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This lesson will explore the applications of the Law of Sines and the Law of Cosines to solve different types of triangles. Furthermore, real-life applications of these ideas will also be explored.
### Catch-Up and Review

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

Take a look at the following triangles. Think whether they can be solved by using the Law of Sines or the Law of Cosines.

The following figure shows a circle circumscribed around a non-right triangle. Notice that none of the triangle's sides correspond to the diameter of the circle. What is the area of the circle?

Trigonometric ratios are often used to solve right triangles, but they cannot be used to solve non-right, or *oblique*, triangles. For these triangles, the Law of Sines and the Law of Cosines are particularly useful because they can be used to solve *any* triangle, right or oblique. To solve a triangle using the Law of Sines or the Law of Cosines, three pieces of information must be known.

Case | Given Information | Law | Strategy |
---|---|---|---|

$1$ | Two angles and a side length | Law of Sines | The Triangle Angle Sum Theorem can be used to find the missing angle measure. Then the Law of Sines can be used to find the unknown side lengths. |

$2$ | Two side lengths and a non-included angle | Law of Sines | The Law of Sines can be used to solve for one of the unknown angle measures. Then the Triangle Angle Sum Theorem can be used to find the third angle measure. Finally, the Law of Sines can be applied one more time to find the unknown side length. |

$3$ | Three side lengths | Law of Cosines | The Law of Cosines can be used to find any of the unknown angle measures. Then, either the Law of Sines or the Law of Cosines can be used to find another missing angle measure. Finally, the Triangle Angle Sum Theorem can be used to find the third angle measure. |

$4$ | Two side lengths and their included angle | Law of Cosines | The Law of Cosines can be used to find the missing side length. Then, either the Law of Cosines or the Law of Sines can be used used to find a missing angle measure. Finally, the Triangle Angle Sum Theorem can be used to find the last angle measure. |

Zain is vacationing in Italy. They were in Pisa to see the famous Leaning Tower when a question came across their mind. What would the tower's height be if it was not a *leaning* tower? Zain's distance to the tower is $80$ meters and they can measure an angle of elevation of $37_{∘}$ to the tower's top. Furthermore, the guidebook states that the tower's inclination is about $4_{∘}.$

Remembering a Geometry lesson, they realize that the situation can be modeled using a non-right triangle. Help Zain calculate the height of the upright tower. Write the answer rounded to one decimal place.

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The inclination angle and its adjacent angle are complementary angles.

The situation can be modeled using a non-right triangle.

The tower's inclination angle and $∠B$ are complementary angles, so the sum of their measures adds to $90_{∘}.$ The measure of $∠B$ can now be calculated.$4_{∘}+m∠B=90_{∘}⇕m∠B=86_{∘} $

Now that two angles and their included side are known, it is possible to use the Law of Sines. To do so, the measure of $∠A$ must be found. This will be done by using the Triangle Angle Sum Theorem. $m∠A+86_{∘}+37_{∘}=180_{∘}⇕m∠A=57_{∘} $

Finally, the Law of Sines will be used to write a proportion. This equation will be solved for $c,$ the height of the upright tower.
$sinAa =sinCc $

SubstituteValues

Substitute values

$sin57_{∘}80 =sin37_{∘}c $

Solve for $c$

MultEqn

$LHS⋅sin37_{∘}=RHS⋅sin37_{∘}$

$sin57_{∘}80 (sin37_{∘})=c$

UseCalc

Use a calculator

$57.406571…=c$

RearrangeEqn

Rearrange equation

$c=57.406571…$

RoundDec

Round to $1$ decimal place(s)

$c≈57.4$

Magdalena is doing some research in the forest for a biology project. She has a device that allows her to measure angles. She can also measure the distance from a tree to her device. To help Magdalena complete her research project, find the lengths of different trees, rounded to one decimal place.

As mentioned before, the Law of Sines and the Law of Cosines are valid for all types of triangles, including both right and non-right triangles. However, the definitions of the sine and cosine of an angle are given in terms of the ratios of a right triangle's sides.

Therefore, these definitions do not seem to be compatible with obtuse angles. Nevertheless, they can be extended to deal with obtuse angles by considering the following identities.

The sine of supplementary angles are equal. Conversely, the cosine of supplementary angles are opposite values.

$sin(180_{∘}−θ)cos(180_{∘}−θ) =sinθ=-cosθ $

These identities can also be used when the angle is given in radians.

$sin(π−θ)cos(π−θ) =sinθ=-cosθ $

Ignacio's grandparent wants to construct a fence for a quadrilateral piece of land. To find the perimeter of the land, he starts measuring its sides using an old trundle wheel. Unfortunately, after measuring just two sides, the trundle wheel breaks.

Ignacio wants to help his grandparent and, in an attempt to simplify the problem, he divides the land into two triangles. Then, by using a compass, he is able to measure the angles of these triangles.

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Since two side lengths and the measure of their included angle are known in $△ABC,$ the Law of Cosines can be used to solve for the missing side length.

Note that two side lengths and all the angle measurements are known in $△ABC.$

In particular, since two side lengths and the measure of their included angle are known, the Law of Cosines can be used to solve for the missing side length.$b_{2}=a_{2}+c_{2}−2accosB$

SubstituteValues

Substitute values

$b_{2}=400_{2}+560_{2}−2(400)(560)cos80_{∘}$

$b≈629$

Now, $△ACD$ will be considered.

Since one side length and all the angle measures are known, the missing side lengths can be found by using the Law of Sines. The length of $DC$ will be calculated. Therefore, the length of $DC$ is about $260$ meters. Next, the length of $AD$ can be calculated by following the same procedure. Now, all the sides of the piece of land are known. Finally, the perimeter will be calculated by adding all the side lengths.$P=457m+560m+400m+260m⇕P=1677m $

The perimeter of the piece of land is about $1677$ meters.
Kriz is setting up for a free shots on an empty goal. When considering their distance to both goal posts, they realize that the Law of Cosines can be used to calculate the top measure of the angle in which they must kick the ball in order to score. They are practicing with a standard $7.3$ meter net. Help Kriz calculate this angle and score the goal! Write the answer rounded to one decimal place.

External credits: @kdekiara

A burglar robbed a store and took the cashier's smartphone. In an attempt to outsmart the police, the burglar turned off the phone's GPS. Once at their secret location, the burglar felt safe and made a call to plan their next move. However, the smartphone signal was detected by two nearby towers, estimating their distance to the phone in use.
### Hint

### Solution

Now that the measure of $∠A$ is known, the horizontal and vertical distance from the left tower to the smartphone, $H$ and $V,$ respectively, can be calculated. To do this, a right triangle formed by the left tower and the phone will be considered. Then, trigonometric ratios will be used.
The vertical distance to the smartphone $V$ is the opposite leg to the angle whose measure is $26_{∘}.$ Therefore, the length of the leg can be found by using the sine ratio.
Similarly, the horizontal distance $H$ is the adjacent leg to the angle whose measure is $26_{∘}.$ Therefore, the cosine ratio can be used to find the value of $H.$
The horizontal distance from the left tower to the burglar's location is about $4$ blocks, and the vertical distance is about $2$ blocks.

With the distance from the towers to the location of the smartphone and knowing that the towers are $6$ blocks apart, the police think they have enough information to locate the burglar. Help the police find the burglar's location! Write the horizontal and vertical distance in blocks with respect to the phone tower at the left. Round the answers to the nearest integer.

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The situation can be modeled using a triangle with three known side lengths.

The distance from the tower at the left to the smartphone is about $4.5$ blocks. The distance from the tower at the right to the phone is about $2.8$ blocks. The towers are $6$ blocks apart. Therefore, the situation can be modeled using a triangle with three known sides lengths.

Even if no angle is known at first, knowing the three side lengths allows the use of the Law of Cosines to solve for any angle. In this case, it is convenient to solve for $∠A,$ which corresponds to the vertex represented by the left tower.$a_{2}=b_{2}+c_{2}−2bccosA$

SubstituteValues

Substitute values

$2.8_{2}=6_{2}+4.5_{2}−2(6)(4.5)cosA$

Solve for $A$

CalcPow

Calculate power

$7.84=36+20.25−2(6)(4.5)cosA$

AddTerms

Add terms

$7.84=56.25−2(6)(4.5)cosA$

Multiply

Multiply

$7.84=56.25−54cosA$

SubEqn

$LHS−56.25=RHS−56.25$

$-48.41=-54cosA$

DivEqn

$LHS/(-54)=RHS/(-54)$

$0.896481…=cosA$

RearrangeEqn

Rearrange equation

$cosA=0.896481…$

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

$A=cos_{-1}0.896481…$

UseCalc

Use a calculator

$A=26.300639…_{∘}$

RoundInt

Round to nearest integer

$A≈26_{∘}$

$sinθ=HypotenuseOpposite Leg $

SubstituteValues

Substitute values

$sin26_{∘}=4.5V $

Solve for $V$

MultEqn

$LHS⋅4.5=RHS⋅4.5$

$(4.5)sin26_{∘}=V$

UseCalc

Use a calculator

$1.972670…=V$

RoundInt

Round to nearest integer

$2≈V$

RearrangeEqn

Rearrange equation

$V≈2$

$cosθ=HypotenuseAdjacent Leg $

SubstituteValues

Substitute values

$cos26_{∘}=4.5H $

Solve for $H$

MultEqn

$LHS⋅4.5=RHS⋅4.5$

$(4.5)cos26_{∘}=H$

UseCalc

Use a calculator

$4.044573…=H$

RoundInt

Round to nearest integer

$4≈H$

RearrangeEqn

Rearrange equation

$H≈4$

Two radar stations located $30$ kilometers apart detect a passing helicopter. The first station measures an angle of elevation to the helicopter of $40_{∘},$ while the second station measures an angle of elevation of $45_{∘}.$ At what altitude, rounded to one decimal place, is the helicopter flying?
### Hint

### Solution

The distance from Station $1$ to the helicopter is about $21.3$ kilometers. This distance is the hypotenuse of the right triangle formed by considering the helicopter and Station $1$ as vertices. Furthermore, in this right triangle, the opposite leg to the angle that measures $40_{∘}$ represents the helicopter's altitude.
Finally, the sine ratio can be used to find the altitude of the helicopter.
The helicopter is flying at an altitude of about $13.7$ kilometers.

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The missing angle can be found by using the Triangle Angle Sum Theorem.

Because two angles and the included side are known, this problem can be approached by using the Law of Sines to find the distance between the helicopter and one of the radar stations. Once one of the missing side lengths is determined, trigonometric ratios can be used to find the altitude, or height, of the helicopter. The first thing to do is to find the unknown angle.

To do this, the Triangle Angle Sum Theorem can be used. The measure of $∠A$ will now be found.$m∠A+40_{∘}+45_{∘}=180_{∘}⇕m∠A=95_{∘} $

Now that the opposite angle to the known side length $a$ has been found, the Law of Sines can be used to write a proportion and solve for any of the missing sides. For instance, $b$ can be calculated.
$sinAa =sinBb $

SubstituteValues

Substitute values

$sin95_{∘}30 =sin45_{∘}b $

Solve for $b$

MultEqn

$LHS⋅sin45_{∘}=RHS⋅sin45_{∘}$

$sin95_{∘}30 (sin45_{∘})=b$

UseCalc

Use a calculator

$21.294234…=b$

RearrangeEqn

Rearrange equation

$b=21.294234…$

RoundDec

Round to $1$ decimal place(s)

$b≈21.3$

$sinθ=hypotenuseopposite leg $

SubstituteValues

Substitute values

$sin40_{∘}=21.3h $

Solve for $h$

MultEqn

$LHS⋅21.3=RHS⋅21.3$

$sin40_{∘}(21.3)=h$

UseCalc

Use a calculator

$13.691376…=h$

RearrangeEqn

Rearrange equation

$h=13.691376…$

RoundDec

Round to $1$ decimal place(s)

$h≈13.7$

The Law of Sines states that for any triangle, the ratio of the sine of an angle to the length of its opposite side is constant. However, this is not just *any* constant. In fact it has an important geometrical interpretation.

The following goes for any triangle. The diameter of a triangle's circumcircle is equal to the ratio of a side length to the sine of its opposite angle.

Now, consider the above figure and let $D$ be the diameter of the circle. With the given information, the following equation holds true.

$sinAa =sinBb =sinCc =D$

The proof can be completed in two parts. Part I shows how the sides of a triangle are proportional to the sines of the opposite angles. Part II shows this proportion is equal to the diameter of the circle circumscribed around the triangle.

Consider a triangle $ABC$ with side lengths $a,$ $b,$ and $c,$ and angle measures $A,$ $B,$ and $C.$

By the Law of Sines, the ratio of the side length of the triangle to the sine of the opposite angle is the same for all sides.$sinAa =sinBb =sinCc $

Draw the circumcircle of the triangle $ABC$ with its circumcenter $O.$

By the Inscribed Angle Theorem, the measure of $A$ is half of the measure of the intercepted arc $BC.$

Now, draw the central angle with the intercepted arc $BC.$ Recall that a central angle and its intercepted arc have the same measure. Therefore, the measure of $BC$ is equal to the measure of $m∠COB.$

By the Substitution Property of Equality, the measure of $∠A$ is half of the measure of $∠COB.$$⎩⎪⎨⎪⎧ m∠A=21 mBCmBC=m∠COB ⇓m∠A=21 m∠COB $

Let $m∠A$ be $α.$ Then $m∠COB$ becomes $2α.$
Note that $OB$ and $OC$ are both equal to the radius $R.$ Hence, $△COB$ is an isosceles triangle with two congruent sides $OB$ and $OC.$

Now, focus on $△COB.$ Draw its altitude from the vertex angle $O.$ Since $△COB$ is an isosceles triangle, the altitude bisects both the vertex angle and the opposite side.

The sine of $α$ — the ratio of the opposite side to the hypotenuse — can be written using the right triangle $CIO.$$sinα =R2a =2Ra $

Rearrange $sinα$ by substituting $α=A.$
Since $R$ represents the radius, $2R$ is equal to the diameter $D$ of the circle.
$sinAa =2R⇒sinAa =D $

$⎩⎪⎪⎪⎨⎪⎪⎪⎧ sinAa =sinBb =sinCc sinAa =D ⇓sinAa =sinBb =sinCc =D $

The challenge presented at the beginning of this lesson can be solved by using a combination of the Law of Sines, the Law of Cosines, and the extended form of the Law of Sines.

The question here was to find the area of the circle. This will be answered by first finding the ratio of the triangle's side lengths to the sine of their opposite angles.

a What is the ratio of a side length to the sine of its opposite angle? Write the answer rounded to one decimal place.

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b What is the area of the circumscribed circle? Write the answer rounded to one decimal place.

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a Two side lengths and the measure of the included angle are known.

b The area of a circle is $A=πr_{2},$ where $r$ is the radius.

Since two side lengths and the included angle are known, the Law of Cosines can be used to find the missing side length. Because the missing side is opposite to the known angle, finding the side length will allow to calculate the desired ratio.

The Law of Cosines will be used to find the missing side $a.$$a_{2}=b_{2}+c_{2}−2bccosA$

SubstituteValues

Substitute values

$a_{2}=4.9_{2}+4.6_{2}−2(4.9)(4.6)cos44_{∘}$

$a≈3.6$

$Radius:25.2 =2.6units $

With this information, the area of the circle can be calculated. $A=πr_{2}$

Substitute

$r=2.6$

$A=π(2.6)_{2}$

Simplify right-hand side

CalcPow

Calculate power

$A=π(6.76)$

CommutativePropMult

Commutative Property of Multiplication

$A=6.76π$

UseCalc

Use a calculator

$A=21.237166…$

RoundDec

Round to $1$ decimal place(s)

$A≈21.2$