Chapter 3: Laws of Sines and Cosines
3.2 The Law of Sines
Algebra Refresher
Algebra Refresher
Convert to a decimal fraction.
- 5 inches [latex]=\underline \qquad[/latex]foot
- 10 ounces [latex]=\underline \qquad[/latex] pound
- 24 minutes [latex]=\underline \qquad[/latex]hour
- 35 seconds [latex]=\underline \qquad[/latex]minute
- 16 minutes [latex]=\underline \qquad[/latex] °
- [latex]4^{\prime} =\underline \qquad[/latex]°
- [latex]2^{\prime\prime} =\underline \qquad[/latex]°
- [latex]1^{\prime} ~5^{\prime\prime}=\underline \qquad[/latex]°
Algebra Refresher Answers
- [latex]0.41\overline{6}[/latex]
- [latex]0.625[/latex]
- [latex]0.4[/latex]
- [latex]0.58\overline{3}[/latex]
- [latex]0.2\overline{6}[/latex]
- [latex]0.0\overline{6}[/latex]
- [latex]0.000\overline{5}[/latex]
- [latex]0.0180\overline{5}[/latex]
Learning Objectives
- Use the Law of sines to find a side #1-6
- Use the Law of sines to find an angle #7-12
- Use the Law of sines to solve an oblique triangle #13-18
- Solve problems using the Law of sines #19-28
- Compute distances using parallax #29-32
- Solve problems involving the ambiguous case #33-46
Law of Sines
We have learned to use the trigonometric ratios to solve right triangles. But the trig ratios are only valid for the sides of right triangles. Can we find unknown sides or angles in an oblique triangle?
In this section and the next, we find relationships among the sides and angles of oblique triangles. These relationships are called the law of sines and the law of cosines. To derive these new rules, we use what we already know about right triangles.
Note 3.21.
Reducing a new problem to an earlier one is a frequently used technique in mathematics.
Consider the oblique triangle below. By drawing in the altitude [latex]h[/latex] of the triangle, we create two right triangles, [latex]\triangle BCD[/latex] and [latex]\triangle ABD{,}[/latex] as shown in the figure. Now we can write expressions in terms of [latex]h[/latex] for [latex]\sin A[/latex] and for [latex]\sin C{.}[/latex]
Looking at [latex]\triangle BCD{,}[/latex] we see that [latex]~~~\dfrac{h}{a} = \sin C.[/latex]
Looking at [latex]\triangle ABD{,}[/latex] we see that [latex]~~\dfrac{h}{c} = \sin A.[/latex]
Now we solve each of these equations for [latex]h{:}[/latex]
[latex]\dfrac{h}{a} = \sin C[/latex] and [latex]\dfrac{h}{c} = \sin A[/latex] Solve each equation for [latex]h[/latex].
[latex]h = a \sin C[/latex] and [latex]h = c \sin A[/latex] Equate the expressions for [latex]h[/latex].
[latex]a \sin C = c \sin A[/latex] Divide both sides by [latex]ac[/latex].
[latex]\dfrac {\sin C}{c} = \dfrac {\sin A}{a}[/latex]
We have derived a relationship between the angles [latex]\angle A[/latex] and [latex]\angle C[/latex] and their opposite sides, [latex]a[/latex] and [latex]c{.}[/latex] If we know any three of these quantities, we can find the fourth.
In a similar way, by drawing in the altitude from the vertex [latex]\angle C{,}[/latex] we can show that
[latex]\dfrac {\sin A}{a} = \dfrac {\sin B}{b}[/latex]
Putting both results together, we have the law of sines. The law of sines is true for any triangle, whether it is acute, right, or obtuse.
Law of Sines.
If the angles of a triangle are [latex]\angle A, \angle B{,}[/latex] and [latex]\angle C{,}[/latex] and the opposite sides are, respectively, [latex]a, b,[/latex] and [latex]c{,}[/latex] then
[latex]{\dfrac {\sin \angle A}{a} = \dfrac {\sin \angle B}{b} = \dfrac {\sin \angle C}{c}}[/latex]
or equivalently,
[latex]{\dfrac {a}{\sin \angle A} = \dfrac {b}{\sin \angle B} = \dfrac {c}{\sin \angle C}}[/latex]
Finding a Side
In the next example, we use the law of sines to find a distance.
Example 3.22.
Two observers onshore sight a ship at an unknown distance from the shore. The observers are [latex]400[/latex] yards apart at points [latex]A[/latex] and [latex]B{,}[/latex] and they each measure the angle from the shoreline to the ship, as shown below. How far is the ship from the observer at [latex]A{?}[/latex]
Solution
First note that [latex]\triangle ABC[/latex] is not a right triangle, so we cannot use the trig ratios directly to find the sides of this triangle.
The unknown distance [latex]d[/latex] is the side opposite [latex]\angle B = 79.4°{.}[/latex] In order to use the law of sines, we must know another angle and the side opposite that angle. We know that side [latex]c = 400{,}[/latex] and we can compute the angle at the ship, [latex]\angle C{.}[/latex] [latex]\angle C = 180° -(79.4° + 83.2°) = 17.4°[/latex]. Now we apply the law of sines, using angles [latex]B[/latex] and [latex]C{.}[/latex] [latex]\dfrac{b}{\sin \angle B} = \dfrac{c}{\sin \angle C}[/latex]. Substitute the given values. [latex]\dfrac{d}{sin 79.4°} = \dfrac{400}{sin 17.4°}[/latex]. Evaluate sines. [latex]\dfrac{d}{0.9829} = \dfrac{400}{0.2990}[/latex]. Multiply both sides by 0.9829. [latex]d = 1315[/latex]
The ship is about 1315 yards from the observer at [latex]A{.}[/latex]
Checkpoint 3.23.
Delbert and Francine are [latex]40[/latex] feet apart on one side of a river. They make angle measurements to a pine tree on the opposite shore as shown at right. What is the distance from Francine to the pine tree?
Solution
About [latex]114.8[/latex] feet
Solving Triangles with the Law of Sines
In order to apply the law of sines to find a side, we must know one angle of the triangle and its opposite side either [latex]a[/latex] and [latex]\angle A{,}[/latex] or [latex]b[/latex] and [latex]\angle B{,}[/latex] or [latex]c[/latex] and [latex]\angle C[/latex], and one other angle. Then we can find the side opposite that angle.
Example 3.24.
In the triangle shown at right, [latex]\angle A = 37°, \angle B = 54°{,}[/latex] and [latex]a = 11{.}[/latex]
Find [latex]b{.}[/latex]
Solve the triangle.
Solution
We use the law of sines with [latex]a[/latex] and angle [latex]\angle A[/latex] to find [latex]b{.}[/latex]
[latex]\dfrac{a}{\sin \angle A} = \dfrac{b}{\sin \angle B} \ \ \ \text{Substitute the given values.} \\ \dfrac{11}{\sin 37°} = \dfrac{b}{\sin 54°} \ \ \ \text{Multiply both sides by sin 54°.} \\ \dfrac{11}{\sin 37°} \cdot \sin 54° = b[/latex]
Evaluating this expression with a calculator, we find that [latex]b \approx 14.79{.}[/latex]
The angle [latex]\angle C = 180° -(37° + 54°) = 89°{.}[/latex] Now we use the law of sines to find side [latex]c{.}[/latex] Note that it is safer to use the given side, [latex]a{,}[/latex] rather than the value we calculated for [latex]b{.}[/latex]
[latex]\dfrac{a}{\sin \angle A} = \dfrac{c}{\sin \angle C} \ \ \text{Substitute the known values.}\\ \dfrac{11}{\sin 37°} = \dfrac{c}{\sin 89°} \ \ \text{Multiply both sides by} \sin 89°.\\ \dfrac{11}{\sin 37°}\cdot \sin 89° = c[/latex]
so [latex]c \approx 18.28.[/latex]
Checkpoint 3.25.
Solve the triangle with: [latex]\angle A = 65°, \angle C = 42°{,}[/latex] and [latex]c = 16{.}[/latex] (Hint: Which side will you find first?)
Solution
[latex]\angle B = 73°{,}[/latex] [latex]~b = 22.87{,}[/latex] [latex]~a = 21.68[/latex]
Finding an Angle
We can also use the law of sines to find an unknown angle of a triangle. We must know two sides of the triangle and the angle opposite one of them.
Example 3.26.
In the triangle shown at right, [latex]\angle B = 55°[/latex], [latex]a = 5{,}[/latex] and [latex]b = 11{.}[/latex] Solve the triangle.
Solution
We must find the three remaining parts of the triangle, [latex]\angle A[/latex], [latex]\angle C{,}[/latex] and [latex]c{.}[/latex] First, we use the law of sines to find [latex]\sin \angle A{.}[/latex]
[latex]\dfrac{\sin \angle A}{a} = \dfrac{\sin \angle B}{b} \ \text{Substitute the known values.} \\ \dfrac{\sin \angle A}{5} = \dfrac{\sin 55°}{11} \ \text{Multiply both sides by 5.} \\ \sin \angle A = 5 \cdot \dfrac{\sin 55°}{11} \approx 0.3723[/latex]
So [latex]\angle A = \sin^{-1}(0.3723) \approx 21.9°{.}[/latex] Now we know two angles, and we can find angle [latex]\angle C{.}[/latex]
[latex]\angle C = 180° - (\angle B + \angle A) = 180° - 55° - 21.9° = 103.1°[/latex]
Finally, we use the law of sines to find side [latex]c{.}[/latex]
[latex]\dfrac{b}{\sin \angle B} = \dfrac{c}{\sin \angle C} \ \text{Substitute the known values.}\\ \dfrac{11}{\sin 55°} = \dfrac{c}{\sin 103.1°} \ \text{Multiply both sides by sin 103.1°.}\\ \dfrac{11}{\sin 55°} \cdot \sin 103.1° = c[/latex]
Evaluating this expression with a calculator gives [latex]c \approx 13.1.[/latex]
Checkpoint 3.27.
Sketch a triangle with [latex]\angle C = 93°[/latex], [latex]a = 7{,}[/latex] and [latex]c = 11{.}[/latex]
Use the law of sines to find another angle of the triangle.
Solve the triangle and label your sketch with the results.
Solution
[latex]\angle A = 39.5°[/latex]
[latex]b = 8.13[/latex], [latex]\angle B = 47.5°[/latex]
Caution 3.28.
In the last example, we used the value of [latex]\sin \angle A[/latex] to find angle [latex]\angle A{.}[/latex] But there are two angles that satisfy [latex]\sin \angle A = 0.3723{,}[/latex] one acute [latex]21.9°[/latex] and one obtuse [latex]168.1°[/latex]. How did we know which one to choose?
In this case, the obtuse angle is too big to fit in the triangle, because angle [latex]\angle B = 55°{.}[/latex] Sometimes both angles will produce different triangles, and sometimes only the acute angle will work. You should always check whether both angles provide solutions.
Note 3.29.
Because there can be more than one solution for a triangle in which we know two sides and an angle opposite one of them, this situation is called the ambiguous case for the law of sines. (See Homework Problems 33–42 for more about the ambiguous case.)
Example 3.30.
Find two triangles in which [latex]\angle B = 14.4°[/latex], [latex]a = 8{,}[/latex] and [latex]b = 3{,}[/latex] and sketch both triangles.
Solution
Using the law of sines, we have
[latex]\dfrac{\sin \angle A}{a} = \dfrac{\sin \angle B}{b} \ \text{Substitute the known values.} \\ \dfrac{\sin \angle A}{8} = \dfrac{\sin 14.4°}{3} \ \text{Multiply both sides by} 8.\\ \sin \angle A = 8 \cdot \dfrac{\sin 14.4°}{3} \approx 0.6632[/latex]
There are two angles with sine [latex]0.6632{:}[/latex]
the acute angle [latex]\angle A = \sin^{-1}(0.6632) = 41.5°,[/latex]
or its supplement, [latex]\angle A^{ \prime} = 180° - 41.5° = 138.5°{.}[/latex]
Each of these angles produces a different solution triangle, because the angle [latex]\angle C[/latex] and side [latex]c[/latex] will be different also.
In the first case, angle [latex]\angle C = 180° - (14.4° + 138.5°) = 27.1°{,}[/latex] and we have the triangle shown in figure [latex]a[/latex].
In the second case, [latex]\angle C = 180° - (14.4° + 41.5°) = 124.1°{,}[/latex] which gives us the triangle shown in figure [latex]b[/latex].
Checkpoint 3.31.
Suppose that [latex]\angle C = 29.7°[/latex], [latex]b = 8{,}[/latex] and [latex]c = 5{.}[/latex]
Find two possible values for angle [latex]\angle B{.}[/latex]
Solve the triangle for both values of [latex]\angle B{,}[/latex] and sketch both solutions.
Solution
[latex]\angle B = 52.4°[/latex] or [latex]\angle B = 127.6°[/latex]
[latex]\angle A = 97.9°[/latex], [latex]a = 10[/latex]
or [latex]\angle A = 22.7°[/latex], [latex]a = 3.9[/latex]
Applications
In the next example, we use two triangles to solve the problem.
Example 3.32.
Richard wants to measure the height of a castle controlled by hostile forces. When he is as close as he can get to the castle, the angle of elevation to the top of the wall is [latex]18.5°{.}[/latex] He then retreats [latex]20[/latex] yards and measures the angle of elevation again; this time it is [latex]15.9°{.}[/latex] How tall is the castle?
Solution
Notice that [latex]h[/latex] is one side of the right triangle [latex]ADC{.}[/latex] If we can find its hypotenuse, labeled [latex]r[/latex] in the figure, we can use the sine ratio to find [latex]h{.}[/latex] To find [latex]r{,}[/latex] we consider a second triangle, [latex]\triangle ABC{,}[/latex] as shown below.
In this triangle, we know side [latex]BC = 20[/latex] and would like to find side [latex]AC = r{.}[/latex] We can use the law of sines to find [latex]r{,}[/latex] but first we must calculate the other angles of the triangle.
Now, the angle opposite [latex]r, \angle ABC{,}[/latex] is the complement of [latex]18.5°{,}[/latex] so
[latex]\angle ABC = 180° - 18.5° = 161.5°[/latex]
The angle opposite the 20-yard side, [latex]\angle BAC{,}[/latex] is
[latex]\angle BAC = 180° - (161.5° + 15.9°) = 2.6°[/latex]
Now we can apply the law of sines to find [latex]r{.}[/latex] We have [latex]\dfrac{r}{sin 161.5°} = \dfrac{20}{sin 2.6°}[/latex] Solve for [latex]r{.}[/latex]
[latex]r = \sin 161.5° \cdot \dfrac{20}{sin 2.6°} \approx 139.9[/latex]
So [latex]r[/latex] is about 139.9 yards. Finally, using the right triangle [latex]ADC[/latex] and the definition of sine, we can write
[latex]\dfrac{h}{r} = \sin 15.9°[/latex]. Solve for [latex]h[/latex].
[latex]h = r \cdot \sin 15.9° \approx 38.33[/latex]
The castle is about 38.33 yards tall.
Checkpoint 3.33.
Solve the problem in the previous example again, but instead of finding [latex]r{,}[/latex] find the length [latex]AB[/latex] and then use [latex]\triangle ABD[/latex] to find [latex]h{.}[/latex]
Solution
[latex]AB = 120.79[/latex]; the castle is about [latex]38.33[/latex] yards tall.
Measuring Astronomical Distances
If you look at a nearby object and alternately close your left and right eyes, the object seems to jump in position. This apparent change occurs because your eyes are viewing the object from two different positions spaced several centimeters apart. If the object at point [latex]O[/latex] is straight ahead of one eye, it appears to be at some angle [latex]p[/latex] away from the line of sight of the other eye. The angle [latex]p[/latex] is called the parallax of the object.
Use the figure below to see that [latex]p[/latex] is also the angle between the directions to your two eyes when viewed from point [latex]O{.}[/latex] (What fact from geometry justifies this statement?)
Astronomers use parallax to determine the distance from Earth to stars and other celestial objects. Two observers on Earth at a known distance apart both measure the direction to the star. The difference in angle between those two directions is the parallax.
Example 3.34.
Astronomers [latex]1000[/latex] kilometers apart observe an asteroid with a parallax of [latex]0.001°{.}[/latex] How far is the asteroid from Earth?
Solution
We let [latex]x[/latex] represent the distance to the asteroid. The asteroid and the two observers make an isosceles triangle with base 1000 km and equal sides of length [latex]x{,}[/latex] as shown below.
The base angles of the triangle are both [latex]\dfrac{180° - 0.001°}{2} = 89.999°{.}[/latex] By the law of sines,
[latex]\dfrac{x}{sin 89.999°} = \dfrac{1000}{sin 0.001°}[/latex] Solve for x.
[latex]x = \sin 89.999° \cdot \dfrac{1000}{sin 0.001°}[/latex]
[latex]x \approx 57,000,000[/latex]
The asteroid is about 57 million kilometers from Earth (roughly one-third of the distance to the Sun).
Checkpoint 3.35.
Two observers 800 kilometers apart observe an object with a parallax of [latex]0.0005°{.}[/latex] How far is the object from Earth?
Solution
About [latex]91,673,247[/latex] km
Small Angles: Minutes and Seconds
To obtain the most accurate parallax measurements, the distance between the two observers should be as large as possible. But even measured from opposite sides of Earth’s orbit, stars outside the solar system have parallaxes much smaller than [latex]0.0001°{.}[/latex]
In order to handle such small angles, we divide °s into smaller units called minutes and seconds. One minute is [latex]\dfrac{1}{60}[/latex] of a °, and 1 second is [latex]\dfrac{1}{60}[/latex] of a minute, or [latex]\dfrac{1}{3600}[/latex] of a °. We use the following notation for minutes and seconds.
Fractions of a °.
One minute: [latex]1 ^{ \prime} = \dfrac{1°}{60}[/latex]
One second: [latex]1 ^{ \prime \prime} = \dfrac{1^{\prime}}{60} = \dfrac{1°}{3600}[/latex]
When describing large distances, astronomers sometimes use the distance from Earth to the Sun, about 93 million miles, as the unit of measurement. This distance is called 1 astronomical unit, or 1 AU. For example, an object that is three times as far away as the Sun would be at a distance of 3 AU.
Example 3.36.
The star Wolf 359 has a parallax of [latex]0.85^{\prime\prime}[/latex] when observed from opposite sides of Earth’s orbit. How far away is the star?
Solution
In the figure below, the star Wolf 359 is located at point [latex]O{.}[/latex] Our Sun is located at point [latex]C{,}[/latex] halfway between the two observation points at [latex]A[/latex] and [latex]B{.}[/latex] Thus, the distance between the observation points is twice the distance from the Earth to the Sun, or 2 AU.
The altitude from [latex]O[/latex] to side [latex]\overline{AB}[/latex] forms a right triangle [latex]\triangle ACO[/latex] and bisects the angle at [latex]O{.}[/latex] Thus
[latex]\angle AOC = \dfrac{1}{2}(0.85^{\prime} = 0.425^{\prime}[/latex]
We use the definition of tangent to find
[latex]\tan 0.425^{\prime \prime} = \dfrac{opposite}{adjacent}[/latex]
[latex]\tan (0.425° / 3600) = \dfrac{1}{x}[/latex]. Solve for x.
[latex]x = \dfrac{1}{\tan 0.425° / 3600} \approx 485,000[/latex]
The star is approximately [latex]485,000[/latex] AU from Earth, or nearly half a million times the distance from Earth to the Sun.
Checkpoint 3.37.
Two observers are 1 AU (astronomical unit) apart. They find that the parallax to a distant star is [latex]1^{\prime\prime}{.}[/latex] What is the distance to the star, in astronomical units?
(This distance is called a parsec. In other words, a parsec is the distance at which the parallax from observations 1 AU apart is [latex]1^{\prime\prime}{.}[/latex] In this exercise, you are calculating the number of astronomical units in 1 parsec.)
Solution
1 parsec [latex]\approx 206,265[/latex] AU
Section 3.2 Summary
Vocabulary
Concepts
- Law of sines.
If the angles of a triangle are [latex]\angle A[/latex], [latex]\angle B{,}[/latex] and [latex]\angle C{,}[/latex] and the opposite sides are, respectively, [latex]a[/latex], [latex]b,[/latex] and [latex]c{,}[/latex] then [latex]\dfrac {\sin A}{a} = \dfrac {\sin B}{b} = \dfrac {\sin C}{c}[/latex]
or equivalently, [latex]\dfrac {a}{\sin A} = \dfrac {b}{\sin B} = \dfrac {c}{\sin C}[/latex]. - We can use the law of sines to find an unknown side in an oblique triangle. We must know the angle opposite the unknown side and another side-angle pair.
- We can also use the law of sines to find an unknown angle of a triangle. We must know two sides of the triangle and the angle opposite one of them.
- Remember that there are two angles with a given sine. When using the law of sines, we must check whether both angles result in possible triangles.
- We use minutes and seconds to measure very small angles. Fractions of a °.
One minute: [latex]1^{\prime} = \dfrac{1°}{60}[/latex]
One second: [latex]1^{\prime\prime} = \dfrac{1^{\prime}}{60} = \dfrac{1°}{3600}[/latex]
Study Questions
- Can we use the law of sines to solve a right triangle?
- Explain why we cannot use the law of sines to solve the triangle with [latex]a = 8,~b = c[/latex] and [latex]\angle C = 35°{.}[/latex]
- Francine says, “I’m thinking of an angle whose sine is [latex]0.3420[/latex] (rounded to four decimal places).” Delbert says, “The angle must be [latex]20°[/latex] (rounded to the nearest °).” Is he correct? Why or why not?
- Sketch two possible triangles with [latex]\angle A=25°[/latex], [latex]b = 18{,}[/latex] and [latex]a = 10{.}[/latex]
- Try to sketch a triangle with [latex]\angle A=65°[/latex], [latex]b = 18{,}[/latex] and [latex]a = 10{.}[/latex] What went wrong?
Parallax is the apparent shift or difference in the position of an object when viewed from different positions. In other words, it is the displacement or difference in the apparent position of an object seen along two different lines of sight.
The minute is a unit of time measurement that is equal to 60 seconds, or 1/60th of an hour
The second is a unit of time measurement that is equal to 1/60th of a minute or 1/3,600th of an hour.