Understanding the Law of Cosine and Sine Law in Applications

Substitute values

80/sin 57 ^(∘) = c/sin 37 ^(∘)

▼

Solve for c

LHS * sin 37 ^(∘)=RHS* sin 37 ^(∘)

80/sin 57 ^(∘) (sin 37 ^(∘))= c

Use a calculator

57.406571... = c

Rearrange equation

c = 57.406571...

Round to 1 decimal place(s)

c ≈ 57.4

Therefore, the height of the Leaning Pisa Tower would be around 57.4 meters if it was standing straight.

Pop Quiz

Measuring Trees Using the Law of Sines

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.

Interactive graph showing different trees and triangles

Discussion

Extending the Definition of Sine and Cosine to Obtuse Angles

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.

Rule

Sine and Cosine of Supplementary Angles

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

sin (180 ^(∘) - θ) & = sin θ cos (180 ^(∘) - θ) & = - cos θ

The following graph verifies both identities for different angles.

Graph comparing the sine and cosine of two supplementary angles

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

sin (π - θ) & = sin θ cos (π - θ) & = - cos θ

As shown, these identities allow the definition of the value of the sine and cosine of obtuse angles and the application the Law of Sines and the Law of Cosines to any triangle.

Example

Calculating the Perimeter of a Piece of Land

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.

Help Ignacio find the perimeter of the piece of land, rounded to nearest integer.

Hint

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.

Solution

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^2-2accos B

Substitute values

b^2 = 400^2+ 560^2-2(400)( 560)cos 80^(∘)

▼

Solve for b

Calculate power

b^2 = 160 000+ 313 600-2(400)(560)cos 80^(∘)

AddTerms

Add terms

b^2 = 473 600 -2(400)(560)cos 80^(∘)

Multiply

b^2 = 473 600 - 448 000 cos 80^(∘)

SqrtEqn

sqrt(LHS)=sqrt(RHS)

b = sqrt(473 600 - 448 000 cos 80^(∘))

Use a calculator

b =629.130842 ...

Round to nearest integer

b ≈ 629

When solving the above equation, only the principal root was considered because side lengths are always positive. Therefore, the length of AC is about 629 meters.

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.

a/sin A = d/sin D

Substitute values

a/sin 21 ^(∘) = 629/sin 120 ^(∘)

▼

Solve for a

LHS * sin 21^(∘)=RHS* sin 21^(∘)

a = 629/sin 120 ^(∘) (sin 21 ^(∘))

Use a calculator

a = 260.285020 ...

Round to nearest integer

a ≈ 260

Therefore, the length of DC is about 260 meters. Next, the length of AD can be calculated by following the same procedure.

c/sin C = d/sin D

Substitute values

c/sin 39 ^(∘) = 629/sin 120 ^(∘)

▼

Solve for c

LHS * sin 39^(∘)=RHS* sin 39^(∘)

c = 629/sin 120 ^(∘)(sin 39 ^(∘))

Use a calculator

c = 457.079577...

Round to nearest integer

c ≈ 457

Now, all the sides of the piece of land are known.

Finally, the perimeter will be calculated by adding all the side lengths. P = 457 m+560 m + 400 m + 260 m ⇕ P = 1677 m The perimeter of the piece of land is about 1677 meters.

Pop Quiz

Law of Cosines in Soccer Practice

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.

triangle with three known sides formed by soccerplayer and the goal net

External credits: @kdekiara

Example

Locating of a Smartphone Using Triangulation

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.

Iteractive diagram showing houses and cellphone towers working

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.

Hint

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

Solution

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 -2bc cos A

Substitute values

2.8^2 = 6^2 + 4.5^2 -2( 6)( 4.5) cos A

▼

Solve for A

Calculate power

7.84 = 36 + 20.25 -2(6)(4.5) cos A

AddTerms

Add terms

7.84 = 56.25 -2(6)(4.5) cos A

Multiply

7.84 = 56.25 -54 cos A

SubEqn

LHS-56.25=RHS-56.25

- 48.41 = - 54 cos A

DivEqn

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

0.896481... = cos A

Rearrange equation

cos A = 0.896481...

cos^(-1)(LHS) = cos^(-1)(RHS)

A=cos ^(- 1)0.896481...

Use a calculator

A=26.300639... ^(∘)

Round to nearest integer

A≈ 26 ^(∘)

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.

sinθ = Opposite Leg/Hypotenuse

Substitute values

sin 26 ^(∘) = V/4.5

▼

Solve for V

LHS * 4.5=RHS* 4.5

(4.5) sin 26 ^(∘) = V

Use a calculator

1.972670... = V

Round to nearest integer

2 ≈ V

Rearrange equation

V ≈ 2

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.

cosθ = Adjacent Leg/Hypotenuse

Substitute values

cos 26 ^(∘) = H/4.5

▼

Solve for H

LHS * 4.5=RHS* 4.5

(4.5) cos 26 ^(∘) = H

Use a calculator

4.044573... = H

Round to nearest integer

4 ≈ H

Rearrange equation

H ≈ 4

The horizontal distance from the left tower to the burglar's location is about 4 blocks, and the vertical distance is about 2 blocks.

Example

Finding the Altitude of a Helicopter

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?

Oblique triangle with a helicopter and two antennas on its vertices

Hint

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

Solution

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.

a/sin A= b/sin B

Substitute values

30/sin 95^(∘)= b/sin 45^(∘)

▼

Solve for b

LHS * sin 45^(∘)=RHS* sin 45^(∘)

30/sin 95^(∘)(sin 45^(∘))= b

Use a calculator

21.294234... = b

Rearrange equation

b = 21.294234...

Round to 1 decimal place(s)

b ≈ 21.3

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.

sin θ = opposite leg/hypotenuse

Substitute values

sin 40^(∘) = h/21.3

▼

Solve for h

LHS * 21.3=RHS* 21.3

sin 40^(∘)(21.3) = h

Use a calculator

13.691376 ... = h

Rearrange equation

h = 13.691376 ...

Round to 1 decimal place(s)

h ≈ 13.7

The helicopter is flying at an altitude of about 13.7 kilometers.

Discussion

Connection Between the Law of Sines and the Circumcircle of a Triangle

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.

Rule

Extended Form of the Law of Sines

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.

a/sin A=b/sin B=c/sin C = D

Proof

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.

Part I: Proportional Sides

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. a/sin A=b/sin B=c/sin C

Part II: Diameter of the Circumscribed Circle

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=1/2 mBC mBC=m∠COB ⇓ m∠A=1/2m∠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. sin α & = a/2/R [0.5em] & = a/2 R Rearrange sin α by substituting α= A.

sin α = a/2R

Substitute

α= A

sin A = a/2R

▼

Solve for 2R

LHS * 2R=RHS* 2R

2Rsin A = a

DivEqn

.LHS /sin A .=.RHS /sin A .

2R=a/sin A

Rearrange equation

a/sin A=2R

Since R represents the radius, 2 R is equal to the diameter D of the circle. a/sin A=2 R ⇒ a/sin A= D

Conclusion

Combining the results of Part I and Part II, the extended form of the Law of Sines can be obtained. a/sin A=b/sin B=c/sin C a/sin A= D [0.5em] ⇓ [0.5em] a/sin A=b/sin B=c/sin C= D

Closure

Calculating the Area of a Circle Circumscribed Around a Non-Right Triangle

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.

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

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

Hint

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

The area of a circle is A = π r^2, where r is the radius.

Solution

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 -2bccos A

Substitute values

a^2 = 4.9^2 + 4.6^2 -2( 4.9)( 4.6)cos 44^(∘)

▼

Solve for a

SqrtEqn

sqrt(LHS)=sqrt(RHS)

a = sqrt(4.9^2+ 4.6^2 - 2(4.9)(4.6) cos 44^(∘))

Calculate power

a = sqrt(24.01+ 21.16 - 2(4.9)(4.6) cos 44^(∘))

AddTerms

Add terms

a = sqrt(45.17 - 2(4.9)(4.6) cos 44^(∘))

Multiply

a = sqrt(45.17 - 45.08 cos 44^(∘))

Use a calculator

a = 3.569616 ...

Round to 1 decimal place(s)

a ≈ 3.6

When solving the above equation, only the principal root was considered. This is because a side length is always positive. Therefore, the missing side length is about 3.6 units.

Now that the measure of ∠ A and the length of its opposite side a are known, the desired ratio can be calculated.

a/sin A

SubstituteII

a= 3.6, A= 44 ^(∘)

3.6/sin 44 ^(∘)

▼

Evaluate

Use a calculator

5.182403...

Round to 1 decimal place(s)

5.2

The ratio of a side length to the sine of its opposite angle is about 5.2. Therefore, by the extended form of the Law of Sines, the length of the circle's diameter is about 5.2. Recall that the radius of a circle is half its diameter. Radius: 5.2/2=2.6 units With this information, the area of the circle can be calculated.

A = π r^2

Substitute

r= 2.6

A=π ( 2.6)^2

▼

Simplify right-hand side

Calculate power

A=π (6.76)

CommutativePropMult

Commutative Property of Multiplication

A=6.76 π

Use a calculator

A = 21.237166...