{{ item.displayTitle }}

No history yet!

equalizer

rate_review

{{ r.avatar.letter }}

{{ u.avatar.letter }}

+

{{ item.displayTitle }}

{{ item.subject.displayTitle }}

{{ searchError }}

{{ courseTrack.displayTitle }} {{ statistics.percent }}% Sign in to view progress

{{ printedBook.courseTrack.name }} {{ printedBook.name }} {{ greeting }} {{userName}}

{{ 'ml-article-collection-banner-text' | message }}

{{ 'ml-article-collection-your-statistics' | message }}

{{ 'ml-article-collection-overall-progress' | message }}

{{ 'ml-article-collection-challenge' | message }} Coming Soon!

{{ 'ml-article-collection-randomly-challenge' | message }}

{{ 'ml-article-collection-community' | message }}

{{ 'ml-article-collection-follow-latest-updates' | message }}

In previous lessons, it has been seen how trigonometric ratios can be used to solve right triangles. It has also been seen how the Law of Sines can be used to solve non-right triangles. However, there are cases in which this law is not useful. This lesson will explore the relationship between the side lengths and an angle measure of different triangles.
### Catch-Up and Review

**Here is a bundle of recommended readings before getting started with this lesson.**

Try your knowledge on these topics.

Consider the triangle with vertices A, B, and C.

a Find the measure of the missing angle.

{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":["x"],"constants":["PI"]}},"text":"<span class=\"mlmath-simple\"><\/span>"},"formTextBefore":"<span class=\"mlmath-simple\"><span class=\"variable\">m<\/span>\u2220<span class=\"variable\">BCA<\/span><span class=\"space big-before big-after\">=<\/span><\/span>","formTextAfter":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.674115em;vertical-align:0em;\"><\/span><span class=\"mord\"><span><\/span><span class=\"msupsub\"><span class=\"vlist-t\"><span class=\"vlist-r\"><span class=\"vlist\" style=\"height:0.674115em;\"><span style=\"top:-3.063em;margin-right:0.05em;\"><span class=\"pstrut\" style=\"height:2.7em;\"><\/span><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mbin mtight\">\u2218<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span>","answer":{"text":["115"]}}

b Find the length of the longest side. If necessary, round the answer to the nearest integer.

{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":["x"],"constants":["PI"]}},"text":"<span class=\"mlmath-simple\"><\/span>"},"formTextBefore":null,"formTextAfter":null,"answer":{"text":["21"]}}

c If the altitude of the triangle is 6, find AD and DB. If necessary, round the answer to the nearest integer.

{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":["x"],"constants":["PI"]}},"text":"<span class=\"mlmath-simple\"><\/span>"},"formTextBefore":"<span class=\"mlmath-simple\"><span class=\"variable\">AD<\/span><span class=\"space big-before big-after\">=<\/span><\/span>","formTextAfter":null,"answer":{"text":["7"]}}

{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":["x"],"constants":["PI"]}},"text":"<span class=\"mlmath-simple\"><\/span>"},"formTextBefore":"<span class=\"mlmath-simple\"><span class=\"variable\">DB<\/span><span class=\"space big-before big-after\">=<\/span><\/span>","formTextAfter":null,"answer":{"text":["14"]}}

On a June night, Zosia observes an astronomical asterism, which is called the Summer Triangle.

From an observer's point of view, the apparent distances of celestial objects in the sky are measured in angular distance. The apparent distance from Altair to Deneb is 38 angular units and 34 angular units from Altair to Vega. Furthermore, the angle measure between these sides is $38_{∘}.$ Help Zosia find the apparent distance between Deneb and Vega.

The applet below shows two different cases for a triangle with vertices A, B, and C. Case I shows the lengths of two sides and the measure of their included angle. Case II shows the lengths of all three sides.

What does the Law of Sines state? Is it possible to find the missing side lengths and angle measures by using the Law of Sines? Explain.

When the Law of Sines cannot be used to solve triangles, the Law of Cosines may be applied.

Consider △ABC with sides of length a, b, and c, which are respectively opposite the angles with measures A, B, and C.

The following equations hold true with regard to △ABC.

The first equation will be proven. The other two equations can be proven by following the same procedure.
Begin by drawing the altitude from B to its opposite side AC.
By the definition of an altitude, both △ADB and △CDB are right triangles. By applying the Pythagorean Theorem to △CDB and △ADB, two equations can be obtained.
Now, the x-term in this equation can be written using the cosine ratio.
In △ADB, the cosine of A is the ratio of x to c.
Finally, $ccosA$ can be substituted for x into a2=b2+c2−2bx. By doing so, the formula for the Law of Cosines is obtained.

$Equation I:Equation II: a_{2}=(b−x)_{2}+h_{2}c_{2}=x_{2}+h_{2} $

The binomial in Equation I can be expanded.
Notice that Equation II says that x2+h2 is equal to c2. Therefore, the expanded form of Equation I can be rewritten by using the Substitution Property of Equality.
a2=b2−2bx+x2+h2

Substitute

x2+h2=c2

a2=b2−2bx+c2

CommutativePropAdd

Commutative Property of Addition

a2=b2+c2−2bx

The first equation will be proven for obtuse angles. The remaining equations can be proven similarly.
Consider △ABC with side lengths of a, b, and c, respectively opposite the angles with measures A, B, and C, such that m∠A is greater than $90_{∘}.$
From Equation II, h2+x2 is equal to c2. Therefore, the expanded form of Equation I can be rewritten by using the Substitution Property of Equality.
Note that A and $180_{∘}−A$ are supplementary angles. Using the cosine ratio of $180_{∘}−A$ then gives an expression for the x-term.
In △BDA, the cosine of $180_{∘}−A$ is the ratio of x to c.
By the Sine and Cosine of Supplementary Values Angles, $cos(180_{∘}−A)$ and $cosA$ have opposite values.
Finally, by substituting $-ccosA$ for x into a2=b2+c2+2bx, the Law of Cosines is obtained.

The altitude of the triangle is the perpendicular segment from B to the extension of the base AC. Let D be the endpoint of this segment and x be the distance from D to A.

From the definition of an altitude, it follows that △BDA and △BDC are right triangles. Two equations can be obtained by applying the Pythagorean Theorem to both triangles.$Equation I:Equation II: a_{2}=h_{2}+(b+x)_{2}c_{2}=h_{2}+x_{2} $

Expand the binomial in Equation I.
a2=h2+(b+x)2

ExpandPosPerfectSquare

(a+b)2=a2+2ab+b2

a2=h2+b2+2bx+x2

CommutativePropAdd

Commutative Property of Addition

a2=h2+x2+b2+2bx

a2=b2+c2+2bx

Substitute

$x=-ccosA$

$a_{2}=b_{2}+c_{2}+2b(-ccosA)$

MultPosNeg

a(-b)=-a⋅b

$a_{2}=b_{2}+c_{2}−2bccosA$

In a triangle, when the lengths of two sides and the measure of their included angle are known, the missing side length can be found by applying the Law of Cosines.

Kriz wants to determine the distance between two trees on the other side of the river. Kriz uses a tool that measures the distances to objects. The tool is able to find the distances to each tree as 4 meters and 3.5 meters.

The angle between these sides, from where the Kriz stands with the measuring tool, measures $67_{∘}.$ Find the distance between the trees, and round the answer to the nearest tenth of a meter.

{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":["x"],"constants":["PI"]}},"text":"<span class=\"mlmath-simple\"><\/span>"},"formTextBefore":null,"formTextAfter":"m","answer":{"text":["4.2"]}}

Start by naming the vertices and sides of the triangle.

For simplicity, the vertices and sides of the triangle will be named.

The Law of Cosines relates the lengths of the sides and the cosine of one of the angles. Therefore, the missing side length of the triangle can be found by using this law.$a_{2}=b_{2}+c_{2}−2bccosA$

SubstituteValues

Substitute values

$a_{2}=4_{2}+3.5_{2}−2(4)(3.5)cos67_{∘}$

$a≈4.2$

When all the three side lengths of a triangle are known, the Law of Cosines can be used to find the measure of the angles.

Ramsha lives near a lighthouse. As she likes to observe the landscape, she notices that the light rays coming out of the lighthouse create an angle. She decides to ask the lighthouse keeper, but he insists on not telling her the measure of the angle. She sees a blueprint on the desk behind him, and quickly writes the lengths shown in the diagram before the grumpy keeper blocks her view!
Help Ramsha calculate the measure of each angle of the triangle formed by the light rays. Round the answers to the nearest degree.

{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":["x"],"constants":["PI"]}},"text":"<span class=\"mlmath-simple\"><\/span>"},"formTextBefore":"<span class=\"mlmath-simple\"><span class=\"variable\">m<\/span>\u2220<span class=\"variable\">L<\/span><\/span>","formTextAfter":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.674115em;vertical-align:0em;\"><\/span><span class=\"mord\"><span><\/span><span class=\"msupsub\"><span class=\"vlist-t\"><span class=\"vlist-r\"><span class=\"vlist\" style=\"height:0.674115em;\"><span style=\"top:-3.063em;margin-right:0.05em;\"><span class=\"pstrut\" style=\"height:2.7em;\"><\/span><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mbin mtight\">\u2218<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span>","answer":{"text":["12"]}}

{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":["x"],"constants":["PI"]}},"text":"<span class=\"mlmath-simple\"><\/span>"},"formTextBefore":"<span class=\"mlmath-simple\"><span class=\"variable\">m<\/span>\u2220<span class=\"variable\">M<\/span><\/span>","formTextAfter":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.674115em;vertical-align:0em;\"><\/span><span class=\"mord\"><span><\/span><span class=\"msupsub\"><span class=\"vlist-t\"><span class=\"vlist-r\"><span class=\"vlist\" style=\"height:0.674115em;\"><span style=\"top:-3.063em;margin-right:0.05em;\"><span class=\"pstrut\" style=\"height:2.7em;\"><\/span><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mbin mtight\">\u2218<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span>","answer":{"text":["39"]}}

{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":["x"],"constants":["PI"]}},"text":"<span class=\"mlmath-simple\"><\/span>"},"formTextBefore":"<span class=\"mlmath-simple\"><span class=\"variable\">m<\/span>\u2220<span class=\"variable\">K<\/span><\/span>","formTextAfter":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.674115em;vertical-align:0em;\"><\/span><span class=\"mord\"><span><\/span><span class=\"msupsub\"><span class=\"vlist-t\"><span class=\"vlist-r\"><span class=\"vlist\" style=\"height:0.674115em;\"><span style=\"top:-3.063em;margin-right:0.05em;\"><span class=\"pstrut\" style=\"height:2.7em;\"><\/span><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mbin mtight\">\u2218<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span>","answer":{"text":["129"]}}

To find m∠L, use the Law of Cosines. Then, the measures of other two angles can be found by using either the Law of Cosines or the Law of Sines.

As the diagram indicates, the light rays form a triangle. In this triangle, the three side lengths are known.

The measure of the three angles will be found one at a time.

$ℓ_{2}=k_{2}+m_{2}−2kmcosL$

SubstituteValues

Substitute values

$35_{2}=105_{2}+130_{2}−2(105)(130)cosL$

Solve for $cosL$

CalcPow

Calculate power

$1225=11025+16900−2(105)(130)cosL$

Multiply

Multiply

$1225=11025+16900−27300cosL$

AddTerms

Add terms

$1225=27925−27300cosL$

SubEqn

LHS−27925=RHS−27925

$-26700=-27300cosL$

DivEqn

$LHS/(-27300)=RHS/(-27300)$

$-27300-26700 =cosL$

DivNegNeg

$-b-a =ba $

$2730026700 =cosL$

RearrangeEqn

Rearrange equation

$cosL=2730026700 $

ReduceFrac

$ba =b/300a/300 $

$cosL=9189 $

$cosL=9189 $

$cos_{-1}(LHS)=cos_{-1}(RHS)$

$L=cos_{-1}(9189 )$

$L≈12_{∘}$

$35sin12_{∘} =105sinM $

Solve for M

MultEqn

LHS⋅105=RHS⋅105

$105⋅35sin12_{∘} =sinM$

RearrangeEqn

Rearrange equation

$sinM=105⋅35sin12_{∘} $

$sin_{-1}(LHS)=sin_{-1}(RHS)$

$M=sin_{-1}(105⋅35sin12_{∘} )$

UseCalc

Use a calculator

$M=38.589405…_{∘}$

RoundInt

Round to nearest integer

$M≈39_{∘}$

In △ABC, all three side lengths and the measure of the angle at C are given. Examine how the length of AB changes as the measure of ∠C varies.

Calculate the sum of the squares of AC and BC. Compare it with the square of AB when ∠ACB is an acute, right, and obtuse angle. What can be concluded about the cosine ratio of obtuse angles?

Consider △ABC, in which all three sides lengths and the measure of the angle at C are known. Let a, b, and c be the lengths of the sides opposite A, B, and C, respectively. In the following applet, the values of a2+b2 and c2 are shown. Move the slider to change the measure of ∠C.

Note that a calculator can be used to verify that the cosine of an acute angle is greater than 0, that the cosine of a right angle is equal to 0, and that the cosine of an obtuse angle is less than 0.

The table below shows the equation for the Law of Cosines when ∠ACB is an acute, right, and obtuse angle. In the table it is also shown the relationship between a2+b2 and c2 for these values, and conclusions are made.

m∠C | $c_{2}=a_{2}+b_{2}−2abcosC$ | Relationship between a2+b2 and c2 | Conclusion |
---|---|---|---|

$m∠C<90_{∘}$ | $c_{2}=a_{2}+b_{2}−2ab>0cosC $ | c2<a2+b2 | If ∠C is acute, not too many conclusions can be made. The opposite side to ∠C can be the largest side, the shortest side, or none. |

$m∠C=90_{∘}$ | $c_{2}=a_{2}+b_{2}−2ab=0cos90_{∘} $ | c2=a2+b2 | The cosine of $90_{∘}$ is 0. In this case, the Law of Cosines becomes the Pythagorean Theorem. This means that the opposite side to ∠C is the largest side of the triangle. |

$90_{∘}<m∠C<180_{∘}$ | $c_{2}=a_{2}+b_{2}−2ab<0cosC $ | c2>a2+b2 | If ∠C is obtuse, its measure is greater than the measures of ∠A and ∠C. Therefore, its opposite side is the largest side of the triangle. |

Once Diego completed a grueling month-long shift as a lighthouse keeper, he decided to fly from San Juan to New York. After flying for 3 hours on a straight path, he felt that the pilot made a course correction, then continued to fly for about 2 more hours on a path still toward New York. On Diego's return flight, the pilot flew on a straight path, without any change in direction, from New York to San Juan.
### Hint

### Solution

### Finding SD and DN

The average speed is the distance traveled divided by the amount of time spent traveling.

### Finding SD

Since the plane was deflected $10_{∘}$ from the first route, the measure of the angle SDN is $170_{∘}.$
Substitute 660 for s, 990 for n, and $170_{∘}$ for D.
Since a length cannot be negative, only the principal root is considered here. Therefore, the distance from San Juan to New York is about 1600 miles.

If the plane is deflected $10_{∘}$ and its average speed is 330 miles per hour, what is the distance from San Juan to New York? Round the answer to the nearest hundred miles.

{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":["x"],"constants":["PI"]}},"text":"<span class=\"mlmath-simple\"><\/span>"},"formTextBefore":null,"formTextAfter":"miles","answer":{"text":["1600"]}}

Use the speed formula to calculate the distance traveled before the change in direction SD, and the distance traveled after the change DN.

First, SD and DN will be found. Then, the Law of Cosines will be used to find NS, the distance between New York and San Juan.

$Speed=TimeDistance $

The flight from S to D takes 3 hours and the speed of the plane is 330 miles per hour. Substitute these values into the formula and solve for SD.
The distance covered in the first 3 hours of the flight is 990 miles. Similarly, the distance covered in the next 2 hours can be calculated. $Speed=TimeDistance $ | ||
---|---|---|

Length | SD | DN |

Substitution | $330=3SD $ | $330=2DN $ |

Calculation | SD=990mi | DN=660mi |

Now, in △SDN, the lengths of two sides and the measure of their included angle are known. Therefore, the Law of Cosines can be used. Let d, s, and n be the lengths of the sides opposite D, S, and N, respectively.

$d_{2}=s_{2}+n_{2}−2sncosD$

SubstituteValues

Substitute values

$d_{2}=660_{2}+990_{2}−2(660)(990)cos170_{∘}$

$d≈1600$

In this course, the use of the Law of Cosines in solving any type of triangle has been studied. By using this law, the challenge presented at the beginning can be solved.

Zosia knows the lengths of two sides of a triangle and the measure of their included angle. Let A, V, and D denote the vertices of the Summer Triangle.

What is the angular distance between Deneb and Vega? Round the answer to the nearest integer.{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":[],"constants":[]}},"text":"<span class=\"mlmath-simple\"><\/span>"},"formTextBefore":null,"formTextAfter":null,"answer":{"text":["24"]}}

The Law of Cosines states the relationship between the side lengths of a triangle and the cosine of one of the angles.

Let a, v, and d be the lengths of the sides opposite A, V, and D, respectively. By the Law of Cosines, the following equations holds true for △AVD.
Since the values of v, d, and A are given, the first equation will give the length of the third side.
Substitute these values and solve for a.
The angular distance between Deneb and Vega is about 24 angular units.

$a_{2}=v_{2}+d_{2}−2vdcosA$

SubstituteValues

Substitute values

$a_{2}=38_{2}+34_{2}−2(38)(34)cos38_{∘}$

Solve for a

$a≈24$

{{ 'mldesktop-placeholder-grade' | message }} {{ article.displayTitle }}!

{{ focusmode.exercise.exerciseName }}