What Is the Value of Z, Rounded to the Nearest Tenth? Use the Law of Sines to Find the Answer.

Learning Objectives

In this section, you will:

• Use the Police force of Sines to solve oblique triangles.
• Find the area of an oblique triangle using the sine function.
• Solve practical problems using the Constabulary of Sines.

Suppose 2 radar stations located 20 miles autonomously each detect an aircraft between them. The angle of peak measured by the first station is 35 degrees, whereas the bending of summit measured by the second station is 15 degrees. How tin can we determine the altitude of the aircraft? Nosotros run across in Figure 1 that the triangle formed by the aircraft and the 2 stations is not a right triangle, so we cannot use what nosotros know about right triangles. In this section, we will find out how to solve problems involving non-right triangles.

Figure ane

Using the Law of Sines to Solve Oblique Triangles

In whatever triangle, we can describe an altitude, a perpendicular line from one vertex to the contrary side, forming ii correct triangles. Information technology would exist preferable, still, to accept methods that we can utilise directly to not-right triangles without get-go 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 3 sides. To do so, nosotros need to offset with at least three of these values, including at least one of the sides. We will investigate 3 possible oblique triangle trouble situations:

1. ASA (angle-side-bending) We know the measurements of two angles and the included side. Encounter Figure ii.

   

   Figure two

2. AAS (angle-angle-side) We know the measurements of ii angles and a side that is not between the known angles. See Figure 3.

   

   Figure iii

3. SSA (side-side-angle) We know the measurements of ii sides and an angle that is non between the known sides. Come across Effigy iv.

   

   Figure iv

Knowing how to approach each of these situations enables the states to solve oblique triangles without having to drop a perpendicular to course 2 right triangles. Instead, we can use the fact that the ratio of the measurement of ane of the angles to the length of its contrary side will be equal to the other two ratios of angle measure out to opposite side. Let's run into how this statement is derived by considering the triangle shown in Effigy 5.

Figure 5

Using the right triangle relationships, nosotros know that $\sin \alpha =\frac{h}{b}$ and $\sin \beta =\frac{h}{a}.$ Solving both equations for $h$ gives 2 different expressions for $h.$

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

We then set the expressions equal to each other.

$$\begin{array}{ll}b\sin \alpha =a\sin \beta \hfill & \hfill \\ \left(\frac{\mathrm{i}}{ab}\right)(b\sin \alpha )=(a\sin \beta )\left(\frac{\mathrm{one}}{ab}\right)\begin{array}{cccc}& & & \end{array}\hfill & \text{Multiply both sides past}\frac{\mathrm{i}}{ab}.\hfill \\ \frac{\sin \alpha }{a}=\frac{\sin \beta }{b}\hfill & \hfill \end{array}$$

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 chosen the Police force of Sines.

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

Note the standard way of labeling triangles: angle $\alpha$ (alpha) is reverse side $a;$ angle $\beta$ (beta) is opposite side $b;$ and angle $\gamma$ (gamma) is opposite side $c.$ Meet Figure 6.

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

Figure 6

Law of Sines

Given a triangle with angles and reverse sides labeled as in Figure half dozen, the ratio of the measurement of an bending to the length of its reverse side volition be equal to the other 2 ratios of angle measure to opposite side. All proportions volition exist equal. The Constabulary of Sines is based on proportions and is presented symbolically ii ways.

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

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

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

Case 1

Solving for Ii Unknown Sides and Angle of an AAS Triangle

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

Figure 7

Endeavor Information technology #1

Solve the triangle shown in Effigy eight to the nearest tenth.

Figure 8

Using The Constabulary of Sines to Solve SSA Triangles

We can utilize the Law of Sines to solve whatever oblique triangle, but some solutions may not exist straightforward. In some cases, more than one triangle may satisfy the given criteria, which we describe every bit an ambiguous instance. Triangles classified as SSA, those in which we know the lengths of two sides and the measurement of the angle opposite 1 of the given sides, may result in 1 or two solutions, or even no solution.

Possible Outcomes for SSA Triangles

Oblique triangles in the category SSA may take four dissimilar outcomes. Figure nine illustrates the solutions with the known sides $a$ and $b$ and known angle $\alpha .$

Figure 9

Case ii

Solving an Oblique SSA Triangle

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

Effigy 10

Endeavour It #2

Given $\alpha =\mathrm{lxxx°},a=120,$ and $b=121,$ find the missing side and angles. If there is more than i 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. Circular your answers to the nearest 10th.

Figure 13

Try It #3

Given $\alpha =\mathrm{fourscore°},a=100,b=10,$ find the missing side and angles. If at that place is more than one possible solution, prove both. Round your answers to the nearest 10th.

Example 4

Finding the Triangles That Encounter the Given Criteria

Discover all possible triangles if ane side has length 4 opposite an angle of l°, and a 2d side has length ten.

Endeavor It #4

Determine the number of triangles possible given $a=31,$ $b=26,$ $\beta =\mathrm{48°}.$

Finding the Area of an Oblique Triangle Using the Sine Part

Now that we can solve a triangle for missing values, nosotros can use some of those values and the sine function to find the area of an oblique triangle. Call back that the area formula for a triangle is given as $\text{Area}=\frac{1}{2}bh,$ where $b$ is base and $h$ is top. For oblique triangles, we must find $h$ before we can utilize the area formula. Observing the ii triangles in Figure 15, one acute and one birdbrained, we can drop a perpendicular to represent the top and then employ the trigonometric holding $\sin \alpha =\frac{\text{contrary}}{\text{hypotenuse}}$ to write an equation for area in oblique triangles. In the astute triangle, we take $\sin \alpha =\frac{h}{c}$ 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 correct triangle. The angle used in calculation is ${\alpha }^{\prime },$ or $180-\alpha .$

Figure 15

Thus,

$$\text{Area}=\frac{\mathrm{ane}}{2}\left(\text{base}\right)\left(\text{height}\right)=\frac{1}{\mathrm{two}}b\left(c\sin \alpha \right)$$

Similarly,

$$\text{Expanse}=\frac{1}{\mathrm{two}}a\left(b\sin \gamma \right)=\frac{\mathrm{one}}{\mathrm{two}}a\left(c\sin \beta \right)$$

Area of an Oblique Triangle

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

$$\begin{array}{l}\text{Surface area}=\frac{1}{\mathrm{two}}bc\sin \alpha \hfill \\ =\frac{1}{2}ac\sin \beta \hfill \\ =\frac{1}{\mathrm{two}}ab\sin \gamma \hfill \end{array}$$

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

Example 5

Finding the Area of an Oblique Triangle

Find the surface area of a triangle with sides $a=\mathrm{ninety},b=52,$ and angle $\gamma =\mathrm{102°}.$ Round the area to the nearest integer.

Try It #5

Find the expanse of the triangle given $\beta =\mathrm{42°},$ $a=\mathrm{7.ii}\text{ft},$ $c=3.4\text{ft}.$ Round the surface area to the nearest tenth.

Solving Applied Problems Using the Law of Sines

The more than we study trigonometric applications, the more nosotros discover that the applications are countless. Some are flat, diagram-type situations, but many applications in calculus, applied science, and physics involve 3 dimensions and motility.

Example six

Finding an Altitude

Find the altitude of the aircraft in the problem introduced at the offset of this section, shown in Figure 16. Round the altitude to the nearest 10th of a mile.

Figure 16

Attempt Information technology #half dozen

The diagram shown in Effigy 17 represents the height of a blimp flying over a football stadium. Find the height of the blimp if the angle of meridian at the southern terminate zone, bespeak A, is 70°, the angle of summit from the northern end zone, point $B,$ is 62°, and the altitude between the viewing points of the two end zones is 145 yards.

Figure 17

8.1 Section Exercises

Verbal

i .

Draw the altitude of a triangle.

2 .

Compare right triangles and oblique triangles.

3 .

When can y'all use the Constabulary of Sines to find a missing angle?

4 .

In the Police of Sines, what is the human relationship between the angle in the numerator and the side in the denominator?

5 .

What type of triangle results in an ambiguous case?

Algebraic

For the post-obit exercises, assume $\alpha$ is opposite side $a,\beta$ is contrary side $b,$ and $\gamma$ is opposite side $c.$ Solve each triangle, if possible. Round each answer to the nearest tenth.

6 .

$\alpha =\mathrm{43°},\gamma =\mathrm{69°},a=\mathrm{xx}$

7 .

$\alpha =\mathrm{35°},\gamma =\mathrm{73°},c=\mathrm{twenty}$

8 .

$\alpha =\mathrm{60°},$ $\beta =\mathrm{60°},$ $\gamma =\mathrm{sixty°}$

9 .

$a=4,$ $\alpha =\mathrm{60°},$ $\beta =\mathrm{100°}$

10 .

$b=\mathrm{ten},$ $\beta =\mathrm{95°},\gamma =\mathrm{30°}$

For the following exercises, use the Law of Sines to solve for the missing side for each oblique triangle. Circular each answer to the nearest hundredth. Assume that angle $A$ is opposite side $a,$ bending $B$ is opposite side $b,$ and angle $C$ is reverse side $c.$

xi .

Find side $b$ when $A=\mathrm{37°},$ $B=\mathrm{49°},$ $c=5.$

12 .

Find side $a$ when $A=\mathrm{132°},C=\mathrm{23°},b=10.$

xiii .

Find side $c$ when $B=\mathrm{37°},C=\mathrm{21°},$ $b=23.$

For the post-obit exercises, presume $\alpha$ is contrary side $a,\beta$ is opposite side $b,$ and $\gamma$ is contrary side $c.$ Determine whether in that location is no triangle, i triangle, or two triangles. Then solve each triangle, if possible. Round each reply to the nearest 10th.

fourteen .

$\alpha =\mathrm{119°},a=14,b=26$

fifteen .

$\gamma =\mathrm{113°},b=\mathrm{ten},c=32$

16 .

$b=3.5,$ $c=5.3,$ $\gamma =\mathrm{80°}$

17 .

$a=12,$ $c=17,$ $\alpha =\mathrm{35°}$

eighteen .

$a=20.5,$ $b=35.0,$ $\beta =\mathrm{25°}$

19 .

$a=\mathrm{vii},$ $c=9,$ $\alpha =\mathrm{43°}$

20 .

$a=\mathrm{seven},b=\mathrm{iii},\beta =\mathrm{24°}$

21 .

$b=\mathrm{thirteen},c=\mathrm{five},\gamma =\mathrm{10°}$

22 .

$a=2.3,c=\mathrm{one.eight},\gamma =\mathrm{28°}$

23 .

$\beta =\mathrm{119°},b=8.2,a=11.3$

For the following exercises, use the Constabulary of Sines to solve, if possible, the missing side or angle for each triangle or triangles in the cryptic example. Round each answer to the nearest tenth.

24 .

Detect angle $A$ when $a=24,b=5,B=\mathrm{22°.}$

25 .

Find angle $A$ when $a=\mathrm{thirteen},b=\mathrm{vi},B=\mathrm{20°.}$

26 .

Find angle $B$ when $A=\mathrm{12°},a=2,b=\mathrm{ix.}$

For the following exercises, find the area of the triangle with the given measurements. Round each answer to the nearest 10th.

27 .

$a=5,c=\mathrm{six},\beta =\mathrm{35°}$

28 .

$b=\mathrm{xi},c=8,\alpha =\mathrm{28°}$

29 .

$a=32,b=24,\gamma =\mathrm{75°}$

thirty .

$a=\mathrm{vii.2},b=\mathrm{four.5},\gamma =\mathrm{43°}$

Graphical

For the following exercises, observe the length of side $x.$ Circular to the nearest tenth.

32 .

34 .

36 .

For the following exercises, find the mensurate of bending $\mathrm{10},$ if possible. Circular to the nearest tenth.

38 .

forty .

41 .

Detect that $\mathrm{ten}$ is an obtuse bending.

42 .

For the following exercise, solve the triangle. Round each answer to the nearest tenth.

44 .

For the post-obit exercises, detect the expanse of each triangle. Round each answer to the nearest tenth.

46 .

48 .

Extensions

50 .

Discover the radius of the circle in Figure 18. Circular to the nearest 10th.

Figure 18

51 .

Find the bore of the circle in Figure nineteen. Circular to the nearest tenth.

Figure nineteen

52 .

Find $\mathrm{yard}\angle ADC$ in Figure 20. Round to the nearest tenth.

Figure xx

53 .

Find $AD$ in Figure 21. Round to the nearest tenth.

Effigy 21

54 .

Solve both triangles in Effigy 22. Round each answer to the nearest tenth.

Figure 22

55 .

Detect $AB$ in the parallelogram shown in Figure 23.

Figure 23

56 .

Solve the triangle in Figure 24. (Hint: Depict a perpendicular from $H$ to $JK).$ Round each respond to the nearest 10th.

Figure 24

57 .

Solve the triangle in Figure 25. (Hint: Depict a perpendicular from $N$ to $LK).$ Round each answer to the nearest tenth.

Figure 25

58 .

In Figure 26, $ABCD$ is not a parallelogram. $\angle m$ is obtuse. Solve both triangles. Circular each answer to the nearest tenth.

Figure 26

Real-Globe Applications

59 .

A pole leans abroad from the sun at an bending of $\mathrm{7°}$ to the vertical, equally shown in Effigy 27. When the elevation of the sun is $\mathrm{55°},$ the pole casts a shadow 42 feet long on the level ground. How long is the pole? Round the answer to the nearest tenth.

Figure 27

60 .

To determine how far a boat is from shore, two radar stations 500 feet apart find the angles out to the boat, as shown in Effigy 28. Determine the distance of the boat from station $A$ and the distance of the boat from shore. Round your answers to the nearest whole foot.

Effigy 28

61 .

Figure 29 shows a satellite orbiting Earth. The satellite passes straight over ii tracking stations $A$ and $B,$ which are 69 miles apart. When the satellite is on i side of the 2 stations, the angles of elevation at $A$ and $B$ are measured to be $\mathrm{83.9°}$ and $\mathrm{86.2°},$ respectively. How far is the satellite from station $A$ and how high is the satellite higher up the basis? Round answers to the nearest whole mile.

Effigy 29

62 .

A communications tower is located at the top of a steep hill, as shown in Figure thirty. The angle of inclination of the hill is $\mathrm{67°}.$ A guy wire is to be fastened to the top of the tower and to the footing, 165 meters downhill from the base of the tower. The angle formed by the guy wire and the hill is $\mathrm{xvi°.}$ Discover the length of the cable required for the guy wire to the nearest whole meter.

Figure 30

63 .

The roof of a house is at a $\mathrm{twenty°}$ angle. An 8-foot solar panel is to be mounted on the roof and should exist angled $\mathrm{38°}$ relative to the horizontal for optimal results. (See Figure 31). How long does the vertical support belongings up the back of the console demand to be? Round to the nearest 10th.

Figure 31

64 .

Like to an bending of elevation, an angle of depression is the acute angle formed past a horizontal line and an observer'due south line of sight to an object below the horizontal. A pilot is flying over a direct highway. He determines the angles of depression to two mileposts, 6.6 km autonomously, to exist $\mathrm{37°}$ and $\mathrm{44°},$ as shown in Effigy 32. Find the distance of the airplane from bespeak $A$ to the nearest 10th of a kilometer.

Effigy 32

65 .

A pilot is flying over a straight highway. He determines the angles of depression to two mileposts, 4.iii km apart, to be 32° and 56°, equally shown in Figure 33. Detect the distance of the airplane from point $A$ to the nearest tenth of a kilometer.

Figure 33

66 .

In guild to estimate the peak of a building, two students stand at a sure distance from the building at street level. From this point, they find the angle of elevation from the street to the top of the building to exist 39°. They and then move 300 anxiety closer to the building and discover the bending of elevation to be fifty°. Assuming that the street is level, estimate the peak of the building to the nearest foot.

67 .

In order to estimate the height of a building, two students stand at a sure distance from the building at street level. From this point, they discover the bending of elevation from the street to the elevation of the building to exist 35°. They then move 250 feet closer to the building and discover the angle of elevation to exist 53°. Bold that the street is level, judge the height of the building to the nearest foot.

68 .

Points $A$ and $B$ are on opposite sides of a lake. Signal $C$ is 97 meters from $A.$ The measure of angle $BAC$ is adamant to be 101°, and the mensurate of angle $ACB$ is determined to be 53°. What is the distance from $A$ to $B,$ rounded to the nearest whole meter?

69 .

A homo and a woman standing $3\frac{\mathrm{one}}{\mathrm{ii}}$ miles apart spot a hot air balloon at the same time. If the angle of elevation from the man to the balloon is 27°, and the angle of elevation from the woman to the airship is 41°, find the altitude of the balloon to the nearest foot.

70 .

Two search teams spot a stranded climber on a mountain. The first search team is 0.v miles from the 2d search team, and both teams are at an altitude of ane mile. The angle of elevation from the first search team to the stranded climber is xv°. The angle of superlative from the second search team to the climber is 22°. What is the distance of the climber? Round to the nearest tenth of a mile.

71 .

A street light is mounted on a pole. A 6-foot-tall man is continuing on the street a short distance from the pole, casting a shadow. The angle of elevation from the tip of the man's shadow to the top of his head of 28°. A 6-pes-tall woman is standing on the same street on the reverse side of the pole from the man. The angle of elevation from the tip of her shadow to the top of her head is 28°. If the homo and adult female are 20 feet apart, how far is the street light from the tip of the shadow of each person? Circular the distance to the nearest tenth of a foot.

72 .

Three cities, $A,B,$ and $C,$ are located and then that city $A$ is due east of city $B.$ If metropolis $C$ is located 35° west of north from city $B$ and is 100 miles from city $A$ and 70 miles from metropolis $B,$ how far is city $A$ from city $B?$ Round the distance to the nearest 10th of a mile.

73 .

Two streets see at an 80° bending. At the corner, a park is being built in the shape of a triangle. Find the area of the park if, along i road, the park measures 180 anxiety, and forth the other road, the park measures 215 anxiety.

74 .

Brian's house is on a corner lot. Observe the area of the front end m if the edges measure out 40 and 56 feet, as shown in Effigy 34.

Figure 34

75 .

The Bermuda triangle is a region of the Atlantic Ocean that connects Bermuda, Florida, and Puerto Rico. Notice the area of the Bermuda triangle if the distance from Florida to Bermuda is 1030 miles, the distance from Puerto Rico to Bermuda is 980 miles, and the bending created past the two distances is 62°.

76 .

A yield sign measures 30 inches on all iii sides. What is the expanse of the sign?

77 .

Naomi bought a modernistic dining table whose acme is in the shape of a triangle. Observe the surface area of the table meridian if two of the sides measure 4 feet and 4.v feet, and the smaller angles measure 32° and 42°, as shown in Effigy 35.

Figure 35

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Source: https://openstax.org/books/precalculus/pages/8-1-non-right-triangles-law-of-sines

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