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Mathematics LibreTexts

10.2: Non-right Triangles - Law of Sines

Skills to Develop

  • 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.

Suppose two radar stations located 20 miles apart each detect an aircraft between them. The angle of elevation measured by the first station is 35 degrees, whereas the angle of elevation measured by the second station is 15 degrees. How can we determine the altitude of the aircraft? We see in Figure 10.2.1 that the triangle formed by the aircraft and the two stations is not a right triangle, so we cannot use what we know about right triangles. In this section, we will find out how to solve problems involving non-right triangles.

A diagram of a triangle where the vertices are the first ground station, the second ground station, and the airplane in the air between them. The angle between the first ground station and the plane is 15 degrees, and the angle between the second station and the airplane is 35 degrees. The side between the two stations is of length 20 miles. There is a dotted line perpendicular to the ground side connecting the airplane vertex with the ground - an altitude line.

Figure 10.2.1

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 Figure 10.2.2.

An oblique triangle consisting of angles alpha, beta, and gamma. Alpha and gamma's values are known, as is the side opposite beta, between alpha and gamma.

Figure 10.2.2

AAS (angle-angle-side) We know the measurements of two angles and a side that is not between the known angles. See Figure 10.2.3.

An oblique triangle consisting of angles alpha, beta, and gamma. Alpha and gamma are known, as is the side opposite alpha, between beta and gamma.

Figure 10.2.3

SSA (side-side-angle) We know the measurements of two sides and an angle that is not between the known sides. See Figure 10.2.4.

An oblique triangle consisting of angles alpha, beta, and gamma. Alpha is the only angle known. Two sides are known. The first is opposite alpha, between beta and gamma, and the second is opposite gamma, between alpha and beta.

Figure 10.2.4

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 Figure 10.2.5.

An oblique triangle consisting of sides a, b, and c, and angles alpha, beta, and gamma. Side c is opposide angle gamma and is the horizontal base of the triangle. Side b is opposite angle beta, and side a is opposite angle alpha. There is a dotted perpendicular line - an altitude - from the gamma angle to the horizontal base c.

Figure 10.2.5

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

\(h=b \sin \alpha\) and \(h=a \sin \beta\)

We then set the expressions equal to each other.         

\(b \sin \alpha=a \sin \beta\) 

\((\dfrac{1}{ab})(b \sin \alpha)=(a \sin \beta)(\dfrac{1}{ab})\)  Multiply both sides by \(\dfrac{1}{ab}\).              

\(\dfrac{\sin \alpha}{a}=\dfrac{\sin \beta}{b}\)

Similarly, we can compare the other ratios.

\(\dfrac{\sin \alpha}{a}=\dfrac{\sin \gamma}{c}\) and \(\dfrac{\sin \beta}{b}=\dfrac{\sin \gamma}{c}\)

Collectively, these relationships are called the Law of Sines.

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

Note the standard way of labeling triangles: angle \(\alpha\) (alpha) is opposite side \(a\); angle \(\beta\) (beta) is opposite side \(b\); and angle \(\gamma\) (gamma) is opposite side \(c\). See Figure 10.2.6.

While calculating angles and sides, be sure to carry the exact values through to the final answer. Generally, final answers are rounded to the nearest tenth, unless otherwise specified.

A triangle with standard labels.

Figure 10.2.6

A General Note: LAW OF SINES

Given a triangle with angles and opposite sides labeled as in Figure 10.2.6, the ratio of the measurement of an angle to the length of its opposite side will be equal to the other two ratios of angle measure to opposite side. All proportions will be equal. The Law of Sines is based on proportions and is presented symbolically two ways.

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

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

To solve an oblique triangle, use any pair of applicable ratios.

Example

Solving for Two Unknown Sides and Angle of an AAS Triangle

Solve the triangle shown in Figure 10.2.7 to the nearest tenth.

An oblique triangle with standard labels. Angle alpha is 50 degrees, angle gamma is 30 degrees, and side a is of length 10. Side b is the horizontal base.

Figure 10.2.7

Solution:

The three angles must add up to 180 degrees. From this, we can determine that

\[\beta =180°−50°−30°\]    

\[=100°\]

To find an unknown side, we need to know the corresponding angle and a known ratio. We know that angle \(\alpha=50°\)and its corresponding side \(a=10\). We can use the following proportion from the Law of Sines to find the length of \(c\). 

\(\dfrac{\sin(50°)}{10}=\dfrac{\sin(30°)}{c}\)

\(c\dfrac{\sin(50°)}{10}=\sin(30°)\)  Multiply both sides by \(c\).                   

\(c=\sin(30°)\dfrac{10}{\sin(50°)}\)  Multiply by the reciprocal to isolate \(c\).                   

\(c≈6.5\)

Similarly, to solve for \(b\), we set up another proportion.

\(\dfrac{\sin(50°)}{10}=\dfrac{\sin(100°)}{b}\)  

\(b \sin(50°)=10 \sin(100°)\)  Multiply both sides by \(b\).               

\(b=\dfrac{10 \sin(100°)}{\sin(50°)}\)  Multiply by the reciprocal to isolate \(b\).               

\(b≈12.9\)

Therefore, the complete set of angles and sides is

\(\alpha=50°\)               \(a=10\)

\(\beta=100°\)              \(b≈12.9\)

\(\gamma=30°\)                \(c≈6.5\)

Exercise 

Solve the triangle shown in Figure 10.2.8 to the nearest tenth.

An oblique triangle with standard labels. Angle alpha is 98 degrees, angle gamma is 43 degrees, and side b is of length 22. Side b is the horizontal base.

Figure 10.2.8

Solution:

\(\alpha=98°\)            \(a=34.6\)

\(\beta=39°\)            \(b=22\)

\(\gamma=43°\)              \(c=23.8\)

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.

A General Note: POSSIBLE OUTCOMES FOR SSA TRIANGLES

Oblique triangles in the category SSA may have four different outcomes. Figure 10.2.9 illustrates the solutions with the known sides \(a\) and \(b\) and known angle \(\alpha\).

Four attempted oblique triangles are in a row, all with standard labels. Side c is the horizontal base. In the first attempted triangle, side a is less than the altitude height. Since side a cannot reach side c,  there is no triangle. In the second attempted triangle, side a is equal to the length of the altitude height, so side a forms a right angle with side c. In the third attempted triangle, side a is greater than the altitude height and less than side b, so side a can form either an acute or obtuse angle with side c. In the fourth attempted triangle, side a is greater than or equal to side b, so side a forms an acute angle with side c.

Figure 10.2.9

Example

Solving an Oblique SSA Triangle

Solve the triangle in Figure 10.2.10 for the missing side and find the missing angle measures to the nearest tenth.

An oblique triangle with standard labels where side a is of length 6, side b is of length 8, and angle alpha is 35 degrees.

Figure 10.2.10

Solution:

Use the Law of Sines to find angle \(\beta\) and angle \(\gamma\), and then side \(c\). Solving for \(\beta\), we have the proportion

\[\dfrac{\sin \alpha}{a}=\dfrac{\sin \beta}{b}\]

\[\dfrac{\sin(35°)}{6}=\dfrac{\sin \beta}{8}\]

\[\dfrac{8 \sin(35°)}{6}=\sin \beta\] 

\[0.7648≈\sin \beta\] 

\[{\sin}^{−1}(0.7648)≈49.9°\]

\[\beta≈49.9°\]

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 Figure 10.2.11. It appears that there may be a second triangle that will fit the given criteria.

An oblique triangle built from the previous with standard prime labels. Side a is of length 6, side b is of length 8, and angle alpha prime is 35 degrees. An isosceles triangle is attached, using side a as one of its congruent legs and the angle supplementary to angle beta as one of its congruent base angles. The other congruent angle is called beta prime, and the entire new horizontal base, which extends from the original side c, is called c prime. There is a dotted altitude line from angle gamma prime to side c prime.

Figure 10.2.11

The angle supplementary to \(\beta\) is approximately equal to \(49.9°\), which means that \(\beta=180°−49.9°=130.1°\). (Remember that the sine function is positive in both the first and second quadrants.) Solving for \(\gamma\), we have

\[\gamma=180°−35°−130.1°≈14.9°\]

We can then use these measurements to solve the other triangle. Since \(\gamma′\) is supplementary to \(\gamma\), we have

\[\gamma′=180°−35°−49.9°≈95.1°\]

Now we need to find \(c\) and \(c′\).

We have

\[\dfrac{c}{\sin(14.9°)}=\dfrac{6}{\sin(35°)}\]              

\[c=\dfrac{6 \sin(14.9°)}{\sin(35°)}≈2.7\]

Finally,

\[\dfrac{c′}{\sin(95.1°)}=\dfrac{6}{\sin(35°)}\]             

\[c′=\dfrac{6 \sin(95.1°)}{\sin(35°)}≈10.4\]

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 Figure 10.2.12.

There are two triangles with standard labels. Triangle a is the orginal triangle. It has angles alpha of 35 degrees, beta of 130.1 degrees, and gamma of 14.9 degrees. It has sides a = 6, b = 8, and c is approximately 2.7. Triangle b is the extended triangle. It has angles alpha prime = 35 degrees, angle beta prime = 49.9 degrees, and angle gamma prime = 95.1 degrees. It has side a prime = 6, side b prime = 8, and side c prime is approximately 10.4.

Figure 10.2.12

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 Figure 10.2.12.

Exercise 

Given \(\alpha=80°\), \(a=120\), and \(b=121\), find the missing side and angles. If there is more than one possible solution, show both.

Solution:

Solution 1

\(\alpha=80°\)             \(a=120\)

\(\beta≈83.2°\)          \(b=121\)

\(\gamma≈16.8°\)          \(c≈35.2\)

Solution 2

\(\alpha′=80°\)                 \(a′=120\)

\(\beta′≈96.8°\)             \(b′=121\)

\(\gamma′≈3.2°\)               \(c′≈6.8\)

Example

Solving for the Unknown Sides and Angles of a SSA Triangle

In the triangle shown in Figure 10.2.13, solve for the unknown side and angles. Round your answers to the nearest tenth.

An oblique triangle with standard labels. Side b is 9, side c is 12, and angle gamma is 85. Angle alpha, angle beta, and side a are unknown.

Figure 10.2.13

Solution:

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, \(\gamma=85°\), and its corresponding side \(c=12\), and we know side \(b=9\). We will use this proportion to solve for \(\beta\).

\(\dfrac{\sin(85°)}{12}=\dfrac{\sin \beta}{9}\)  Isolate the unknown. 

\(\dfrac{9 \sin(85°)}{12}=\sin \beta\)

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.

\[\beta={\sin}^{−1}(\dfrac{9 \sin(85°)}{12})\]

\[\beta≈{\sin}^{−1} (0.7471)\]

\[\beta≈48.3°\]

In this case, if we subtract \(\beta\) from \(180°\), we find that there may be a second possible solution. Thus, \(\beta=180°−48.3°≈131.7°\). To check the solution, subtract both angles, \(131.7°\) and \(85°\), from \(180°\). This gives

\[\alpha=180°−85°−131.7°≈−36.7°\]

which is impossible, and so \(\beta≈48.3°\).

To find the remaining missing values, we calculate \(\alpha=180°−85°−48.3°≈46.7°\). Now, only side \(a\) is needed. Use the Law of Sines to solve for \(a\) by one of the proportions.

\[\dfrac{\sin(85°)}{12}=\dfrac{\sin(46.7°)}{a}\]

\[a\dfrac{\sin(85°)}{12}=\sin(46.7°)\]                 

\[a=\dfrac{12\sin(46.7°)}{\sin(85°)}≈8.8\]

The complete set of solutions for the given triangle is

\(\alpha≈46.7°\)       \(a≈8.8\)

\(\beta≈48.3°\)       \(b=9\)

\(\gamma=85°\)           \(c=12\)

Exercise 

Given \(\alpha=80°\), \(a=100\), \(b=10\), find the missing side and angles. If there is more than one possible solution, show both. Round your answers to the nearest tenth.

Solution:

\(\beta≈5.7°\), \(\gamma≈94.3°\), \(c≈101.3\)

Example

 

Finding the Triangles That Meet the Given Criteria

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 Figure 10.2.14.

\[\dfrac{\sin \alpha}{10}=\dfrac{\sin(50°)}{4}\]  

\[\sin \alpha=\dfrac{10 \sin(50°)}{4}\]  

\[\sin \alpha≈1.915\]

An incomplete triangle. One side has length 4 opposite a 50 degree angle, and a second side has length 10 opposite angle a. The side of length 4 is too short to reach the side of length 10, so there is no third angle.

Figure 10.2.14

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.

Exercise 

Determine the number of triangles possible given \(a=31\), \(b=26\), \(\beta=48°\).

Solution:

two

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 \(Area=\dfrac{1}{2}bh\), 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 Figure 10.2.15, one acute and one obtuse, we can drop a perpendicular to represent the height and then apply the trigonometric property \(\sin \alpha=\dfrac{opposite}{hypotenuse}\) to write an equation for area in oblique triangles. In the acute triangle, we have \(\sin \alpha=\dfrac{h}{c}\) 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 \(180−\alpha\).

Two oblique triangles with standard labels. Both have a dotted altitude line h extended from angle beta to the horizontal base side b. In the first, which is an acute triangle, the altitude is within the triangle. In the second, which is an obtuse triangle, the altitude h is outside of the triangle.

Figure 10.2.15

Thus,

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

Similarly,

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

A General Note: AREA OF AN OBLIQUE TRIANGLE

The formula for the area of an oblique triangle is given by

\[Area=\dfrac{1}{2}bc \sin \alpha\]            

\[=\dfrac{1}{2}ac \sin \beta\]            

\[=\dfrac{1}{2}ab \sin \gamma\]

This is equivalent to one-half of the product of two sides and the sine of their included angle.

Example

Finding the Area of an Oblique Triangle

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

Solution:

Using the formula, we have

\[Area=\dfrac{1}{2}ab \sin \gamma\]

\[Area=\dfrac{1}{2}(90)(52) \sin(102°)\]

\[Area≈2289  square  units\]

Exercise 

Find the area of the triangle given \(\beta=42°\), \(a=7.2 ft\), \(c=3.4 ft\). Round the area to the nearest tenth.

Solution:

about 8.2  square feet

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

Finding an Altitude

Find the altitude of the aircraft in the problem introduced at the beginning of this section, shown in Figure 10.2.16. Round the altitude to the nearest tenth of a mile.

A diagram of a triangle where the vertices are the first ground station, the second ground station, and the airplane in the air between them. The angle between the first ground station and the plane is 15 degrees, and the angle between the second station and the airplane is 35 degrees. The side between the two stations is of length 20 miles. There is a dotted altitude line perpendicular to the ground side connecting the airplane vertex with the ground.

Figure 10.2.16

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.

\[\dfrac{\sin(130°)}{20}=\dfrac{\sin(35°)}{a}\]

\[a \sin(130°)=20 \sin(35°)\]               

\[a=\dfrac{20 \sin(35°)}{\sin(130°)}\]               

\[a≈14.98\]

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 \(h\).

\[\sin(15°)=\dfrac{opposite}{hypotenuse}\]

\[\sin(15°)=\dfrac{h}{a}\]

\[\sin(15°)=\dfrac{h}{14.98}\]           

\[h=14.98 \sin(15°)\]          

\[h≈3.88\]

The aircraft is at an altitude of approximately \(3.9\) miles.

Exercise 

The diagram shown in Figure 10.2.17 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, pointB, B, is 62°, and the distance between the viewing points of the two end zones is 145 yards.

An oblique triangle formed from three vertices A, B, and C. Verticies A and B are points on the ground, and vertex C is the blimp in the air between them. The distance between A and B is 145 yards. The angle at vertex A is 70 degrees, and the angle at vertex B is 62 degrees.

Figure 10.2.17

Solution:

161.9 yd.

Media

Access these online resources for additional instruction and practice with trigonometric applications.

Key Equations

Law of Sines

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

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

Area for oblique triangles

\[Area=\dfrac{1}{2}bc \sin \alpha\]       

\[=\dfrac{1}{2} ac \sin \beta\]       

\[=\dfrac{1}{2} ab \sin \gamma\]

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 Example.
  • 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. SeeExample and Example.
  • The Law of Sines can be used to solve triangles with given criteria. See Example.
  • The general area formula for triangles translates to oblique triangles by first finding the appropriate height value. See Example.
  • 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 Example.