obtuse angle . In this case, there may be one or two triangles determined. If , then the is not a solution.
In this problem, , which means that there are no solutions to 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.
Example Question #61 : Law Of Cosines And Law Of Sines
If c=70, a=50, and find to the nearest degree.
and
and
no solution
no solution
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 , we get . In this equation, if , no that satisfies the triangle can be found. If and there is a right triangle determined. Finally, if , two measures of can be calculated: an acute and an obtuse angle . In this case, there may be one or two triangles determined. If , then the is not a solution.
In this problem, , which means that there are no solutions to 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.