Efficiently Solve the Triangle Using the Law of Sines

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.

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

An obtuse triangle with a base of 3 and a height of 12. The obtuse angle is at the bottom-left vertex.

Challenge

Investigate Triangles With Two Sides and an Angle Known

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

Triangle ABC has sides AC and BC measuring 15 and 20 units respectively. The angle A of the triangle measures 70 degrees.

Explore

Relationship Between a Triangle's Angles and Their Opposite Sides

△ A B C,

all the side lengths and angle measures are given. Calculate the ratio of the sine of each angle to the length of its opposite side.

What conclusions can be made?

Discussion

Law of Sines

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.

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

$\frac{sin A}{a} = \frac{sin B}{b} = \frac{sin C}{c}$

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

$\frac{a}{sin A} = \frac{b}{sin B} = \frac{c}{sin C}$

Proof

Acute Triangles

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.

sin A = \frac{h}{b} sin B = \frac{h}{a} \Leftrightarrow h = b sin A \Leftrightarrow h = a sin B

By the Transitive Property of Equality, it can be stated that

b sin A

and

a sin B

are equal.

{h = b sin A h = a sin B \Rightarrow b sin A = a sin B

Finally, the obtained equation can be rearranged to obtain the desired result.

b sin A = a sin B

Simplify

DivEqn

$LHS / a b = RHS / a b$

\frac{b sin A}{a b} = \frac{a sin B}{a b}

CancelCommonFac

Cancel out common factors

\frac{b sin A}{a b} = \frac{a sin B}{a b}

SimpQuot

Simplify quotient

\frac{sin A}{a} = \frac{sin B}{b}

By following the same procedure but drawing the height from vertex

A,

it can be shown that

\frac{s i n B}{b} = \frac{s i n C}{c} .

Putting these two results together, the Law of Sines is obtained.

$\frac{sin A}{a} = \frac{sin B}{b} = \frac{sin C}{c}$

Obtuse Triangles

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.

sin A = \frac{h _{1}}{c} sin C = \frac{h _{1}}{a} \Leftrightarrow h_{1} = c sin A \Leftrightarrow h_{1} = a sin C

By the Transitive Property of Equality, it can be stated that

c sin A

and

a sin C

are equal.

{h_{1} = c sin A h_{1} = a sin C \Rightarrow c sin A = a sin C

The obtained equation can be rearranged to obtain the desired result.

c sin A = a sin C

Simplify

DivEqn

$LHS / a c = RHS / a c$

\frac{c sin A}{a c} = \frac{a sin C}{a c}

CancelCommonFac

Cancel out common factors

\frac{c sin A}{a c} = \frac{a sin C}{a c}

SimpQuot

Simplify quotient

\frac{sin A}{a} = \frac{sin C}{c}

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

\overline{A B} .

Here,

∠ A B C

and

∠ C B E

form a linear pair and are therefore supplementary angles. Because the sine of supplementary angles is the same, the sine of

∠ A B C

equals the sine of

∠ C B E .

Also, using the sine ratio on

△ B C E,

it can be stated that the sine of

∠ C B E

is the ratio of

h_{2}

a .

sin A B C = sin C B E and sin C B E = \frac{h _{2}}{a}

By the Transitive Property of Equality, the sine of

∠ A B C

can be written in terms of

h_{2}

and

a .

⎩ ⎪ ⎨ ⎪ ⎧ sin A B C = s i n C B E s i n C B E = \frac{h _{2}}{a} ⇓ sin A B C = \frac{h _{2}}{a}

Now

△ A C E

will be considered. By using the sine ratio, it follows that the sine of

∠ A

is the ratio of

h_{2}

b .

Now,

h_{2}

can be written in terms of

sin A B C

and

a,

and in terms of

sin A

and

b .

sin A B C = \frac{h _{2}}{a} sin A = \frac{h _{2}}{b} \Leftrightarrow h_{2} = a sin A B C \Leftrightarrow h_{2} = b sin A

The Transitive Property of Equality can be used one more time.

{h_{2} = a sin A B C h_{2} = b sin A ⇓ a sin A B C = b sin A

If only

△ A B C

is considered, then

∠ A B C

can be named as

∠ B .

This allows

sin A B C

to be written as

sin B .

Therefore,

a sin B = b sin A,

and this equation can be rearranged to obtain the desired formula.

a sin B = b sin A

Simplify

DivEqn

$LHS / a b = RHS / a b$

\frac{a sin B}{a b} = \frac{b sin A}{a b}

CancelCommonFac

Cancel out common factors

\frac{a sin B}{a b} = \frac{b sin A}{a b}

SimpQuot

Simplify quotient

\frac{sin B}{b} = \frac{sin A}{a}

Finally, this result can be combined with the previous determination

\frac{s i n A}{a} = \frac{s i n C}{c}

to derive the Law of Sines.

$\frac{sin A}{a} = \frac{sin B}{b} = \frac{sin C}{c}$

The Law of Sines can be used to find side lengths and angle measures of any triangle.

Example

Using the Law of Sines To Find the Side of a Triangle

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.

Hint

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

Solution

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.

\frac{a}{sin A} = \frac{c}{sin C}

The values

a = 3,

A = 5 5^{\circ},

and

C = 6 0^{\circ}

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.

\frac{a}{sin A} = \frac{c}{sin C}

SubstituteValues

Substitute values

\frac{3}{sin 5 5 ^{\circ}} = \frac{c}{sin 6 0 ^{\circ}}

Solve for

c

MultEqn

$LHS \cdot sin 6 0^{\circ} = RHS \cdot sin 6 0^{\circ}$

\frac{3}{sin 5 5 ^{\circ}} (sin 6 0^{\circ}) = c

RearrangeEqn

Rearrange equation

c = \frac{3}{sin 5 5 ^{\circ}} (sin 6 0^{\circ})

UseCalc

Use a calculator

c = 3.171665 \dots

RoundSigDig

Round to $3$ significant digit(s)

c \approx 3.17

It was found that the distance between the golf course and Magdalena's house is

3.17

kilometers, rounded to three significant figures.

Pop Quiz

Practice Finding Sides Using the Law of Sines

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.

Example

Using the Law of Sines To Find Angles of a Triangle

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

4 4^{\circ} .

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.

Hint

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

Solution

The angles of the triangle will be labeled

A,

B,

and

S .

Similarly, the sides will be labeled

a,

b,

and

s .

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.

\frac{sin S}{s} = \frac{sin A}{a}

In the above formula,

a = 1140,

s = 1040,

and

S = 4 4^{\circ}

can be substituted. Then the resulting equation can be solved for

A,

the measure of the angle formed at Asunción.

\frac{sin S}{s} = \frac{sin A}{a}

SubstituteValues

Substitute values

\frac{sin 4 4 ^{\circ}}{1 0 4 0} = \frac{sin A}{1 1 4 0}

Solve for

A

MultEqn

$LHS \cdot 1140 = RHS \cdot 1140$

\frac{sin 4 4 ^{\circ}}{1 0 4 0} (1140) = sin A

RearrangeEqn

Rearrange equation

sin A = \frac{sin 4 4 ^{\circ}}{1 0 4 0} (1140)

$sin^{- 1} (LHS) = sin^{- 1} (RHS)$

A = sin^{- 1} (\frac{sin 4 4 ^{\circ}}{1 0 4 0} (1140))

UseCalc

Use a calculator

A = 49.592410 \dots^{\circ}

RoundInt

Round to nearest integer

A \approx 5 0^{\circ}

Rounded to the nearest degree, it was found that the angle formed at Asunción measures

5 0^{\circ} .

Finally, to find the measure of the angle at Buenos Aires, the Triangle Angle Sum Theorem can be used.

4 4^{\circ} + 5 0^{\circ} + B = 18 0^{\circ} ⇕ B = 8 6^{\circ}

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.

\frac{b}{sin B} = \frac{s}{sin S}

Next,

s = 1040,

S = 4 4^{\circ},

and

B = 8 6^{\circ}

will be substituted into the above equation.

\frac{b}{sin B} = \frac{s}{sin S}

SubstituteValues

Substitute values

\frac{b}{sin 8 6 ^{\circ}} = \frac{1 0 4 0}{sin 4 4 ^{\circ}}

Solve for

b

MultEqn

$LHS \cdot sin 8 6^{\circ} = RHS \cdot sin 8 6^{\circ}$

b = \frac{1 0 4 0}{sin 4 4 ^{\circ}} (sin 8 6^{\circ})

UseCalc

Use a calculator

b = 1493.491846 \dots

RoundInt

Round to nearest integer

b \approx 1493

Emily has all she needs to have a great journey!

Pop Quiz

Practice Finding Angles Using the Law of Sines

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.

Example

Using the Law of Sines To Find Measurements of a Triangle

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

4 1^{\circ}

and

5 9^{\circ},

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.

Hint

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.

Solution

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

Perimeter

The perimeter of a triangle is the sum of its three side lengths. Therefore, the missing side lengths need to be found. To make things clearer, the angles and sides will first be labeled.

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 + 5 9^{\circ} + 4 1^{\circ} = 18 0^{\circ} ⇕ D = 8 0^{\circ}

The measure of the angle at Dublin is

8 0^{\circ} .

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.

\frac{ℓ}{s i n L} = \frac{d}{s i n D}

Next,

L = 4 1^{\circ},

D = 8 0^{\circ},

and

d = 530

can be substituted in the above equation, and

ℓ

can be isolated.

\frac{ℓ}{sin L} = \frac{d}{sin D}

SubstituteValues

Substitute values

\frac{ℓ}{s i n 4 1 ^{\circ}} = \frac{5 3 0}{s i n 8 0 ^{\circ}}

Solve for

ℓ

MultEqn

$LHS \cdot sin 4 1^{\circ} = RHS \cdot sin 4 1^{\circ}$

ℓ = \frac{5 3 0}{sin 8 0 ^{\circ}} (sin 4 1^{\circ})

UseCalc

Use a calculator

ℓ = 353.075292 \dots

RoundInt

Round to nearest integer

ℓ \approx 353

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
$\frac{ℓ}{s i n L} = \frac{d}{s i n D}$	$\frac{ℓ}{s i n 4 1 ^{\circ}} = \frac{5 3 0}{s i n 8 0 ^{\circ}}$	$ℓ \approx 353$
$\frac{e}{s i n E} = \frac{d}{s i n D}$	$\frac{e}{s i n 5 9 ^{\circ}} = \frac{5 3 0}{s i n 8 0 ^{\circ}}$	$e \approx 461$

The distance between London and Dublin is about

461

kilometers.

The perimeter of the triangle formed by the three cities can now be calculated.

Perimeter P = 530 + 353 + 461 ⇕ P = 1344 km

Area

The area of a triangle can be found by calculating half the product of two side lengths and the sine of their included angle. Since all angle measures and sides lengths are known, any of them can be used. In this case,

E,

d,

and

ℓ

will be arbitrarily used.

Area : \frac{1}{2} d ℓ sin E

The respective values can be substituted to find the area of the triangle.

Area = \frac{1}{2} d ℓ sin E

SubstituteValues

Substitute values

Area = \frac{1}{2} (530) (353) sin 5 9^{\circ}

Evaluate right-hand side

Multiply

Area = \frac{1}{2} (187090) sin 5 9^{\circ}

MoveRightFacToNumOne

$\frac{1}{b} \cdot a = \frac{a}{b}$

Area = \frac{1 8 7 0 9 0}{2} sin 5 9^{\circ}

CalcQuot

Calculate quotient

Area = 93545 sin 5 9^{\circ}

UseCalc

Use a calculator

Area = 80183.715144 \dots

Approximate to the nearest thousand

Area \approx 80000

The area of the triangle formed by London, Dublin, and Edinburgh is about

80000

square kilometers. Thanks for helping Emily plan her travels!

Discussion

The Ambiguous Case of the Law of Sines

The Law of Sines is useful to find missing angle measures and side lengths for any type of triangle.

Consider a triangle with vertices

A,

B,

and

C,

where the measure of the angle at

C

3 4^{\circ},

and the lengths of

\overline{A B}

and

\overline{B C}

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.

\frac{sin A}{1 0} = \frac{sin 3 4 ^{\circ}}{6}

Solve for

A

MultEqn

$LHS \cdot 10 = RHS \cdot 10$

sin A = \frac{sin 3 4 ^{\circ}}{6} (10)

CommutativePropMult

Commutative Property of Multiplication

sin A = 10 (\frac{sin 3 4 ^{\circ}}{6})

MoveLeftFacToNum

$a \cdot \frac{b}{c} = \frac{a \cdot b}{c}$

sin A = \frac{1 0 sin 3 4 ^{\circ}}{6}

ReduceFrac

$\frac{a}{b} = \frac{a / 2}{b / 2}$

sin A = \frac{5 sin 3 4 ^{\circ}}{3}

$sin^{- 1} (LHS) = sin^{- 1} (RHS)$

A = sin^{- 1} (\frac{5 sin 3 4 ^{\circ}}{3})

UseCalc

Use a calculator

A = 68.746886 \dots^{\circ}

RoundInt

Round to nearest integer

A \approx 6 9^{\circ}

Using the Law of Sines, it was found that the measure of the angle at

A

is about

6 9^{\circ} .

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.

\frac{sin 6 9 ^{\circ}}{1 0} = \frac{sin 3 4 ^{\circ}}{6}

Recall now that the sine of an angle is equal to the sine of the supplement of the angle.

sin θ = sin (18 0^{\circ} - θ)

With this information, it can be stated that the sine of an angle whose measure is

6 9^{\circ}

is equal to the sine of an angle whose measure is

18 0^{\circ} - 6 9^{\circ} = 11 1^{\circ} .

Therefore, by the Substitution Property of Equality, a new proportion can be derived.

⎩ ⎪ ⎨ ⎪ ⎧ \frac{s i n 6 9 ^{\circ}}{1 0} = \frac{sin 3 4 ^{\circ}}{6} s i n 6 9^{\circ} = s i n 11 1^{\circ} ⇓ \frac{s i n 1 1 1 ^{\circ}}{1 0} = \frac{sin 3 4 ^{\circ}}{6}

The obtained proportion implies a new triangle that satisfies the initial conditions.

Therefore, in the given context, there are two valid measures for the angle at

A,

which are

6 9^{\circ}

and

11 1^{\circ} .

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.

Closure

Calculating an Angle of a Triangle Using the Law of Sines

The challenge presented at the beginning of this lesson can be solved by using the Law of Sines.

Find the measure of the angle at

B .

Write the answer rounded to the nearest degree.

Hint

In a triangle, the ratio of the sine of an angle to the length of its opposite side is constant.

Solution

Since the ratio of the sine of an angle to the length of its opposite side is constant, a proportion can be written.

\frac{sin B}{A C} = \frac{sin A}{B C} \Rightarrow \frac{sin B}{1 5} = \frac{sin 7 0 ^{\circ}}{2 0}

The above equation can be solved for

B .

\frac{sin B}{1 5} = \frac{sin 7 0 ^{\circ}}{2 0}

Solve for

B

MultEqn

$LHS \cdot 15 = RHS \cdot 15$

sin B = \frac{sin 7 0 ^{\circ}}{2 0} (15)

$sin^{- 1} (LHS) = sin^{- 1} (RHS)$

B = sin^{- 1} (\frac{sin 7 0 ^{\circ}}{2 0} (15))

UseCalc

Use a calculator

B = 44.810922 \dots^{\circ}

RoundInt

Round to nearest integer

B \approx 4 5^{\circ}

The measure of the angle at

B

is about

4 5^{\circ} .

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Catch-Up and Review

Challenge

Investigate Triangles With Two Sides and an Angle Known

Explore

Relationship Between a Triangle's Angles and Their Opposite Sides

Discussion

Law of Sines

Proof

Acute Triangles

Obtuse Triangles

Example

Using the Law of Sines To Find the Side of a Triangle

Hint

Solution

Pop Quiz

Practice Finding Sides Using the Law of Sines

Example

Using the Law of Sines To Find Angles of a Triangle

Hint

Solution

Pop Quiz

Practice Finding Angles Using the Law of Sines

Example

Using the Law of Sines To Find Measurements of a Triangle

Hint

Solution

Perimeter

Area

Discussion

The Ambiguous Case of the Law of Sines

Closure

Calculating an Angle of a Triangle Using the Law of Sines

Hint

Solution

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