Learning activity 4.2: Exploring the sine law

Acute triangles

The trigonometric ratios from the previous learning activity can only be applied to solve problems that involve right triangles. For situations that involve oblique triangle (a triangle that does not contain a right angle), different methods are required that involve a trigonometric equation traditionally called the sine law.

In this learning activity you will learn how the sine law is used to solve acute triangles. Begin by comparing acute triangles to obtuse triangles.

An acute triangle is a non-right triangle in which all three angles are less than 90°. The following are examples of acute triangles.

Image of three different acute triangles.

The following are not acute triangles. You will notice that in each triangle there is one angle that is larger than 90°. The following are examples of obtuse triangles.

Image of two different obtuse triangles.

History of Trigonometry

Trigonometry is derived from Greek words trigonon meaning “triangle” and metron meaning “to measure”. To learn more about the the historical concepts and importance of it’s applications explore the information below.

3000 BCE

The Egyptians and Babylonians believed the triangle was a part of a rectangle and knew the relationship between the different sides in a rectangle. They used the base number 60 as a unit of measure which we call 60 degrees and divided the unit of measure.

190 -120 BCE

Hipparchus, a Greek mathematician, adopted the degree measurements and compiled a table of chords. Chords are line segments joining two points whose ends lie on a circle.
Claudius Ptolemy used the trigonometry table to observe planetary motion.

476 – 550 CE

An Indian astronomer Aryabhata discovered the circumference of the earth to the value of pi $π$ to four decimal places. He was also responsible for discovering the sine and cosine.

780 – 850 CE

Persian mathematician Musa al-Khwarizmi discovered the tangent and demonstrated how to solve quadratic equations by completing the square.

940 – 988 CE

Buzjani, a Persian mathematician, discovered the secant, cotangent, and cosecant also known as the trigonometric ratios.

Investigating sine law

Investigation 1: Relating angle measures and side lengths

In this section, you’ll investigate relationships between the angles and side lengths of acute triangles. You are encouraged to follow the instructions and answer the related questions. The solutions to the investigations are provided so that you may assess your answers and table entries.

Try it!

Consider the following triangle. Enter its angle measures and side lengths into a table like the one beside and check your answers.

Diagram of triangle A B C. Angle A is 47 degrees; Angle C is 68 degrees and Angle B is 65 degrees. Side AC is 45 metres; Side BC is 36.3 metres and side AB is 46 metres. — Press here for long description(Open in new window)

Angle (degrees)	Side length (m)
A = $47 °$	a = $36.3$
B = $65 °$	b = $45.0$
C = $68 °$	c = $46.0$

What relationship do you notice between the side length and the angle measure?

The bigger angle, the bigger the side length across from it.

You might own a circular or semicircular protractor. A protractor is a device used to measure and draw angles. Beside is a protractor with $360 °$ .

Circular protractor for complete 360 degree measurements.

Notebook

Draw your own acute triangle in your notebook. Label the vertices (or angles) $P$ , $Q$ , and $R$ . Use a ruler to measure each side to the nearest centimetre. Use a protractor to measure the angles to the nearest degree.

Angle (degrees)	Side length (m)
$P =$	$p =$
$Q =$	$q =$
$R =$	$r =$

Explore this!

If you need to refresh yourself on how to measure angles.

Think

Does the relationship you described for the previous table also hold true for triangle ABC?

Yes, the larger the angle, the larger the side length opposite that angle.

Create a conjecture (a hypothesis or prediction) about the positions of the largest angle and the smallest angle in a triangle as related to the side lengths.

One of the student’s response is: “The shortest side is opposite the smallest angle. The next shortest side is opposite the next smallest angle. The longest side is opposite the largest angle. The same holds true for $∆ P Q R$ .”

Investigation 2: Comparing sine ratios and side length ratios

Use the angles and side lengths of $∆ A B C$ to complete a table like the following.

Sine ratios		Side length ratios
$\frac{s i n A}{s i n B} =$	$0.807$	$\frac{a}{b} =$	$0.807$
$\frac{s i n B}{s i n C} =$	$0.977$	$\frac{b}{c} =$	$0.978$
$\frac{s i n A}{s i n C} =$	$0.788$	$\frac{a}{c} =$	$0.788$

Think

What relationship do you notice between the ratios in the first column and the ratios in the second column?

The ratios of side lengths are equal to the ratios of the sine of the corresponding angles. Small differences are due to rounding of measurements on the diagram.

Create a conjecture about the relationship between the sine ratios (in the first column) and the corresponding side length ratios (in the second column).

Conjecture: The ratio of the sine of the angles equals the ratio of the sides opposite the angles. The same holds true for any triangle.

In general,

$\frac{s i n A}{s i n B} = \frac{a}{b}$ , $\frac{s i n B}{s i n C} = \frac{b}{c}$ , $\frac{s i n A}{s i n C} = \frac{a}{c}$

Investigation 3: Comparing ratios of sine and side lengths

Use the angles and side lengths of $∆ A B C$ to complete a table like the following.

Sine vs side ratios		Side vs sine ratios
$\frac{s i n A}{a} =$	$0.0201$	$\frac{a}{s i n A} =$	$50$
$\frac{s i n B}{b} =$	$0.0201$	$\frac{b}{s i n B} =$	$50$
$\frac{s i n C}{c} =$	$0.0202$	$\frac{c}{s i n C} =$	$50$

Think

What relationship do you notice between the ratios in the first column?

The ratios in the first column are equal. Any small differences that you notice are due to rounding of measurements on the diagram.

What relationship do you notice between the ratios in the second column?

The ratios in the second column are equal. Any small differences that you notice are due to rounding of measurements on the diagram.

Consider a conjecture about the relationship between the ratios in the first column.

Conjecture: The ratios in the first column are equal.

Consider a conjecture about the relationship between the ratios in the second column.

Conjecture: The ratios in the second column are equal.

How are the two columns related?

Conjecture: The ratios in the second column are the reciprocals of the ratios in the first column.

In general,

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

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

The sine law

For any triangle $∆ A B C$ ,

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

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

You can use the sine law when the measure of one angle and the length of its opposite side are known in an acute triangle. If two angles and the contained side (the side between the two angles) are known, then first find the third angle and then the sine law may be used.

Using sine law to solve triangles

Think

The following exercise demonstrates how to use the sine law to find the length of unknown sides.

1. Determine the length of the indicated sides ( $a$ and $b$ ) and unknown angle $(B)$ for the triangle beside. Round to one decimal place.

Angle B

Before you use the sine law, find the measure of $B$ .

Since the sum of the angles in triangle is $180 °$ , then

$B = 180 ° - (69 ° + 61 °)$

$B = 180 ° - 130 °$

$B = 50 °$

Side a

Substitute the known values into the sine law.

$\frac{s i n 69 °}{a} = \frac{s i n 50 °}{b} = \frac{s i n 61 °}{21.5}$

$\frac{s i n 69 °}{a} = \frac{s i n 50 °}{b} = \frac{s i n 61 °}{21.5}$ Solve for $a$ .

$\frac{s i n 61 °}{21.5} = \frac{s i n 69 °}{a}$

$a ∙ s i n 61 ° = 21.5 ∙ s i n 69 °$

$a = \frac{21.5 ∙ s i n 69 °}{s i n 61 °}$

$a = 22.94$

Therefore $a = 22.9 ° c m$ .

Side b

Substitute the known values into the sine law.

$\frac{s i n 69 °}{a} = \frac{s i n 50 °}{b} = \frac{s i n 61 °}{21.5}$

Solve for $b$ .

$\frac{s i n 61 °}{21.5} = \frac{s i n 50 °}{b}$

$b ∙ s i n 61 ° = 21.5 ∙ s i n 50 °$

$b = \frac{21.5 ∙ s i n 50 °}{s i n 61 °}$

$b = 18.8$

Therefore $b = 18.8 ° c m$

the following exercise demonstrates how to use the sine law to find the measure of an unknown angle.

In $∆ A B C$ , $a = 8 ° c m$ , $c = 5 ° c m$ , and $∠ A = 70 °$ . Determine the measures of $∠ C$ and $∠ B$ and round to one decimal place.

Step 1: Draw a diagram

First, draw a diagram of the triangle, similar to the following, labelling the known quantities.

Triangle A B C. Angle A is 70 degrees, side a is 8 centimetres in length, and side c is 5 centimetres in length. — Press here for long description(Open in new window)

Step 2: Find $∠ C$

$\frac{s i n 70 °}{8} = \frac{s i n C}{5}$

$8 ∙ s i n C = 5 ∙ s i n 70 °$

$s i n C = \frac{5 ∙ s i n 70 °}{8}$

$C = {s i n}^{- 1} (\frac{5 ∙ s i n 70 °}{8})$

$C = 36.0 °$

Therefore, $C$ is $36.0 °$ .

Step 3: Find $∠ B$

You need to find $C$ first.

$B = 180 ° - (36 ° + 70 °)$

$B = 180 ° - 106 °$

$B = 74 °$

Therefore, $B = 74 °$ .

3. In $∆ X Y Z$ , $X Y = 35 ° m$ , $∠ X = 63 °$ , and $∠ Z = 50 °$ . Determine the length of the altitude(the height of the triangle) from $Z$ to $X Y$ . Draw a diagram as your first step.

Step 1: Draw a diagram

Draw the diagram similar to the following and label the given information:

Note that in a triangle, the altitude is the height and so intersects at a $90 °$ angle. Therefore, the altitude from $Z$ to $X Y$ intersects $X Y$ at a $90 °$ angle.
Draw the altitude from $Z$ to $X Y$ .
Let $P$ be the point of intersection of the altitude from $Z$ to $X Y$ .
Let $h$ represent the length of the altitude.

Triangle X Y Z is depicted. Angle Z is 50 degrees; angle X is 63 degrees and angle Y is unknown. Side XY is 35 metres. Altitude ZP is also indicated, with an unknown height and right angle indicated. — Press here for long description(Open in new window)

Step 2: Find $∠ Y$

$Y = 180 ° - (63 ° + 50 °)$

$Y = 180 ° - 113 °$

$Y = 67 °$

Therefore $Y = 67 °$ .

Step 3: Find the length of $X Z$

Let $X Z = y$ and $Y X = z$ ,

Find the length of $y$ using the sine law in $∆ X Y Z$ .

You know $Z = 50 °$ and the opposite side $z = 35 ° m$ .

$\frac{s i n Y}{y} = \frac{s i n Z}{z}$ therefore $\frac{s i n 67 °}{y} = \frac{s i n 50 °}{35}$

Solve for y.

$\frac{s i n 67 °}{y} = \frac{s i n 50 °}{35}$

$y ∙ s i n 50 ° = 35 ∙ s i n 67 °$

$y = \frac{35 ∙ s i n 67 °}{s i n 50 °}$ $y = 42.1$

$y = 42.1$

The length of $X Z$ is $42.1 ° m$

Step 4: Find the length of altitude ( $h$ )

The value of $h$ can be determined by using the right triangle $∆ X P Z$ .

Diagram of right triangle X P Z. Known angle is 63 degrees, opposite side is labelled h, and hypotenuse is 42.1 metres in length. — Press here for long description(Open in new window)

As you learned in Learning Activity 4.1, you can solve for unknown lengths of right triangles when given another side length and an angle by using the primary trigonometric ratios.

$sin 63 ° = \frac{o p p o s i t e}{h y p o t e n u s e}$

$sin 63 ° = \frac{h}{42.1}$

$h = 42.1 ∙ sin 63 °$

$h = 37.5$

The length of the altitude is $37.5 ° m$ .

Applications of sine law

It is important to notice that the sine law is used when the degree measure of an angle and the length of its opposite side are known. Sometimes two angles and the contained side (side between the two angles) are given. In this case, the third angle must be found first and then the sine law can be applied.

Notebook

You can use your notebook to complete each of the following questions using sine law. Compare your work with the suggested answers to check your understanding.

1. A chandelier is suspended from the ceiling by two chains.

One chain is $40 ° c m$ long and forms an angle of $55 °$ with the ceiling. The other chain is $52 ° c m$ long. Determine the measure of the angle that the $52 ° c m$ chain makes with the ceiling. Round your answer to two decimal places.

Step 1: Draw a diagram

Let $A$ and $B$ represent the points in the ceiling where the chains are attached. Let $C$ represent the chandelier as depicted in the following diagram.

Diagram of triangle A B C. Angle B is 55 degrees, with angles A and C unknown. Side b is 52 centimetres in length, side a is 40 centimetres in length, and side c is unknown. — Press here for long description(Open in new window)

Step 2: Solve for $∠ A$

You are required to find $A$ .

$\frac{s i n A}{a} = \frac{s i n B}{b}$

$\frac{s i n A}{40} = \frac{s i n 55 °}{52}$

$52 ∙ s i n A = 40 ∙ s i n 55 °$

$s i n A = \frac{40 ∙ s i n 55 °}{52}$

$A = {s i n}^{- 1} (\frac{40 ∙ s i n 55 °}{52})$

$A = 39.06 °$

Therefore, the $52 ° c m$ chain makes an angle of $39.06 °$ with the ceiling.

This exercise illustrates the application of the sine law in a construction problem.

2. An architect designs a cottage that is $15 ° m$ wide. The rafters holding up the roof meet at a $76 °$ angle and are equal in length. The rafters extend $0.5 ° m$ beyond the two exterior supporting walls. Determine the length of the rafters to one decimal place. Check your diagram and following solution.

Step 1: Draw a diagram

Since the two rafters are equal in length, then $∆ A B C$ is isosceles and so $A C = A B$ .

Label each of the $x$ . The length of each rafter is $x + 0.5 ° m$ .

It is also true that $∠ A C B = ∠ A B C$ . Label each of the $θ$ .

Diagram of triangle A B C representing the roof of the cottage. Sides A B and A C are extended by 0.5 metres. Angle A is 76 degrees and side C B is 15 metres in length. — Press here for long description(Open in new window)

Step 2: Find the measure of angle $θ$

$76 ° + 2 θ = 180 °$

$2 θ = 180 ° - 76 °$

$2 θ = 104 °$

$θ = 52 °$

Step 3: Solve for $x$

$\frac{s i n 52 °}{x} = \frac{s i n 76 °}{15}$

$x ∙ s i n 76 ° = 15 ∙ s i n 52 °$

$x = \frac{15 ∙ s i n 52 °}{s i n 76 °}$

$x = 12.2$

Length of the beam $= x + 0.5$

Length of the beam $= 12.2 + 0.5$

Length of the beam $= 12.7$

Therefore, the length of each rafter is $12.7 ° m$ .

The following exercise involves the use of compass measurements.

3. A sightseeing yacht leaves a mainland dock and sails $11.3 ° k m$ to an island located at a point located N 15° W off the mainland. It remains there for two days and then sails to a second island located S 40° W of the first island. If the second island is $10.4 ° k m$ from the mainland, how far apart are the two islands?

Step 1: Draw a diagram

Sketch of the path of the yacht. From the mainland, point M, it travels N 15° W.

It stops at the first island, point A.

From point A it travels S 40° W to the second island.

As shown, $A = 15 ° + 40 ° = 55 °$ .

Let $m$ represent $A B$ .

Let $θ$ represent $∠ B M A$ .

Add this information to your diagram as depicted in the following:

Triangle A B C formed from the sketch of the yacht’s journey.

Step 2: Find B

$\frac{s i n 55 °}{10.4} = \frac{s i n B}{11.3}$

$10.4 ∙ s i n B = 11.3 ∙ s i n 55 °$

$s i n B = \frac{11.3 ∙ s i n 55 °}{10.4}$

$B = {s i n}^{- 1} (\frac{11.3 ∙ s i n 55 °}{10.4})$

$B = 62.9 °$

Therefore, $B = 62.9 °$ .

Step 3: Find $θ$

$θ = 180 ° - (55 ° + 62.9 °)$

$θ = 180 ° - 117.9 °$

$θ = 62.1 °$

Therefore, $θ = 62.1 °$ .

Step 4: Solve for $m$

Solving for m, which is the distance between the two islands:

$\frac{s i n 62.1 °}{m} = \frac{s i n 55 °}{10.4}$

$m ∙ s i n 55 ° = 10.4 ∙ s i n 62.1 °$

$m = \frac{10.4 ∙ s i n 62.1 °}{s i n 55 °}$

$m = 11.2$

Therefore, the distance between the two islands is $11.2 ° k m$ .

Self-check

Rate your level of understanding from 1 (I am still confused) to 5 (I fully understand this concept) based on your results from the questions you just completed.

I am able to:

Statement	1	2	3	4	5
Identify the situations for when to use sine law
Solve triangles using sine law
Apply sine law to solve real-world examples

If there are any criteria where you rated your level of understanding a 3 or below, you should review the concepts before moving on to the next learning activity.

Math journal

At the end of the course, you will fine-tune 8 entries (two from each unit) from your math journal and submit them as your “Culminating Assessment - Math journal”(Opens in new window).

Compare when to use Pythagorean theorem, primary trigonometric ratios, and sine law. You can use a chart like the following one or create a mind map like in Learning activity 1.5.

	Pythagorean theorem	Primary trigonometric ratios	sine law
Equation(s)
When do we use it?
Example of triangle

Once you feel comfortable with the success criteria, complete the following questions to assess your progress.

Assess your understanding of the sine law

The following are some questions for you to try in order to assess your understanding of the sine law.

Try it!

Solve the following triangle. Round your answers to one decimal place.

Diagram of triangle L M N. Angle L is 54 degrees, angle M is 42 degrees, and angle N is unknown. The side opposite angle N is 9 centimetres in length. — Press here for long description(Open in new window)

Part 1: Solve for $∠ N$

$N = 180 ° - (54 ° + 42 °)$

$N = 180 ° - 96 °$

$N = 84 °$

Therefore $N = 84 °$ .

Part 2: Solve for side $l$

To determine the lengths l and m, use the following form of the sine law so that the unknowns are in the numerator:

$\frac{s i n L}{l} = \frac{s i n M}{m} = \frac{s i n N}{n}$

$\frac{s i n 54 °}{l} = \frac{s i n 42 °}{m} = \frac{s i n 84 °}{9}$

Solve for $l$ .

$\frac{s i n 54 °}{l}$ = $\frac{s i n 84 °}{9}$

$l ∙ s i n 84 ° = 9 ∙ s i n 54 °$

$l = \frac{9 ∙ s i n 54 °}{s i n 84 °}$

$l = 7.3$

Therefore, $l$ is $7.3 ° c m$ .

Part 3: Solve for side $∠ m$

$\frac{s i n 42 °}{m}$ = $\frac{s i n 84 °}{9}$

$m ∙ s i n 84 ° = 9 ∙ s i n 42 °$

$m = \frac{9 ∙ s i n 42 °}{s i n 84 °}$

$m = 6.1$

Therefore, $m$ is $6.1 ° c m$ .

Solve the following triangle. Round your answers to one decimal place.

Diagram of triangle P Q R. Angle Q is 75 degrees, with angles P and R unknown. Side r is 8.2 centimetres in length and side q is 14 centimetres in length. — Press here for long description(Open in new window)

Step 1: Solve for $∠ R$

Given that $Q$ and the length of its opposite side are known, use the sine law.

First, find $R$ because the length of its opposite side is known.

$\frac{s i n P}{p}$ = $\frac{s i n Q}{q} = \frac{s i n R}{r}$

$\frac{s i n P}{p}$ = $\frac{s i n 75 °}{14} = \frac{s i n R}{8.2}$

Solving for $R$ ,

$\frac{s i n 75 °}{14}$ = $\frac{s i n R}{8.2}$

$14 ∙ s i n R = 8.2 ∙ s i n 75 °$

$s i n R = \frac{8.2 ∙ s i n 75 °}{14}$

$R = {s i n}^{- 1} (\frac{8.2 ∙ s i n 75 °}{14})$

$R = 34.5 °$

Therefore, $R$ is $34.5 °$ .

Step 2: Solve for $∠ P$

$P = 180 ° - (75 ° + 34.5 °)$

$P = 180 ° - 109.5 °$

$P = 70.5 °$

Therefore, $P = 70.5 °$ .

Step 3: Solve for side $p$

$\frac{s i n 70.5 °}{p}$ = $\frac{s i n 75 °}{14}$

$p ∙ s i n 75 ° = 14 ∙ s i n 70.5 °$

$p = \frac{14 ∙ s i n 70.5 °}{s i n 75 °}$

$p = 13.7$

Therefore $p = 13.7 ° c m$ .

Determine the area of $∆ J K L$ given that $J = 48 °$ , $L = 69 °$ , and $k = 11.2 ° c m$ . Round your answer to one decimal place.

Step 1: Draw a diagram

Draw the diagram and label the given information:

Diagram of triangle J K L. Angle J is 48 degrees, angle L is 69 degrees, and angle K is unknown. Side k is 11.2 centimetres in length. Sides j and l are unknown. — Press here for long description(Open in new window)

Step 2: Solve for $∠ K$

$K = 180 ° - (48 ° + 69 °)$

$K = 180 ° - 117 °$

$K = 63 °$

Recall that the equation for the area of a triangle is $A r e a = \frac{(b a s e) (h e i g h t)}{2}$

$k = 11.2 ° c m$ as the base of the triangle.

Draw height from $K$ to a point on line $L J$ . Call this point $P$ .

Let $h$ represent $K P$ .

Step 3: Solve for side $l$

$\frac{s i n K}{k}$ = $\frac{s i n L}{l}$

$\frac{s i n 63 °}{11.2}$ = $\frac{s i n 69 °}{l}$

$l ∙ s i n 63 ° = 11.2 ∙ s i n 69 °$

$l = \frac{11.2 ∙ s i n 69 °}{s i n 63 °}$

$l = 11.7$

Therefore $l = 11.7 ° c m$ .

Step 4: Solve for side $h$

$sin 48 ° = \frac{o p p o s i t e}{h y p o t e n u s e}$

$sin 48 ° = \frac{h}{11.7}$

$h = 11.7 ∙ sin 48 °$

$h = 8.7$

The height of the triangle is $8.7 ° c m$ .

Step 5: Find the area

$A r e a = \frac{(b a s e) (h e i g h t)}{2}$

$A r e a = \frac{(11.2) (8.7)}{2}$

$A r e a = 48.7$

Therefore, the area of the triangle is $48.7 ° c m^{2}$ .

Assess your understanding of solving real-world problems

The following questions will help you assess your understanding of solving real-world problems.

Try it!

Be sure to try the questions on your own first before comparing them to the suggested answers.

1. A farmer’s field is in the shape of a triangle.

One side of the field is located along a river and measures 720 m. The other two sides are fenced and make angles of 45° and 55° with the river. Determine the area of the field.

Step 1: Draw a diagram

Consider that the area of a triangle is found using the formula

$A r e a = \frac{(b a s e) (h e i g h t)}{2}$

Use $c = 720 ° m$ as the base of the triangle.

Draw the height from $C$ to $A B$ . Call this point $P$ as depicted in the following diagram.

Diagram of triangle A B C with altitude drawn from C to AB labelled P. Angle A is 45 degrees, B is 55 degrees, and angle C is unknown. Side A B is 720 metres in length. Side h runs from point C to point P and makes a right angle with P. — Press here for long description(Open in new window)

Step 2: Find $∠ C$

$C = 180 ° - (45 ° + 55 °)$

$C = 180 ° - 100 °$

$C = 80 °$

Step 3: Find $a$

To determine the length of $h$ , first use the sine law to find the length of $a$ .

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

$\frac{s i n 80 °}{720} = \frac{s i n 45 °}{a}$

$a ∙ s i n 80 ° = 720 ∙ s i n 45 °$

$a = \frac{720 ∙ s i n 45 °}{s i n 80 °}$

$a = 517.0$

Therefore, $a$ is $517.0 ° m$ .

Step 4: Find $h$

$sin 55 ° = \frac{o p p o s i t e}{h y p o t e n u s e}$

$sin 55 ° = \frac{h}{517}$

$h = 517 ∙ sin 55 °$

$h = 423.5$

Therefore, the height of the triangle is $423.5 ° m$ .

Step 5: Find the area

$A r e a = \frac{(b a s e) (h e i g h t)}{2}$

$A r e a = \frac{(720) (423.5)}{2}$

$A r e a = 152 460$

Therefore, the area of the rice field is $152460 m^{2}$ .

2. A traffic light is suspended above a road with two cables, each attached to a horizontal metal beam. One cable is $4.3 ° m$ long and forms an angle of $52 °$ with the metal beam. The other cable is $5.7 ° m$ long. Determine the measure of the angle that the $5.7$ cable makes with the beam. Round your answer to one decimal place.

Step 1: Draw a diagram

Let $A$ and $B$ represent the points on the metal beam where the cables are attached.

Let $C$ represent the traffic light.

Diagram of triangle A B C. Angle B is 52 degrees, with angles A, and C unknown. Side b is 5.7 metres in length, side a is 4.3 metres in length, and side c is unknown. — Press here for long description(Open in new window)

Step 2: Find $∠ A$

Since you know the measure of B and the length of its opposite side, use the sine law.

$\frac{s i n A}{a}$ = $\frac{s i n B}{b}$

$\frac{s i n A}{4.3}$ = $\frac{s i n 52 °}{5.7}$

$5.7 ∙ s i n A = 4.3 ∙ s i n 52 °$

$s i n A = \frac{4.3 ∙ s i n 52 °}{5.7}$

$A = {s i n}^{- 1} (\frac{4.3 ∙ s i n 52 °}{5.7})$

$A = 36.5 °$

Therefore, the $5.7 ° c m$ cable makes an angle of $36.5 °$ with the metal beam.