{{ 'ml-label-loading-course' | message }}

{{ tocSubheader }}

{{ 'ml-toc-proceed-mlc' | message }}

{{ 'ml-toc-proceed-tbs' | message }}

An error ocurred, try again later!

Chapter {{ article.chapter.number }}

{{ article.number }}. # {{ article.displayTitle }}

{{ article.intro.summary }}

Show less Show more Lesson Settings & Tools

| {{ 'ml-lesson-number-slides' | message : article.intro.bblockCount }} |

| {{ 'ml-lesson-number-exercises' | message : article.intro.exerciseCount }} |

| {{ 'ml-lesson-time-estimation' | message }} |

Trigonometric ratios are useful in solving right triangles. Furthermore, the Law of Sines is a fairly used law when solving non-right triangles. However, there are cases in which this law is not applicable. This lesson will explore those cases, where 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=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.69224em;vertical-align:0em;\"><\/span><span class=\"mord mathdefault\">m<\/span><span class=\"mord\">\u2220<\/span><span class=\"mord mathdefault\" style=\"margin-right:0.05017em;\">B<\/span><span class=\"mord mathdefault\" style=\"margin-right:0.07153em;\">C<\/span><span class=\"mord mathdefault\">A<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><span class=\"mrel\">=<\/span><\/span><\/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=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/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=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.68333em;vertical-align:0em;\"><\/span><span class=\"mord mathdefault\">A<\/span><span class=\"mord mathdefault\" style=\"margin-right:0.02778em;\">D<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><span class=\"mrel\">=<\/span><\/span><\/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=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.68333em;vertical-align:0em;\"><\/span><span class=\"mord mathdefault\" style=\"margin-right:0.02778em;\">D<\/span><span class=\"mord mathdefault\" style=\"margin-right:0.05017em;\">B<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><span class=\"mrel\">=<\/span><\/span><\/span><\/span>","formTextAfter":null,"answer":{"text":["14"]}}

Challenge

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.

Explore

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.

Discussion

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 $△ABC.$

$a_{2}=b_{2}+c_{2}−2bccos(A)$

$b_{2}=a_{2}+c_{2}−2accos(B)$

$c_{2}=a_{2}+b_{2}−2abcos(C)$

The first equation will be proven. The other two equations can be proven by following the same procedure.
Notice that Equation II says that $x_{2}+h_{2}$ is equal to $c_{2}.$ Therefore, the expanded form of Equation I can be rewritten by using the Substitution Property of Equality.
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 $c.$

$a_{2}=b_{2}+c_{2}−2bccosA $

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.
$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.
$a_{2}=(b−x)_{2}+h_{2}$

ExpandNegPerfectSquare

$(a−b)_{2}=a_{2}−2ab+b_{2}$

$a_{2}=b_{2}−2bx+x_{2}+h_{2}$

$a_{2}=b_{2}−2bx+x_{2}+h_{2}$

Substitute

$x_{2}+h_{2}=c_{2}$

$a_{2}=b_{2}−2bx+c_{2}$

CommutativePropAdd

Commutative Property of Addition

$a_{2}=b_{2}+c_{2}−2bx$

$cosA=cx ⇔x=ccosA $

Finally, $ccosA$ can be substituted for $x$ into $a_{2}=b_{2}+c_{2}−2bx.$ By doing so, the formula for the Law of Cosines is obtained.
$a_{2}a_{2} =b_{2}+c_{2}−2bx=b_{2}+c_{2}−2bccosA $

The first equation will be proven for obtuse angles. The remaining equations can be proven similarly.
From Equation II, $h_{2}+x_{2}$ is equal to $c_{2}.$ 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 $c.$

$a_{2}=b_{2}+c_{2}−2bccosA $

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 $90_{∘}.$
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.
$a_{2}=h_{2}+(b+x)_{2}$

ExpandPosPerfectSquare

$(a+b)_{2}=a_{2}+2ab+b_{2}$

$a_{2}=h_{2}+b_{2}+2bx+x_{2}$

CommutativePropAdd

Commutative Property of Addition

$a_{2}=h_{2}+x_{2}+b_{2}+2bx$

$cos(180_{∘}−A)=cx ⇕x=ccos(180_{∘}−A) $

By the Sine and Cosine of Supplementary Values Angles, $cos(180_{∘}−A)$ and $cosA$ have opposite values.
$x=ccos(180_{∘}−A)⇕x=-ccosA $

Finally, by substituting $-ccosA$ for $x$ into $a_{2}=b_{2}+c_{2}+2bx,$ the Law of Cosines is obtained. $a_{2}=b_{2}+c_{2}+2bx$

Substitute

$x=-ccosA$

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

MultPosNeg

$a(-b)=-a⋅b$

$a_{2}=b_{2}+c_{2}−2bccosA$

Example

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.

External credits: @kdekiara

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=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/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 $

The next step is to substitute $b=4,$ $c=3.5,$ and $A=67_{∘}$ into the equation. Then, solve for $a.$
$a_{2}=b_{2}+c_{2}−2bccosA$

SubstituteValues

Substitute values

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

▼

Solve for $a$

$a≈4.2$

Example

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=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.69224em;vertical-align:0em;\"><\/span><span class=\"mord mathdefault\">m<\/span><span class=\"mord\">\u2220<\/span><span class=\"mord mathdefault\">L<\/span><\/span><\/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=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.69224em;vertical-align:0em;\"><\/span><span class=\"mord mathdefault\">m<\/span><span class=\"mord\">\u2220<\/span><span class=\"mord mathdefault\" style=\"margin-right:0.10903em;\">M<\/span><\/span><\/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=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.69224em;vertical-align:0em;\"><\/span><span class=\"mord mathdefault\">m<\/span><span class=\"mord\">\u2220<\/span><span class=\"mord mathdefault\" style=\"margin-right:0.07153em;\">K<\/span><\/span><\/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}=m_{2}+k_{2}−2mkcosL $

Substitute $35$ for $ℓ,$ $105$ for $m,$ and $130$ for $k$ into the equation. $ℓ_{2}=k_{2}+m_{2}−2kmcosL$

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 $

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

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

$L≈12_{∘}$

$ℓsinL =msinM ⇒35sin12_{∘} =105sinM $

The equation can be solved for $M.$
$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

$M≈39_{∘}$

$m∠K+12_{∘}+39_{∘}=180_{∘}m∠K=129_{∘} $

Therefore, all measures of $△KLM$ were found.
Explore

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?

Discussion

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 $a_{2}+b_{2}$ and $c_{2}$ 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 $a_{2}+b_{2}$ and $c_{2}$ for these values, and conclusions are made.

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

$m∠C<90_{∘}$ | $c_{2}=a_{2}+b_{2}−2ab>0cosC $ | $c_{2}<a_{2}+b_{2}$ | 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_{∘} $ | $c_{2}=a_{2}+b_{2}$ | 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 $ | $c_{2}>a_{2}+b_{2}$ | 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. |

Example

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 $170_{∘}.$
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=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/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 $

Substitute $660$ for $s,$ $990$ for $n,$ and $170_{∘}$ for $D.$ $d_{2}=s_{2}+n_{2}−2sncosD$

SubstituteValues

Substitute values

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

$d≈1600$

Closure

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=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/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 $△AVD.$
The angular distance between Deneb and Vega is about $24$ angular units.

$a_{2}v_{2}d_{2} =v_{2}+d_{2}−2vdcosA=d_{2}+a_{2}−2dacosV=a_{2}+v_{2}−2avcosD $

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 $a.$
$a_{2}=v_{2}+d_{2}−2vdcosA$

SubstituteValues

Substitute values

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

▼

Solve for $a$

$a≈24$

Loading content