Non-right Triangles: Law of Sines

Learning Objectives

In this section, you will:

• 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.

To ensure the safety of over 5,000 U.S. aircraft flying simultaneously during peak times, air traffic controllers monitor and communicate with them after receiving data from the robust radar beacon system. 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 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.

Figure 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:

1. ASA (angle-side-angle) We know the measurements of two angles and the included side. See Figure 2.Figure 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 3.Figure 3
3. SSA (side-side-angle) We know the measurements of two sides and an angle that is not between the known sides. See Figure 4.Figure 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 5.

Figure 5

Using the right triangle relationships, we know that sinα=hbsin𝛼=ℎ𝑏 and sinβ=ha.sin𝛽=ℎ𝑎. Solving both equations for hℎ gives two different expressions for h.ℎ.

h=bsinαandh=asinβℎ=𝑏sin𝛼andℎ=𝑎sin𝛽

We then set the expressions equal to each other.

bsinα=asinβ(1ab)(bsinα)=(asinβ)(1ab)sinαa=sinβbMultiply both sides by1ab.𝑏sin𝛼=𝑎sin𝛽(1𝑎𝑏)(𝑏sin𝛼)=(𝑎sin𝛽)(1𝑎𝑏)Multiply both sides by1𝑎𝑏.sin𝛼𝑎=sin𝛽𝑏

Similarly, we can compare the other ratios.

sinαa=sinγcandsinβb=sinγcsin𝛼𝑎=sin𝛾𝑐andsin𝛽𝑏=sin𝛾𝑐

Collectively, these relationships are called the Law of Sines.

sinαa=sinβb=sinγcsin𝛼𝑎=sin𝛽𝑏=sin𝛾𝑐

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

Figure 6

LAW OF SINES

Given a triangle with angles and opposite sides labeled as in Figure 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.

sinαa=sinβb=sinγcsin𝛼𝑎=sin𝛽𝑏=sin𝛾𝑐

asinα=bsinβ=csinγ𝑎sin𝛼=𝑏sin𝛽=𝑐sin𝛾

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

EXAMPLE 1

Solving for Two Unknown Sides and Angle of an AAS Triangle

Solve the triangle shown in Figure 7 to the nearest tenth.

Figure 7

Solution

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

β=180°−50°−30°=100°𝛽=180°−50°−30°=100°

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

sin(50°)10=sin(30°)ccsin(50°)10=sin(30°)c=sin(30°)10sin(50°)c≈6.5Multiply both sides byc.Multiply by the reciprocal to isolatec.sin(50°)10=sin(30°)𝑐𝑐sin(50°)10=sin(30°)Multiply both sides by𝑐.𝑐=sin(30°)10sin(50°)Multiply by the reciprocal to isolate𝑐.𝑐≈6.5

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

sin(50°)10=sin(100°)bbsin(50°)=10sin(100°)b=10sin(100°)sin(50°)b≈12.9Multiply both sides byb.Multiply by the reciprocal to isolateb.sin(50°)10=sin(100°)𝑏𝑏sin(50°)=10sin(100°)Multiply both sides by𝑏.𝑏=10sin(100°)sin(50°)Multiply by the reciprocal to isolate𝑏.𝑏≈12.9

Therefore, the complete set of angles and sides is

α=50°a=10β=100°b≈12.9γ=30°c≈6.5𝛼=50°𝑎=10𝛽=100°𝑏≈12.9𝛾=30°𝑐≈6.5

TRY IT #1

Solve the triangle shown in Figure 8 to the nearest tenth.

Figure 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.

POSSIBLE OUTCOMES FOR SSA TRIANGLES

Oblique triangles in the category SSA may have four different outcomes. Figure 9 illustrates the solutions with the known sides a𝑎 and b𝑏 and known angle α.𝛼.

Figure 9

EXAMPLE 2

Solving an Oblique SSA Triangle

Solve the triangle in Figure 10 for the missing side and find the missing angle measures to the nearest tenth.

Figure 10

Solution

Use the Law of Sines to find angle β𝛽 and angle γ,𝛾, and then side c.𝑐. Solving for β,𝛽, we have the proportion

sinαa=sinβbsin(35°)6=sinβ88sin(35°)6=sinβ0.7648≈sinβsin−1(0.7648)≈49.9°β≈49.9°sin𝛼𝑎=sin𝛽𝑏sin(35°)6=sin𝛽88sin(35°)6=sin𝛽0.7648≈sin𝛽sin−1(0.7648)≈49.9°𝛽≈49.9°

However, in the diagram, angle β𝛽 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 β?𝛽? Let’s investigate further. Dropping a perpendicular from γ𝛾 and viewing the triangle from a right angle perspective, we have Figure 11. It appears that there may be a second triangle that will fit the given criteria.

Figure 11

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

γ=180°−35°−130.1°≈14.9°𝛾=180°−35°−130.1°≈14.9°

We can then use these measurements to solve the other triangle. Since γ′𝛾′ is supplementary to the sum of α′𝛼′ and β′,𝛽′, we have

γ′=180°−35°−49.9°≈95.1°𝛾′=180°−35°−49.9°≈95.1°

Now we need to find c𝑐 and c′.𝑐′.

We have

csin(14.9°)=6sin(35°)c=6sin(14.9°)sin(35°)≈2.7𝑐sin(14.9°)=6sin(35°)𝑐=6sin(14.9°)sin(35°)≈2.7

Finally,

c′sin(95.1°)=6sin(35°)c′=6sin(95.1°)sin(35°)≈10.4𝑐′sin(95.1°)=6sin(35°)𝑐′=6sin(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 12.

Figure 12

However, we were looking for the values for the triangle with an obtuse angle β.𝛽. We can see them in the first triangle (a) in Figure 12.

TRY IT #2

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

EXAMPLE 3

Solving for the Unknown Sides and Angles of a SSA Triangle

In the triangle shown in Figure 13, solve for the unknown side and angles. Round your answers to the nearest tenth.

Figure 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 γ=85°,𝛾=85°, and its corresponding side c=12,𝑐=12, and we know side b=9.𝑏=9. We will use this proportion to solve for β.𝛽.

sin(85°)12=sinβ99sin(85°)12=sinβIsolate the unknown.sin(85°)12=sin𝛽9Isolate the unknown.9sin(85°)12=sin𝛽

To find β,𝛽, apply the inverse sine function. The inverse sine will produce a single result, but keep in mind that there may be two values for β.𝛽. It is important to verify the result, as there may be two viable solutions, only one solution (the usual case), or no solutions.

β=sin−1(9sin(85°)12)β≈sin−1(0.7471)β≈48.3°𝛽=sin−1(9sin(85°)12)𝛽≈sin−1(0.7471)𝛽≈48.3°

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

α=180°−85°−131.7°≈−36.7°,𝛼=180°−85°−131.7°≈−36.7°,

which is impossible, and so β≈48.3°.𝛽≈48.3°.

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

sin(85°)12=sin(46.7°)aasin(85°)12=sin(46.7°)a=12sin(46.7°)sin(85°)≈8.8sin(85°)12=sin(46.7°)𝑎𝑎sin(85°)12=sin(46.7°)𝑎=12sin(46.7°)sin(85°)≈8.8

The complete set of solutions for the given triangle is

α≈46.7°a≈8.8β≈48.3°b=9γ=85°c=12𝛼≈46.7°𝑎≈8.8𝛽≈48.3°𝑏=9𝛾=85°𝑐=12

TRY IT #3

Given α=80°,a=100,b=10,𝛼=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.

EXAMPLE 4

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 14.

sinα10=sin(50°)4sinα=10sin(50°)4sinα≈1.915sin𝛼10=sin(50°)4sin𝛼=10sin(50°)4sin𝛼≈1.915

Figure 14

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

TRY IT #4

Determine the number of triangles possible given a=31,𝑎=31, b=26,𝑏=26, β=48°.𝛽=48°.

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=12bh,Area=12𝑏ℎ, where b𝑏 is base and hℎ is height. For oblique triangles, we must find hℎ before we can use the area formula. Observing the two triangles in Figure 15, one acute and one obtuse, we can drop a perpendicular to represent the height and then apply the trigonometric property sinα=oppositehypotenusesin𝛼=oppositehypotenuse to write an equation for area in oblique triangles. In the acute triangle, we have sinα=hcsin𝛼=ℎ𝑐 or csinα=h.𝑐sin𝛼=ℎ. However, in the obtuse triangle, we drop the perpendicular outside the triangle and extend the base b𝑏 to form a right triangle. The angle used in calculation is α′,𝛼′, or 180−α.180−𝛼.

Figure 15

Thus,

Area=12(base)(height)=12b(csinα)Area=12(base)(height)=12𝑏(𝑐sin𝛼)

Similarly,

Area=12a(bsinγ)=12a(csinβ)Area=12𝑎(𝑏sin𝛾)=12𝑎(𝑐sin𝛽)

AREA OF AN OBLIQUE TRIANGLE

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

Area=12bcsinα=12acsinβ=12absinγArea=12𝑏𝑐sin𝛼=12𝑎𝑐sin𝛽=12𝑎𝑏sin𝛾

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

EXAMPLE 5

Finding the Area of an Oblique Triangle

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

Solution

Using the formula, we have

Area=12absinγArea=12(90)(52)sin(102°)Area≈2289squareunitsArea=12𝑎𝑏sin𝛾Area=12(90)(52)sin(102°)Area≈2289squareunits

TRY IT #5

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

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 6

Finding an Altitude

Find the altitude of the aircraft in the problem introduced at the beginning of this section, shown in Figure 16. Round the altitude to the nearest tenth of a mile.

Figure 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,𝑎, and then use right triangle relationships to find the height of the aircraft, h.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.

sin(130°)20=sin(35°)aasin(130°)=20sin(35°)a=20sin(35°)sin(130°)a≈14.98sin(130°)20=sin(35°)𝑎𝑎sin(130°)=20sin(35°)𝑎=20sin(35°)sin(130°)𝑎≈14.98

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

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

sin(15°)=oppositehypotenusesin(15°)=hasin(15°)=h14.98h=14.98sin(15°)h≈3.88sin(15°)=oppositehypotenusesin(15°)=ℎ𝑎sin(15°)=ℎ14.98ℎ=14.98sin(15°)ℎ≈3.88

The aircraft is at an altitude of approximately 3.9 miles.

TRY IT #6

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

Figure 17