All Trigonometry Resources
Example Questions
Example Question #51 : Law Of Cosines And Law Of Sines
If , find the remaining angles and sides.
None of the other answers.
The Law of Sines is a set of ratios that allows one to compute missing angles and side lengths of oblique triangles (non right angle triangles).
The Law of Sines:
.
To find the missing angle A we subtract the sum of the two known angles from 180° as the interior angles of all triangles equal 180°.
The LOS can be rearranged to solve for the missing side.
Example Question #21 : Law Of Sines
If , , and determine the length of side , round to the nearest whole number.
This is a straightforward Law of Sines problem as we are given two angles and a corresponding side:
Substituting the known values:
Solving for the unknown side:
Example Question #91 : Triangles
If , , and determine the measure of , round to the nearest degree.
This is a straightforward Law of Sines problem since we are given one angle and two sides and are asked to determine the corresponding angle.
Substituting the given values:
Now rearranging the equation:
The final step is to take the inverse sine of both sides:
Example Question #21 : Law Of Sines
If , = , and = , find the length of side .
We are given two angles and the length of the corresponding side to one of those angles. Because the problem is asking for the corresponding length of the other angle we can use the Law of Sines to find the length of the side . The equation for the Law of Sines is
If we rearrange the equation to isolate we obtain
Substituting on the values given in the problem
Example Question #24 : Law Of Sines
If , = , and = , find the length of side to the nearest whole number.
Because we are given the two angles and the length of the corresponding side to one of those angles, we can use the Law of Sines to find the length of the side that we need. So we use the equation
Rearranging the equation to isolate gives
Substituting in the values from the problem gives
Example Question #21 : Law Of Sines
If , , and , find to the nearest whole number.
We can use the Law of Sines to find the length of the missing side, because we have its corresponding angle and the length and angle of another side. The equation for the Law of Sines is
Isolating gives us
Finally, substituting in the values of the of from the problem gives
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 .