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{{ printedBook.courseTrack.name }} {{ printedBook.name }} In this lesson, a new relation between side lengths and angle measures of a triangle will be derived and used.
### Catch-Up and Review

**Here are a few recommended readings before getting started with this lesson.**

- Triangle Angle Sum Theorem
- Right triangles
- Pythagorean Theorem
- Trigonometric ratios
- Area of a triangle and area of a triangle using sine

Try your knowledge on these topics.

Find the value of $x.$ If needed, write the answer to the nearest tenth.

a

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b

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c

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Calculate the area of the following triangles. If needed, write the answer correct to the nearest integer.

d

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e

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Can the measure of $∠B$ be found with the given information? If so, what is its value?

One conclusion obtained in the previous exploration can be generalized to all triangles.

For any triangle, the ratio of the sine of an angle to its opposite side is constant.
### Proof

### Acute Triangles

By following the same procedure but drawing the height from vertex $A,$ it can be shown that $bsinB =csinC .$ Putting these two results together, the Law of Sines is obtained.
### Obtuse Triangles

Next, the exterior height $h_{2}$ from vertex $C$ will be drawn. Let $E$ be the point of intersection of this height and the extension of $AB.$
Finally, this result can be combined with the previous determination $asinA =csinC $ to derive the Law of Sines.

Based on the characteristics of the diagram, the following relation can be proven true.

$asinA =bsinB =csinC $

An alternative way to write the Law of Sines is involving the ratio of a side length to the sine of its opposite angle.

$sinAa =sinBb =sinCc $

Consider an acute triangle with height $h$ drawn from one of its vertices. Because $h$ is perpendicular to the base, the original triangle is split into two right triangles.

In these two right triangles, $∠A$ and $∠B$ are both opposite angles to $h.$ Therefore, by applying the definition of the sine ratio to $∠A$ and $∠B,$ it is possible to relate the sine of these angles with the side lengths $a$ and $b.$
Next, $h$ can be isolated and written in terms of the corresponding side length and angle, for both right triangles.
$sinA=bh sinB=ah ⇔h=bsinA⇔h=asinB $ By the Transitive Property of Equality, it can be stated that $bsinA$ and $asinB$ are equal. ${h=bsinAh=asinB ⇒bsinA=asinB $ Finally, the obtained equation can be rearranged to obtain the desired result.

$bsinA=asinB$

Simplify

DivEqn

$LHS/ab=RHS/ab$

$abbsinA =abasinB $

CancelCommonFac

Cancel out common factors

$ab b sinA =a ba sinB $

SimpQuot

Simplify quotient

$asinA =bsinB $

$asinA =bsinB =csinC $

An obtuse triangle will now be considered.

This proof is very similar to the proof for acute triangles, but it uses an interior and an exterior height. First, the height $h_{1}$ from the vertex where the obtuse angle is located will be drawn. Just as before, this generates two right triangles.

The sine ratio will be written for these right triangles.
In the equations, $h_{1}$ can be isolated.
$sinA=ch_{1} sinC=ah_{1} ⇔h_{1}=csinA⇔h_{1}=asinC $ By the Transitive Property of Equality, it can be stated that $csinA$ and $asinC$ are equal. ${h_{1}=csinAh_{1}=asinC ⇒csinA=asinC $ The obtained equation can be rearranged to obtain the desired result.

$csinA=asinC$

Simplify

DivEqn

$LHS/ac=RHS/ac$

$accsinA =acasinC $

CancelCommonFac

Cancel out common factors

$ac c sinA =a ca sinC $

SimpQuot

Simplify quotient

$asinA =csinC $

Here, $∠ABC$ and $∠CBE$ form a linear pair and are therefore supplementary angles. Because the sine of supplementary angles is the same, the sine of $∠ABC$ equals the sine of $∠CBE.$ Also, using the sine ratio on $△BCE,$ it can be stated that the sine of $∠CBE$ is the ratio of $h_{2}$ to $a.$ $sinABC=sinCBEandsinCBE=ah_{2} $ By the Transitive Property of Equality, the sine of $∠ABC$ can be written in terms of $h_{2}$ and $a.$ $⎩⎪⎨⎪⎧ sinABC=sinCBEsinCBE=ah_{2} ⇓sinABC=ah_{2} $ Now $△ACE$ will be considered. By using the sine ratio, it follows that the sine of $∠A$ is the ratio of $h_{2}$ to $b.$

Now, $h_{2}$ can be written in terms of $sinABC$ and $a,$ and in terms of $sinA$ and $b.$ $sinABC=ah_{2} sinA=bh_{2} ⇔h_{2}=asinABC⇔h_{2}=bsinA $ The Transitive Property of Equality can be used one more time. ${h_{2}=asinABCh_{2}=bsinA ⇓asinABC=bsinA $ If only $△ABC$ is considered, then $∠ABC$ can be named as $∠B.$

This allows $sinABC$ to be written as $sinB.$ Therefore, $asinB=bsinA,$ and this equation can be rearranged to obtain the desired formula.$asinB=bsinA$

Simplify

DivEqn

$LHS/ab=RHS/ab$

$abasinB =abbsinA $

CancelCommonFac

Cancel out common factors

$a ba sinB =ab b sinA $

SimpQuot

Simplify quotient

$bsinB =asinA $

$asinA =bsinB =csinC $

As previously stated, the Law of Sines can be used to find side lengths of a triangle.

Magdalena will go golfing after school. The locations of her school, the golf course, and her house form a triangular shape. She knows the measures of two of the triangle's angles, and she knows the distance from her house to the school is $3$ kilometers.
What is the distance between the golf course and Magdalena's house? Write the answer rounded to three significant figures.

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Start by labeling the sides and the angles of the triangle.

The sides of the triangle will be labeled $a,$ $b,$ and $c.$ Similarly, the angles will be labeled $A,$ $B,$ and $C.$

By the Law of Sines, the ratio of a side length to the sine of its opposite angle is constant. With this information, a proportion that relates $A,$ $a,$ $C,$ and $c$ can be written. $sinAa =sinCc $ The values $a=3,$ $A=55_{∘},$ and $C=60_{∘}$ can be substituted into this formula. Then the resulting equation can be solved for $c,$ the distance between Magdalena's house and the golf course.$sinAa =sinCc $

SubstituteValues

Substitute values

$sin55_{∘}3 =sin60_{∘}c $

Solve for $c$

MultEqn

$LHS⋅sin60_{∘}=RHS⋅sin60_{∘}$

$sin55_{∘}3 (sin60_{∘})=c$

RearrangeEqn

Rearrange equation

$c=sin55_{∘}3 (sin60_{∘})$

UseCalc

Use a calculator

$c=3.171665…$

RoundSigDig

Round to $3$ significant ${\textstyle \ifnumequal{3}{1}{\text{digit}}{\text{digits}}}$

$c≈3.17$

In the following applet, $x$ represents the side length of a triangle. By using the Law of Sines and, if needed, the Triangle Angle Sum Theorem, find the value of $x.$ Write the answer rounded to two decimal places.

The Law of Sines can also be used to find angle measures of a triangle.

Emily will go backpacking across South America! She will visit Buenos Aires, Santiago, and Asunción, among other cities. Emily knows that the distances from Buenos Aires to Santiago and Asunción are $1140$ and $1040$ kilometers, respectively. She also knows that the angle whose vertex is located at Santiago and whose sides pass through Buenos Aires and Asunción measures $44_{∘}.$
To optimize her journey, she wants to find the measure of the remaining two angles of the triangle formed by these three cities, as well as the distance between Santiago and Asunción . Help Emily find them! Write the answers to the nearest integer.

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Start by labeling the sides and the angles of the triangle.

The angles of the triangle will be labeled $A,$ $B,$ and $S.$ Similarly, the sides will be labeled $a,$ $b,$ and $s.$
Rounded to the nearest degree, it was found that the angle formed at Asunción measures $50_{∘}.$

By the Law of Sines, the ratio of the sine of an angle to the length of its opposite side is constant. With this information, a proportion that relates $A,$ $a,$ $S,$ and $s$ can be written.
$ssinS =asinA $
In the above formula, $a=1140,$ $s=1040,$ and $S=44_{∘}$ can be substituted. Then the resulting equation can be solved for $A,$ the measure of the angle formed at Asunción.

$ssinS =asinA $

SubstituteValues

Substitute values

$1040sin44_{∘} =1140sinA $

Solve for $A$

MultEqn

$LHS⋅1140=RHS⋅1140$

$1040sin44_{∘} (1140)=sinA$

RearrangeEqn

Rearrange equation

$sinA=1040sin44_{∘} (1140)$

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

$A=sin_{-1}(1040sin44_{∘} (1140))$

UseCalc

Use a calculator

$A=49.592410…_{∘}$

RoundInt

Round to nearest integer

$A≈50_{∘}$

Finally, to find the measure of the angle at Buenos Aires, the Triangle Angle Sum Theorem can be used.
$44_{∘}+50_{∘}+B=180_{∘}⇕B=86_{∘} $
This information can be added to the diagram.

Finally, the Law of Sines can be used again to find the distance between Santiago and Asunción. To do so, recall that the ratio of the length of a side to the sine of its opposite angle is constant.
$sinBb =sinSs $
Next, $s=1040,$ $S=44_{∘},$ and $B=86_{∘}$ will be substituted into the above equation.
Emily has all she needs to have a great journey!

In the following applet, $x$ represents the measure of an angle of a triangle. By using the Law of Sines and, if needed, the Triangle Angle Sum Theorem, find the value of $x.$ Write the answer as a single number rounded to the nearest degree, without the degree symbol.

The Law of Sines can be used to find missing side lengths and angle measures of a triangle. When those side lengths and angle measures are known, the area and perimeter of a triangle can also be found.

Emily is planning to continue her travels. This time, she wants to visit some cities in the UK, as well as Ireland. She would really like to visit London, Edinburgh, and Dublin. It becomes clear to her that these three cities form a triangle. She is able to figure out that the angles at London and Edinburgh measure $41_{∘}$ and $59_{∘},$ respectively. She also knows that the distance between these two cities is $530$ kilometers.
To optimize her journey, Emily wants to find the perimeter and area of the triangle. Help her with her plan by calculating the perimeter and area! Write the perimeter rounded to the nearest integer and the area rounded to the nearest thousand.

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The perimeter of a triangle is the sum of its three side lengths. The area of the triangle can be found by calculating half the product of two side lengths and the sine of their included angle.

The perimeter and the area of the triangle will be calculated one at a time.

Note that the missing angle is opposite to the side of the triangle whose length is known. Therefore, in order to use the Law of Sines to find the missing side lengths, the angle at Dublin must be calculated. To do this, the Triangle Angle Sum Theorem can be used.
$D+59_{∘}+41_{∘}=180_{∘}⇕D=80_{∘} $
The measure of the angle at Dublin is $80_{∘}.$

The Law of Sines can now be used to find the missing side lengths. The law states that, for every triangle, the ratio of the length of a side to the sine of its opposite angle is constant. With this rule in mind, a proportion involving $D,$ $d,$ $L$ and $ℓ$ can be written.
$sinLℓ =sinDd $
Next, $L=41_{∘},$ $D=80_{∘},$ and $d=530$ can be substituted in the above equation, and $ℓ$ can be isolated.
The distance between Dublin and Edinburgh is about $353$ kilometers. By following the same procedure, the distance between London and Dublin can be found.

Law of Sines | Substitute | Simplify |
---|---|---|

$sinLℓ =sinDd $ | $sin41_{∘}ℓ =sin80_{∘}530 $ | $ℓ≈353$ |

$sinEe =sinDd $ | $sin59_{∘}e =sin80_{∘}530 $ | $e≈461$ |

The perimeter of the triangle formed by the three cities can now be calculated.
$PerimeterP=530+353+461⇕P=1344km $

$Area=21 dℓsinE$

SubstituteValues

Substitute values

$Area=21 (530)(353)sin59_{∘}$

Evaluate right-hand side

Multiply

Multiply

$Area=21 (187090)sin59_{∘}$

MoveRightFacToNumOne

$b1 ⋅a=ba $

$Area=2187090 sin59_{∘}$

CalcQuot

Calculate quotient

$Area=93545sin59_{∘}$

UseCalc

Use a calculator

$Area=80183.715144…$

Approximate to the nearest thousand

$Area≈80000$

The Law of Sines is useful to find missing angle measures and side lengths for any type of triangle.
Using the Law of Sines, it was found that the measure of the angle at $A$ is about $69_{∘}.$
*two* valid measures for the angle at $A,$ which are $69_{∘}$ and $111_{∘}.$ Consequently, the measure of a missing angle is *ambiguous*. This shows a limitation of the Law of Sines. To avoid this ambiguity, it must be said whether the missing angle is acute or obtuse, or a diagram of the triangle should be provided.

Consider a triangle with vertices $A,$ $B,$ and $C,$ where the measure of the angle at $C$ is $34_{∘},$ and the lengths of $AB$ and $BC$ are $6$ and $10$ centimeters, respectively. The following diagram represents the triangle that was just described.
The measure of the angle at $A$ can be found by using the Law of Sines.

$10sinA =6sin34_{∘} $

Solve for $A$

MultEqn

$LHS⋅10=RHS⋅10$

$sinA=6sin34_{∘} (10)$

CommutativePropMult

Commutative Property of Multiplication

$sinA=10(6sin34_{∘} )$

MoveLeftFacToNum

$a⋅cb =ca⋅b $

$sinA=610sin34_{∘} $

ReduceFrac

$ba =b/2a/2 $

$sinA=35sin34_{∘} $

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

$A=sin_{-1}(35sin34_{∘} )$

UseCalc

Use a calculator

$A=68.746886…_{∘}$

RoundInt

Round to nearest integer

$A≈69_{∘}$

Is this the only possible measure for the angle at $A?$ To find the measure of this angle, the Law of Sines was applied. That is, for any triangle, the ratio of the sine of an angle to the length of its opposite side is constant. $10sin69_{∘} =6sin34_{∘} $ Recall now that the sine of an angle is equal to the sine of the supplement of the angle. $sinθ=sin(180_{∘}−θ) $ With this information, it can be stated that the sine of an angle whose measure is $69_{∘}$ is equal to the sine of an angle whose measure is $180_{∘}−69_{∘}=111_{∘}.$ Therefore, by the Substitution Property of Equality, a new proportion can be derived. $⎩⎪⎨⎪⎧ 10sin69_{∘} =6sin34_{∘} sin69_{∘}=sin111_{∘} ⇓10sin111_{∘} =6sin34_{∘} $ The obtained proportion implies a new triangle that satisfies the initial conditions.

Therefore, in the given context, there areThe challenge presented at the beginning of this lesson can be solved by using the Law of Sines.

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Since the ratio of the sine of an angle to the length of its opposite side is constant, a proportion can be written.
$ACsinB =BCsinA ⇒15sinB =20sin70_{∘} $
The above equation can be solved for $B.$
The measure of the angle at $B$ is about $45_{∘}.$

$15sinB =20sin70_{∘} $

$B≈45_{∘}$

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