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Non-right Triangles: Law of Sines

Use the Law of Sines to solve oblique triangles.

Using the Law of Sines to Solve Oblique Triangles

In any triangle, we can draw an altitude, a perpendicular line from one vertex to the opposite side, forming two right triangles. It would be preferable, however, to have methods that we can apply directly to non-right triangles without first having to create right triangles.

Any triangle that is not a right triangle is an oblique triangle. Solving an oblique triangle means finding the measurements of all three angles and all three sides. To do so, we need to start with at least three of these values, including at least one of the sides. We will investigate three possible oblique triangle problem situations:

ASA (angle-side-angle) We know the measurements of two angles and the included side. See .
AAS (angle-angle-side) We know the measurements of two angles and a side that is not between the known angles. See .
SSA (side-side-angle) We know the measurements of two sides and an angle that is not between the known sides. See .

Knowing how to approach each of these situations enables us to solve oblique triangles without having to drop a perpendicular to form two right triangles. Instead, we can use the fact that the ratio of the measurement of one of the angles to the length of its opposite side will be equal to the other two ratios of angle measure to opposite side. Let’s see how this statement is derived by considering the triangle shown in .

Using the right triangle relationships, we know that \(\sin \ \alpha =\frac{h}{b}\) and \(\sin \ \beta =\frac{h}{a}.\) Solving both equations for \(h\) gives two different expressions for \(h.\)

\[h=b\sin \ \alpha \ \text{and}\ h=a\sin \ \beta\]

We then set the expressions equal to each other.

\[\begin{array}{lllll}b\sin \ \alpha =a\sin \ \beta & \\ (\frac{1}{ab})(b\sin \ \alpha )=(a\sin \ \beta )(\frac{1}{ab})\begin{array}{llll} & & & \end{array} & \text{Multiply both sides by}\ \frac{1}{ab}. \\ \frac{\sin \ \alpha }{a}=\frac{\sin \ \beta }{b} & \end{array}\]

Similarly, we can compare the other ratios.

\[\frac{\sin \ \alpha }{a}=\frac{\sin \ \gamma }{c}\ \text{and}\ \frac{\sin \ \beta }{b}=\frac{\sin \ \gamma }{c}\]

Collectively, these relationships are called the Law of Sines.

\[\frac{\sin \ \alpha }{a}=\frac{\sin \ \beta }{b}=\frac{\sin \ \gamma }{c}\]

Condensed — the full section is in OpenStax Algebra and Trigonometry 2e.

Using The Law of Sines to Solve SSA Triangles

We can use the Law of Sines to solve any oblique triangle, but some solutions may not be straightforward. In some cases, more than one triangle may satisfy the given criteria, which we describe as an ambiguous case. Triangles classified as SSA, those in which we know the lengths of two sides and the measurement of the angle opposite one of the given sides, may result in one or two solutions, or even no solution.

Example

Try it.

Find all possible triangles if one side has length 4 opposite an angle of 50°, and a second side has length 10.

Solution

Using the given information, we can solve for the angle opposite the side of length 10. See .

\[\begin{array}{l}\frac{\sin \ \alpha }{10}=\frac{\sin (50^{\circ})}{4} \\ \sin \ \alpha =\frac{10\sin (50^{\circ})}{4} \\ \sin \ \alpha \approx 1.915\end{array}\]

We can stop here without finding the value of \(\alpha .\) Because the range of the sine function is \([-1,1],\) it is impossible for the sine value to be 1.915. In fact, inputting \({\sin }^{-1}(1.915)\) in a graphing calculator generates an ERROR DOMAIN. Therefore, no triangles can be drawn with the provided dimensions.

Condensed — the full section is in OpenStax Algebra and Trigonometry 2e.

Finding the Area of an Oblique Triangle Using the Sine Function

Now that we can solve a triangle for missing values, we can use some of those values and the sine function to find the area of an oblique triangle. Recall that the area formula for a triangle is given as \(\text{Area}=\frac{1}{2}bh,\) where \(b\) is base and \(h\) is height. For oblique triangles, we must find \(h\) before we can use the area formula. Observing the two triangles in , one acute and one obtuse, we can drop a perpendicular to represent the height and then apply the trigonometric property \(\sin \ \alpha =\frac{\text{opposite}}{\text{hypotenuse}}\) to write an equation for area in oblique triangles. In the acute triangle, we have \(\sin \ \alpha =\frac{h}{c}\) or \(c\sin \ \alpha =h.\) However, in the obtuse triangle, we drop the perpendicular outside the triangle and extend the base \(b\) to form a right triangle. The angle used in calculation is \({\alpha }^{'},\) or \(180-\alpha .\)

Thus,

\[\text{Area}=\frac{1}{2}(\text{base})(\text{height})=\frac{1}{2}b(c\sin \ \alpha )\]

Similarly,

\[\text{Area}=\frac{1}{2}a(b\sin \ \gamma )=\frac{1}{2}a(c\sin \ \beta )\]

Example

Try it.

Find the area of a triangle with sides \(a=90,b=52,\) and angle \(\gamma =102^{\circ}.\) Round the area to the nearest integer.

Solution

Using the formula, we have

\[\begin{array}{l}\text{Area}=\frac{1}{2}ab\sin \ \gamma \\ \text{Area}=\frac{1}{2}(90)(52)\sin (102^{\circ}) \\ \text{Area}\approx 2289\ \text{square}\ \text{units}\end{array}\]

Solving Applied Problems Using the Law of Sines

The more we study trigonometric applications, the more we discover that the applications are countless. Some are flat, diagram-type situations, but many applications in calculus, engineering, and physics involve three dimensions and motion.

Example

Try it.

Find the altitude of the aircraft in the problem introduced at the beginning of this section, shown in . Round the altitude to the nearest tenth of a mile.

Solution

To find the elevation of the aircraft, we first find the distance from one station to the aircraft, such as the side \(a,\) and then use right triangle relationships to find the height of the aircraft, \(h.\)

Because the angles in the triangle add up to 180 degrees, the unknown angle must be 180°−15°−35°=130°. This angle is opposite the side of length 20, allowing us to set up a Law of Sines relationship.

\[\begin{array}{l}\begin{array}{l}\begin{array}{l} \\ \end{array} \\ \frac{\sin (130^{\circ})}{20}=\frac{\sin (35^{\circ})}{a}\end{array} \\ a\sin (130^{\circ})=20\sin (35^{\circ}) \\ a=\frac{20\sin (35^{\circ})}{\sin (130^{\circ})} \\ a\approx 14.98\end{array}\]

The distance from one station to the aircraft is about 14.98 miles.

Now that we know \(a,\) we can use right triangle relationships to solve for \(h.\)

\[\begin{array}{l}\sin (15^{\circ})=\frac{\text{opposite}}{\text{hypotenuse}} \\ \sin (15^{\circ})=\frac{h}{a} \\ \sin (15^{\circ})=\frac{h}{14.98} \\ h=14.98\sin (15^{\circ}) \\ h\approx 3.88\end{array}\]

The aircraft is at an altitude of approximately 3.9 miles.

Key Equations

Law of Sines	\(\begin{array}{l}\frac{\sin \ \alpha }{a}=\frac{\sin \ \beta }{b}=\frac{\sin \ \gamma }{c} \\ \frac{a}{\sin \ \alpha }=\frac{b}{\sin \ \beta }=\frac{c}{\sin \ \gamma }\end{array}\)
Area for oblique triangles	\(\begin{array}{l}\text{Area}=\frac{1}{2}bc\sin \ \alpha \\ =\frac{1}{2}ac\sin \ \beta \\ =\frac{1}{2}ab\sin \ \gamma \end{array}\)

Key Concepts

The Law of Sines can be used to solve oblique triangles, which are non-right triangles.
According to the Law of Sines, the ratio of the measurement of one of the angles to the length of its opposite side equals the other two ratios of angle measure to opposite side.
There are three possible cases: ASA, AAS, SSA. Depending on the information given, we can choose the appropriate equation to find the requested solution. See .
The ambiguous case arises when an oblique triangle can have different outcomes.
There are three possible cases that arise from SSA arrangement—a single solution, two possible solutions, and no solution. See and .
The Law of Sines can be used to solve triangles with given criteria. See .
The general area formula for triangles translates to oblique triangles by first finding the appropriate height value. See .
There are many trigonometric applications. They can often be solved by first drawing a diagram of the given information and then using the appropriate equation. See .

Using the Law of Sines to Solve Oblique Triangles

ASA (angle-side-angle) We know the measurements of two angles and the included side. See .
AAS (angle-angle-side) We know the measurements of two angles and a side that is not between the known angles. See .
SSA (side-side-angle) We know the measurements of two sides and an angle that is not between the known sides. See .

Using the right triangle relationships, we know that \(\sin \ \alpha =\frac{h}{b}\) and \(\sin \ \beta =\frac{h}{a}.\) Solving both equations for \(h\) gives two different expressions for \(h.\)

\[h=b\sin \ \alpha \ \text{and}\ h=a\sin \ \beta\]

We then set the expressions equal to each other.

Similarly, we can compare the other ratios.

\[\frac{\sin \ \alpha }{a}=\frac{\sin \ \gamma }{c}\ \text{and}\ \frac{\sin \ \beta }{b}=\frac{\sin \ \gamma }{c}\]

Collectively, these relationships are called the Law of Sines.

\[\frac{\sin \ \alpha }{a}=\frac{\sin \ \beta }{b}=\frac{\sin \ \gamma }{c}\]

Condensed — the full section is in OpenStax Precalculus 2e.

Using The Law of Sines to Solve SSA Triangles

Example

Try it.

Find all possible triangles if one side has length 4 opposite an angle of 50°, and a second side has length 10.

Solution

Using the given information, we can solve for the angle opposite the side of length 10. See .

\[\begin{array}{l}\frac{\sin \ \alpha }{10}=\frac{\sin (50^{\circ})}{4} \\ \sin \ \alpha =\frac{10\sin (50^{\circ})}{4} \\ \sin \ \alpha \approx 1.915\end{array}\]

Condensed — the full section is in OpenStax Precalculus 2e.

Finding the Area of an Oblique Triangle Using the Sine Function

Thus,

\[\text{Area}=\frac{1}{2}(\text{base})(\text{height})=\frac{1}{2}b(c\sin \ \alpha )\]

Similarly,

\[\text{Area}=\frac{1}{2}a(b\sin \ \gamma )=\frac{1}{2}a(c\sin \ \beta )\]

Example

Try it.

Find the area of a triangle with sides \(a=90,b=52,\) and angle \(\gamma =102^{\circ}.\) Round the area to the nearest integer.

Solution

Using the formula, we have

\[\begin{array}{l}\text{Area}=\frac{1}{2}ab\sin \ \gamma \\ \text{Area}=\frac{1}{2}(90)(52)\sin (102^{\circ}) \\ \text{Area}\approx 2289\ \text{square}\ \text{units}\end{array}\]

Solving Applied Problems Using the Law of Sines

Example

Try it.

Find the altitude of the aircraft in the problem introduced at the beginning of this section, shown in . Round the altitude to the nearest tenth of a mile.

Solution

The distance from one station to the aircraft is about 14.98 miles.

Now that we know \(a,\) we can use right triangle relationships to solve for \(h.\)

The aircraft is at an altitude of approximately 3.9 miles.

Key Equations

Law of Sines	\(\begin{array}{l}\frac{\sin \ \alpha }{a}=\frac{\sin \ \beta }{b}=\frac{\sin \ \gamma }{c} \\ \frac{a}{\sin \ \alpha }=\frac{b}{\sin \ \beta }=\frac{c}{\sin \ \gamma }\end{array}\)
Area for oblique triangles	\(\begin{array}{l}\text{Area}=\frac{1}{2}bc\sin \ \alpha \\ =\frac{1}{2}ac\sin \ \beta \\ =\frac{1}{2}ab\sin \ \gamma \end{array}\)

Key Concepts

The Law of Sines can be used to solve oblique triangles, which are non-right triangles.
According to the Law of Sines, the ratio of the measurement of one of the angles to the length of its opposite side equals the other two ratios of angle measure to opposite side.
There are three possible cases: ASA, AAS, SSA. Depending on the information given, we can choose the appropriate equation to find the requested solution. See .
The ambiguous case arises when an oblique triangle can have different outcomes.
There are three possible cases that arise from SSA arrangement—a single solution, two possible solutions, and no solution. See and .
The Law of Sines can be used to solve triangles with given criteria. See .
The general area formula for triangles translates to oblique triangles by first finding the appropriate height value. See .
There are many trigonometric applications. They can often be solved by first drawing a diagram of the given information and then using the appropriate equation. See .

Practice (40)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

Solve the triangle shown in to the nearest tenth.

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The three angles must add up to 180 degrees. From this, we can determine that
\[\begin{array}{l}\begin{array}{l} \\ \beta =180^{\circ}-50^{\circ}-30^{\circ}\end{array} \\ =100^{\circ}\end{array}\]
To find an unknown side, we need to know the corresponding angle and a known ratio. We know that angle \(\alpha =50^{\circ}\) and its corresponding side \(a=10.\) We can use the following proportion from the Law of Sines to find the length of \(c.\)
\[\begin{array}{llllll}\frac{\sin (50^{\circ})}{10}=\frac{\sin (30^{\circ})}{c} & & & & & \\ c\frac{\sin (50^{\circ})}{10}=\sin (30^{\circ}) & & & & & \text{Multiply both sides by}\ c. \\ c=\sin (30^{\circ})\frac{10}{\sin (50^{\circ})} & & & & & \text{Multiply by the reciprocal to isolate}\ c. \\ c\approx 6.5 & & & & & \end{array}\]
Similarly, to solve for \(b,\) we set up another proportion.
\[\begin{array}{lllll}\begin{array}{l} \\ \frac{\sin (50^{\circ})}{10}=\frac{\sin (100^{\circ})}{b}\end{array} & \\ b\sin (50^{\circ})=10\sin (100^{\circ}) & \text{Multiply both sides by}\ b. \\ b=\frac{10\sin (100^{\circ})}{\sin (50^{\circ})}\begin{array}{llll} & & & \end{array} & \text{Multiply by the reciprocal to isolate}\ b. \\ b\approx 12.9 & \end{array}\]
Therefore, the complete set of angles and sides is
\[\begin{array}{l}\begin{array}{l} \\ \alpha =50^{\circ}\ a=10\end{array} \\ \beta =100^{\circ}\ b\approx 12.9 \\ \gamma =30^{\circ}\ c\approx 6.5\end{array}\]
Solve the triangle shown in to the nearest tenth.

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\(\begin{array}{l}\alpha ={98}^{∘}\ a=34.6 \\ \beta ={39}^{∘}\ b=22 \\ \gamma ={43}^{∘}\ c=23.8\end{array}\)
Solve the triangle in for the missing side and find the missing angle measures to the nearest tenth.

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Use the Law of Sines to find angle \(\beta\) and angle \(\gamma ,\) and then side \(c.\) Solving for \(\beta ,\) we have the proportion
\[\begin{array}{l}\frac{\sin \ \alpha }{a}=\frac{\sin \ \beta }{b} \\ \frac{\sin (35^{\circ})}{6}=\frac{\sin \ \beta }{8} \\ \frac{8\sin (35^{\circ})}{6}=\sin \ \beta \\ 0.7648\approx \sin \ \beta \\ {\sin }^{-1}(0.7648)\approx 49.9^{\circ} \\ \beta \approx 49.9^{\circ}\end{array}\]
However, in the diagram, angle \(\beta\) appears to be an obtuse angle and may be greater than 90°. How did we get an acute angle, and how do we find the measurement of \(\beta ?\) Let’s investigate further. Dropping a perpendicular from \(\gamma\) and viewing the triangle from a right angle perspective, we have . It appears that there may be a second triangle that will fit the given criteria.
The angle supplementary to \(\beta\) is approximately equal to 49.9°, which means that \(\beta =180^{\circ}-49.9^{\circ}=130.1^{\circ}.\) (Remember that the sine function is positive in both the first and second quadrants.) Solving for \(\gamma ,\) we have
\[\gamma =180^{\circ}-35^{\circ}-130.1^{\circ}\approx 14.9^{\circ}\]
We can then use these measurements to solve the other triangle. Since \({\gamma }^{'}\) is supplementary to the sum of \({\alpha }^{'}\) and \({\beta }^{'},\) we have
\[{\gamma }^{'}=180^{\circ}-35^{\circ}-49.9^{\circ}\approx 95.1^{\circ}\]
Now we need to find \(c\) and \({c}^{'}.\)
We have
\[\begin{array}{l}\frac{c}{\sin (14.9^{\circ})}=\frac{6}{\sin (35^{\circ})} \\ c=\frac{6\sin (14.9^{\circ})}{\sin (35^{\circ})}\approx 2.7\end{array}\]
Finally,
\[\begin{array}{l}\frac{{c}^{'}}{\sin (95.1^{\circ})}=\frac{6}{\sin (35^{\circ})} \\ {c}^{'}=\frac{6\sin (95.1^{\circ})}{\sin (35^{\circ})}\approx 10.4\end{array}\]
To summarize, there are two triangles with an angle of 35°, an adjacent side of 8, and an opposite side of 6, as shown in .
However, we were looking for the values for the triangle with an obtuse angle \(\beta .\) We can see them in the first triangle (a) in .
Given \(\alpha =80^{\circ},a=120,\) and \(b=121,\) find the missing side and angles. If there is more than one possible solution, show both.

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Solution 1
\[\begin{array}{ll}\alpha =80^{\circ} & a=120 \\ \beta \approx 83.2^{\circ} & b=121 \\ \gamma \approx 16.8^{\circ} & c\approx 35.2\end{array}\]
Solution 2
\[\begin{array}{l}{\alpha }^{'}=80^{\circ}\ {a}^{'}=120 \\ {\beta }^{'}\approx 96.8^{\circ}\ {b}^{'}=121 \\ {\gamma }^{'}\approx 3.2^{\circ}\ {c}^{'}\approx 6.8\end{array}\]
In the triangle shown in , solve for the unknown side and angles. Round your answers to the nearest tenth.

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In choosing the pair of ratios from the Law of Sines to use, look at the information given. In this case, we know the angle \(\gamma =85^{\circ},\) and its corresponding side \(c=12,\) and we know side \(b=9.\) We will use this proportion to solve for \(\beta .\)
\[\begin{array}{lllll}\frac{\sin (85^{\circ})}{12}=\frac{\sin \ \beta }{9}\begin{array}{llll} & & & \end{array} & \text{Isolate the unknown}. \\ \frac{9\sin (85^{\circ})}{12}=\sin \ \beta & \end{array}\]
To find \(\beta ,\) apply the inverse sine function. The inverse sine will produce a single result, but keep in mind that there may be two values for \(\beta .\) It is important to verify the result, as there may be two viable solutions, only one solution (the usual case), or no solutions.
\[\begin{array}{l}\beta ={\sin }^{-1}(\frac{9\sin (85^{\circ})}{12}) \\ \beta \approx {\sin }^{-1}(0.7471) \\ \beta \approx 48.3^{\circ}\end{array}\]
In this case, if we subtract \(\beta\) from 180°, we find that there may be a second possible solution. Thus, \(\beta =180^{\circ}-48.3^{\circ}\approx 131.7^{\circ}.\) To check the solution, subtract both angles, 131.7° and 85°, from 180°. This gives
\[\alpha =180^{\circ}-85^{\circ}-131.7^{\circ}\approx -36.7^{\circ},\]
which is impossible, and so \(\beta \approx 48.3^{\circ}.\)
To find the remaining missing values, we calculate \(\alpha =180^{\circ}-85^{\circ}-48.3^{\circ}\approx 46.7^{\circ}.\) Now, only side \(a\) is needed. Use the Law of Sines to solve for \(a\) by one of the proportions.
\[\begin{array}{l}\begin{array}{l} \\ \begin{array}{l} \\ \frac{\sin (85^{\circ})}{12}=\frac{\sin (46.7^{\circ})}{a}\end{array}\end{array} \\ a\frac{\sin (85^{\circ})}{12}=\sin (46.7^{\circ}) \\ a=\frac{12\sin (46.7^{\circ})}{\sin (85^{\circ})}\approx 8.8\end{array}\]
The complete set of solutions for the given triangle is
\[\begin{array}{l}\begin{array}{l} \\ \alpha \approx 46.7^{\circ}\ a\approx 8.8\end{array} \\ \beta \approx 48.3^{\circ}\ b=9 \\ \gamma =85^{\circ}\ c=12\end{array}\]
Given \(\alpha =80^{\circ},a=100,\ b=10,\) find the missing side and angles. If there is more than one possible solution, show both. Round your answers to the nearest tenth.

Cavabı göstər

\(\beta \approx 5.7^{\circ},\gamma \approx 94.3^{\circ},c\approx 101.3\)
Find all possible triangles if one side has length 4 opposite an angle of 50°, and a second side has length 10.

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Using the given information, we can solve for the angle opposite the side of length 10. See .
\[\begin{array}{l}\frac{\sin \ \alpha }{10}=\frac{\sin (50^{\circ})}{4} \\ \sin \ \alpha =\frac{10\sin (50^{\circ})}{4} \\ \sin \ \alpha \approx 1.915\end{array}\]
We can stop here without finding the value of \(\alpha .\) Because the range of the sine function is \([-1,1],\) it is impossible for the sine value to be 1.915. In fact, inputting \({\sin }^{-1}(1.915)\) in a graphing calculator generates an ERROR DOMAIN. Therefore, no triangles can be drawn with the provided dimensions.
Determine the number of triangles possible given \(a=31,\) \(b=26,\) \(\beta =48^{\circ}.\)

Cavabı göstər

two
Find the area of a triangle with sides \(a=90,b=52,\) and angle \(\gamma =102^{\circ}.\) Round the area to the nearest integer.

Cavabı göstər

Using the formula, we have
\[\begin{array}{l}\text{Area}=\frac{1}{2}ab\sin \ \gamma \\ \text{Area}=\frac{1}{2}(90)(52)\sin (102^{\circ}) \\ \text{Area}\approx 2289\ \text{square}\ \text{units}\end{array}\]
Find the area of the triangle given \(\beta =42^{\circ},\) \(a=7.2\ \text{ft},\) \(c=3.4\ \text{ft}.\) Round the area to the nearest tenth.

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about \(8.2\) square feet
Find the altitude of the aircraft in the problem introduced at the beginning of this section, shown in . Round the altitude to the nearest tenth of a mile.

Cavabı göstər

To find the elevation of the aircraft, we first find the distance from one station to the aircraft, such as the side \(a,\) and then use right triangle relationships to find the height of the aircraft, \(h.\)
Because the angles in the triangle add up to 180 degrees, the unknown angle must be 180°−15°−35°=130°. This angle is opposite the side of length 20, allowing us to set up a Law of Sines relationship.
\[\begin{array}{l}\begin{array}{l}\begin{array}{l} \\ \end{array} \\ \frac{\sin (130^{\circ})}{20}=\frac{\sin (35^{\circ})}{a}\end{array} \\ a\sin (130^{\circ})=20\sin (35^{\circ}) \\ a=\frac{20\sin (35^{\circ})}{\sin (130^{\circ})} \\ a\approx 14.98\end{array}\]
The distance from one station to the aircraft is about 14.98 miles.
Now that we know \(a,\) we can use right triangle relationships to solve for \(h.\)
\[\begin{array}{l}\sin (15^{\circ})=\frac{\text{opposite}}{\text{hypotenuse}} \\ \sin (15^{\circ})=\frac{h}{a} \\ \sin (15^{\circ})=\frac{h}{14.98} \\ h=14.98\sin (15^{\circ}) \\ h\approx 3.88\end{array}\]
The aircraft is at an altitude of approximately 3.9 miles.
The diagram shown in represents the height of a blimp flying over a football stadium. Find the height of the blimp if the angle of elevation at the southern end zone, point A, is 70°, the angle of elevation from the northern end zone, point \(B,\) is 62°, and the distance between the viewing points of the two end zones is 145 yards.

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161.9 yd.
Describe the altitude of a triangle.

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The altitude extends from any vertex to the opposite side or to the line containing the opposite side at a 90° angle.
Compare right triangles and oblique triangles.
When can you use the Law of Sines to find a missing angle?

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When the known values are the side opposite the missing angle and another side and its opposite angle.
In the Law of Sines, what is the relationship between the angle in the numerator and the side in the denominator?
What type of triangle results in an ambiguous case?

Cavabı göstər

A triangle with two given sides and a non-included angle.
\(\alpha =43^{\circ},\gamma =69^{\circ},a=20\)
\(\alpha =35^{\circ},\gamma =73^{\circ},c=20\)

Cavabı göstər

\(\beta =72^{\circ},a\approx 12.0,b\approx 19.9\)
\(\alpha =60^{\circ},\) \(\beta =60^{\circ},\) \(\gamma =60^{\circ}\)
\(a=4,\) \(\alpha =\ 60^{\circ},\) \(\beta =100^{\circ}\)

Cavabı göstər

\(\gamma =20^{\circ},b\approx 4.5,c\approx 1.6\)
\(b=10,\) \(\beta =95^{\circ},\gamma =\ 30^{\circ}\)
Find side \(b\) when \(A=37^{\circ},\) \(B=49^{\circ},\) \(c=5.\)

Cavabı göstər

\(b\approx 3.78\)
Find side \(a\) when \(A=132^{\circ},C=23^{\circ},b=10.\)
Find side \(c\) when \(B=37^{\circ},C=21^{\circ},\) \(b=23.\)

Cavabı göstər

\(c\approx 13.70\)
\(\alpha =119^{\circ},a=14,b=26\)
\(\gamma =113^{\circ},b=10,c=32\)

Cavabı göstər

one triangle, \(\alpha \approx 50.3^{\circ},\beta \approx 16.7^{\circ},a\approx 26.7\)
\(b=3.5,\) \(c=5.3,\) \(\gamma =\ 80^{\circ}\)
\(a=12,\) \(c=17,\) \(\alpha =\ 35^{\circ}\)

Cavabı göstər

two triangles, \(\gamma \approx 54.3^{\circ},\beta \approx 90.7^{\circ},b\approx 20.9\) or \({\gamma }^{'}\approx 125.7^{\circ},{\beta }^{'}\approx 19.3^{\circ},{b}^{'}\approx 6.9\)
\(a=20.5,\) \(b=35.0,\) \(\beta =25^{\circ}\)
\(a=7,\) \(c=9,\) \(\alpha =\ 43^{\circ}\)

Cavabı göstər

two triangles, \(\beta \approx 75.7^{\circ},\ \gamma \approx 61.3^{\circ},b\approx 9.9\) or \({\beta }^{'}\approx 18.3^{\circ},{\gamma }^{'}\approx 118.7^{\circ},{b}^{'}\approx 3.2\)
\(a=7,b=3,\beta =24^{\circ}\)
\(b=13,c=5,\gamma =\ 10^{\circ}\)

Cavabı göstər

two triangles, \(\alpha \approx 143.2^{\circ},\beta \approx 26.8^{\circ},a\approx 17.3\) or \({\alpha }^{'}\approx 16.8^{\circ},{\beta }^{'}\approx 153.2^{\circ},{a}^{'}\approx 8.3\)
\(a=2.3,c=1.8,\gamma =28^{\circ}\)
\(\beta =119^{\circ},b=8.2,a=11.3\)

Cavabı göstər

no triangle possible
Find angle \(A\) when \(a=24,b=5,B=22^{\circ}.\)
Find angle \(A\) when \(a=13,b=6,B=20^{\circ}.\)

Cavabı göstər

\(A\approx 47.8^{\circ}\) or \({A}^{'}\approx 132.2^{\circ}\)
Find angle \(B\) when \(A=12^{\circ},a=2,b=9.\)
\(a=5,c=6,\beta =\ 35^{\circ}\)

Cavabı göstər

\(8.6\)
\(b=11,c=8,\alpha =28^{\circ}\)

Symbols used here

\sin,\ \cos,\ \tan

sine, cosine, tangent
Ratios of sides in a right triangle; coordinates on the unit circle.

^\circ

degrees
1/360 of a full turn. 180° = π radians.

\approx

approximately equal
Equal to the precision shown, not exactly.

\pi

pi
Ratio of a circle's circumference to its diameter, 3.14159…

\theta

theta
The usual name for an angle.

\arcsin,\ \sin^{-1}

inverse sine
The angle whose sine is the given value (and likewise arccos, arctan).

How to: Non-right Triangles: Law of Sines

Use the Law of Sines to solve oblique triangles.
Find the area of an oblique triangle using the sine function.
Solve applied problems using the Law of Sines.
The Law of Sines can be used to solve oblique triangles, which are non-right triangles.
According to the Law of Sines, the ratio of the measurement of one of the angles to the length of its opposite side equals the other two ratios of angle measure to opposite side.
There are three possible cases: ASA, AAS, SSA. Depending on the information given, we can choose the appropriate equation to find the requested solution. See
The ambiguous case arises when an oblique triangle can have different outcomes.
There are three possible cases that arise from SSA arrangement—a single solution, two possible solutions, and no solution. See

Questions people ask

Why radians instead of degrees?

A radian is the angle whose arc equals the radius, so it is a pure ratio rather than an arbitrary 1/360 of a turn. With radians the derivative of sin x is exactly cos x; with degrees a factor of π/180 appears everywhere.

Why does sin x = 1/2 have infinitely many solutions?

Sine repeats every full turn, and within one turn it reaches 1/2 twice (at π/6 and 5π/6). Add any whole number of turns to either and it is still true.

How do I remember the exact values?

Two triangles: the 45-45-90 with sides 1, 1, √2 and the 30-60-90 with sides 1, √3, 2. Every value for 30°, 45° and 60° is a ratio of those sides; symmetry gives the rest of the circle.

Özün sına

Parts of this page are adapted from OpenStax Algebra and Trigonometry 2e (CC BY-NC-SA 4.0), OpenStax Precalculus 2e (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

Daha çox Trigonometry

The unit circle Trigonometric equations Trigonometric identities Degrees and radians Right-triangle trigonometry (SOH-CAH-TOA)Law of sines and law of cosines Graphs of sine, cosine and tangent Inverse trigonometric functions