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In this lesson, the concept of similarity will be used to understand that side ratios in right triangles are properties of the angles. This will lead to the definition of trigonometric ratios for acute angles.

Catch-Up and Review

Here are a few recommended readings before getting started.

Similar triangles
Right angle
Right triangles
Pythagorean Theorem

Try your knowledge on these topics.

a Select all the triangles that are similar to

△ A B C .

b Which of the following angles is a right angle?

c Which of the following are right triangles?

d Find the value of

x

by using the Pythagorean Theorem.

Challenge

Investigating a Side of a Right Triangle

The Leaning Tower of Pisa has a tilt of $4$ degrees. Once, a worker maintaining it accidentally dropped a hammer from the top. The hammer landed $3$ meters away from the base of the tower. Luckily, it did not hurt anyone!

What was the vertical distance traveled by the hammer?

Explore

Comparing Ratios of Sides in Similar Right Triangles

Similar triangles have congruent angles and proportional sides. In the following applet, some of the ratios of the side lengths of two similar right triangles are compared.

What conclusion can be made about the ratios of the side lengths of two similar triangles?

Discussion

Analyzing Ratios of Sides in Similar Right Triangles

Because all right angles are congruent, all right triangles have one pair of congruent angles. If they also have one pair of congruent acute angles, then the triangles have two pairs of congruent angles. Therefore, by the Angle-Angle Similarity Theorem, two triangles with one pair of congruent acute angles are similar.

A pair of similar triangles. On the left, right triangle ABC, and on the right, triangle PQR. Both have right angles at A and P, and congruent angles at B and Q.

Since corresponding sides of similar polygons are proportional, the ratios between corresponding sides of similar right triangles are the same.

\frac{A B}{B C} \frac{A C}{B C} \frac{A C}{A B} = \frac{P Q}{Q R} = \frac{P R}{Q R} = \frac{P R}{P Q}

Discussion

Trigonometric Ratios

The ratios between side lengths of right triangles depend on the acute angles of the triangle. Some of these ratios receive a special name.

A trigonometric ratio relates two side lengths of a right triangle. Consider the right triangle $△ A B C .$ One of its acute angles has been named $θ .$

Since it is opposite to the right angle, $\overline{B C}$ is the hypotenuse of the right triangle. The remaining sides — the legs — can be named relative to the marked angle $θ .$ Because $\overline{A B}$ is next to $∠ θ,$ it is called the adjacent side. Conversely, because $\overline{A C}$ lies across from $∠ θ,$ it is called the opposite side.

The names of the three main ratios between side lengths are stated in the following table.

Name	Definition	Notation
Sine of $∠ θ$	$\frac{Length of opposite side to ∠ θ}{Hypotenuse}$	$sin θ = \frac{opp}{hyp}$
Cosine of $∠ θ$	$\frac{Length of adjacent side to ∠ θ}{Hypotenuse}$	$cos θ = \frac{adj}{hyp}$
Tangent of $∠ θ$	$\frac{Length of opposite side to ∠ θ}{Length of adjacent side to ∠ θ}$	$tan θ = \frac{opp}{adj}$

Example

Understanding Trigonometric Ratios in Right Triangles

Dominika is helping Tadeo understand trigonometric ratios. She drew three right triangles for him to write trigonometric ratios with respect to the acute angle $θ .$ Help Tadeo grasp this topic by selecting the correct answers!

A right triangle with legs of 5 and 12 units and a hypotenuse of 13 units. The acute angle formed by the leg of 5 centimeters and the hypotenuse is labeled theta.

Hint

a Identify the hypotenuse of the right triangle and the opposite side to

∠ θ .

b Identify the hypotenuse of the right triangle and the adjacent side to

∠ θ .

c Identify the opposite and adjacent sides to

∠ θ .

Solution

a The sine of

∠ θ

is defined as the ratio of the length of the side

opposite

∠ θ

to the

hypotenuse

of the right triangle. Therefore, Tadeo needs to identify these two sides.

It can be seen above that the hypotenuse of the right triangle is

5

and that the length of the opposite side to

∠ θ

3 .

With this information, the sine of

∠ θ

can be written.

sin θ = \frac{3}{5}

b The cosine of

∠ θ

is the ratio of the length of the

adjacent

side to

∠ θ

to the

hypotenuse

of the right triangle. Therefore, Tadeo needs to identify these two sides.

The hypotenuse of the right triangle is

10

and the length of the adjacent side to

∠ θ

6 .

With this information, the cosine of

∠ θ

can be written.

cos θ = \frac{6}{1 0}

c The tangent of

∠ θ

is the ratio of the length of the side

opposite

∠ θ

to the length of the side

adjacent

∠ θ .

Therefore, Tadeo needs to identify these two sides.

It can be seen above that the lengths of the opposite and adjacent sides to

∠ θ

are

12

and

5,

respectively. With this information, the tangent of

∠ θ

can be written.

tan θ = \frac{1 2}{5}

Example

Explaining Trigonometric Ratios in Right Triangles

Despite the awesome explanations Dominika provided, Tadeo still does not get how to find trigonometric ratios. To help his friend, Dominika thought of one more exercise.

This time, Dominika drew one right triangle and stated its three side lengths. She also labeled one of the triangle's acute angles.

Tadeo has now been asked to find the sine, cosine, and tangent of

∠ θ .

Help him find the answers!

Hint

Start by identifying the hypotenuse of the right triangle. Then identify the opposite and adjacent sides to $∠ θ .$ Finally, recall the definitions of sine, cosine, and tangent of an acute angle of a right triangle.

Solution

To find the sine, cosine, and tangent of

∠ θ,

the

hypotenuse

of the right triangle needs to be identified. The

opposite

and

adjacent

sides to the angle also need to be identified.

It can be seen above that the hypotenuse is

17,

and that the lengths of the opposite and adjacent sides to

∠ θ

are

8

and

15,

respectively. This information can be substituted into the definitions for sine, cosine, and tangent.

Definition	Substitute
$sin θ = \frac{Length of opposite side to ∠ θ}{Hypotenuse}$	$sin θ = \frac{8}{1 7}$
$cos θ = \frac{Length of adjacent side to ∠ θ}{Hypotenuse}$	$cos θ = \frac{1 5}{1 7}$
$tan θ = \frac{Length of opposite side to ∠ θ}{Length of adjacent side to ∠ θ}$	$tan θ = \frac{8}{1 5}$

Example

Using Trigonometric Ratios to Find Side Lengths

Tadeo finally understands the topic! But wait, Dominika wants to level up and has let him know that trigonometric ratios can also be used to find missing side lengths of a right triangle. "Tell me more," Tadeo responds. An acute angle and the hypotenuse of a right triangle are given. To see whether Tadeo masters this topic, Dominika asked him to find the value of $x,$ which is the length of the opposite side to the given angle.

A right triangle has a leg of length x and a hypotenuse of length 10 units. The acute angle between the unlabeled leg and the hypotenuse is 60 degrees.

Help Tadeo find the value of

x .

If necessary, round the answer to three significant figures.

Hint

Identify the trigonometric ratio that should be used according to the given and desired lengths. Then, with the help of a calculator, set and solve an equation.

Solution

The $hypotenuse$ of the right triangle is given, and the length of the $opposite$ side to the given $angle$ is to be found.

The trigonometric ratio that relates these two sides and the acute angle is the sine ratio.

sin θ = \frac{Length of opposite side to ∠ θ}{Hypotenuse} ⇓ sin 6 0^{\circ} = \frac{x}{1 0}

Finally, the equation can be solved for

x .

sin 6 0^{\circ} = \frac{x}{1 0}

Solve for

x

MultEqn

$LHS \cdot 10 = RHS \cdot 10$

sin 6 0^{\circ} \cdot 10 = \frac{x}{1 0} \cdot 10

FracMultDenomToNumber

$\frac{a}{1 0} \cdot 10 = a$

sin 6 0^{\circ} \cdot 10 = x

Multiply

10 sin 6 0^{\circ} = x

RearrangeEqn

Rearrange equation

x = 10 sin 6 0^{\circ}

To find the value of

sin 6 0^{\circ},

a calculator can be used. First, it must be set to degree mode. This is done by pushing

M O D E

and selecting Degree in the third row.

Next the value of $sin 6 0^{\circ}$ can be calculated by pushing $S I N$ followed by the angle measure.

Now the value of

x

can be calculated.

x = 10 sin 6 0^{\circ}

UseCalc

Use a calculator

x = 10 (0.866025 \dots)

Multiply

x = 8.660254 \dots

RoundSigDig

Round to $3$ significant digit(s)

x \approx 8.66

Pop Quiz

Practice Finding Side Lengths Using Trigonometric Ratios

In the right triangles below, one acute angle and one side length are given. By using the corresponding trigonometric ratio, find the length of the side labeled $x .$ Round the answer to one decimal place.

Discussion

Pythagorean Identities

By using trigonometric ratios, an important property of angles can be derived.

For any angle $θ,$ the following trigonometric identities hold true.

sin^{2} θ + cos^{2} θ = 1

Proof

For Acute Angles

Consider a right triangle with a hypotenuse of

1 .

By recalling the sine and cosine ratios, the lengths of the opposite and adjacent sides to

∠ θ

can be expressed in terms of the angle.

	Definition	Substitute	Simplify
$sin θ$	$\frac{Length of opposite side to ∠ θ}{Hypotenuse}$	$\frac{opp}{1}$	$opp$
$cos θ$	$\frac{Length of adjacent side to ∠ θ}{Hypotenuse}$	$\frac{adj}{1}$	$adj$

It can be seen that if the hypotenuse of a right triangle is $1,$ the sine of an acute angle is equal to the length of its opposite side. Similarly, the cosine of the angle is equal to the length of its adjacent side.

By the Pythagorean Theorem, the sum of the squares of the legs of a right triangle is equal to the square of the hypotenuse. Therefore, for the above triangle, the sum of the squares of $s i n θ$ and $c o s θ$ is equal to the square of $1 .$

s i n^{2} θ + c o s^{2} θ ⇓ sin^{2} θ + cos^{2} θ = 1^{2} = 1

Therefore, for any acute angle

θ,

the sum of the squares of its sine and cosine equals

1 .

This property is also valid for any angle. The proof for angles whose measure is greater than or equal to

9 0^{\circ}

will be seen later in this course.

Example

Using Trigonometry to Determine the Cosine of an Angle

The property seen before can be used, among other things, to find the sine or cosine ratio of an acute angle in a right triangle.

Kriz and his friends plan to spend Saturday afternoon playing video games. To optimize the space, they decide to tidy up the basement to ensure the console, snacks, and beverages are placed in the form of a right triangle. Kriz decides to set the snacks and the beverages

3

and

5

meters away from the console, respectively.

The adjacent side to

∠ θ

connects the snacks and the beverages. To guarantee a good flow between the snacks and the beverages, Kriz wants to find the cosine of

∠ θ .

Help the gang have a good time by finding

cos ∠ θ

for them!

Hint

The hypotenuse of the right triangle is $5$ and the measure of the opposite side to $∠ θ$ is $3 .$ With this information, the sine ratio can be found.

Solution

The hypotenuse of the right triangle is

5

and the measure of the opposite side to

∠ θ

3 .

With this information, the sine ratio can be found.

sin θ = \frac{Length of opposite side to θ}{Hypotenuse} ⇓ sin θ = \frac{3}{5}

This value can be substituted in the equation

sin^{2} θ + cos^{2} θ = 1 .

sin^{2} θ + cos^{2} θ = 1

Substitute

$sin θ = \frac{3}{5}$

(\frac{3}{5})^{2} + cos^{2} θ = 1

Solve for

cos θ

PowQuot

$(\frac{a}{b})^{m} = \frac{a ^{m}}{b ^{m}}$

\frac{9}{2 5} + cos^{2} θ = 1

SubEqn

$LHS - \frac{9}{2 5} = RHS - \frac{9}{2 5}$

cos^{2} θ = 1 - \frac{9}{2 5}

OneToFrac

Rewrite $1$ as $\frac{2 5}{2 5}$

cos^{2} θ = \frac{2 5}{2 5} - \frac{9}{2 5}

SubFrac

Subtract fractions

cos^{2} θ = \frac{1 6}{2 5}

SqrtEqn

$LHS = RHS$

cos θ = \frac{1 6}{2 5}

SqrtQuot

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

cos θ = \frac{4}{5}

Note that, when solving the equation for

cos θ,

only the principal root was considered. The reason is that the cosine of

∠ θ

is the ratio between two side lengths, and side lengths are always positive. Therefore, the quotient is also positive.

Explore

Using Inverse Trigonometric Ratios

Trigonometric ratios can also be used to find missing angles. Consider a right triangle $△ A B C$ where the hypotenuse and a leg are given.

A right triangle ABC with right angle at vertex A. The vertical leg AB measures 5 units and the hypotenuse BC is 12 units.

Suppose now that the measure of

∠ C

is desired. Note that, apart from the

hypotenuse,

the side whose length is known is

opposite

∠ C .

The trigonometric ratio that relates the

hypotenuse

and the

opposite

side to an acute angle in a right triangle is the sine ratio.

sin ∠ C = \frac{length of opposite side to ∠ C}{h y p o t e n u s e} ⇓ sin ∠ C = \frac{5}{1 2}

To find the measure of

∠ C,

the inverse of the sine ratio could be used.

sin ∠ C = \frac{5}{1 2} ⇕ m ∠ C = sin^{- 1} \frac{5}{1 2}

Finally, to find the value of

sin^{- 1} \frac{5}{1 2}

and therefore the measure of

∠ C,

a calculator will help. In the following example, it will be shown how to use a calculator to find the value of an inverse trigonometric ratio.

Example

Calculating Angles of Right Triangles

Previously, it was said that apart from being useful to find side lengths of a right triangle, trigonometric ratios can also be used to find missing angle measures.

Before playing video games with his friends, Kriz wants to finish his math homework to have a care-free weekend. He wants to find the measure of an acute angle in three different right triangles. By using the corresponding trigonometric ratios, help Kriz find $m ∠ θ$ in each triangle. Round the answer to the nearest degree.

Hint

a The lengths of the adjacent and opposite sides to

∠ θ

are

12

and

35,

respectively.

b The hypotenuse of the right triangle is

25

and the length of the adjacent side to

∠ θ

7 .

c The hypotenuse of the right triangle is

29

and the length of the opposite side to

∠ θ

20 .

Solution

a In the given diagram, it can be seen that the lengths of the opposite and adjacent sides to

∠ θ

are

35

and

12,

respectively. The trigonometric ratio that relates these two sides is the tangent ratio.

tan θ = \frac{length of opposite side to ∠ θ}{length of adjacent side to ∠ θ} ⇓ tan θ = \frac{3 5}{1 2}

To solve this equation, the inverse of the tangent function could be used.

tan θ = \frac{3 5}{1 2} ⇕ m ∠ θ = tan^{- 1} \frac{3 5}{1 2}

To find the value of

tan^{- 1} \frac{3 5}{1 2},

a calculator should to be used. First, the calculator must be set in degree mode. This is done by pushing

M O D E

and selecting Degree in the third row.

Next the value of $tan^{- 1} \frac{3 5}{1 2},$ can be calculated by pushing $2 N D,$ followed by $T A N,$ and $35 / 12 .$

Thereofre,

m ∠ θ \approx 7 1^{\circ} .

b In the given diagram, it is shown that the length of the adjacent side to

∠ θ

7

and that the hypotenuse of the right triangle is

25 .

The trigonometric ratio that relates these two sides is the cosine ratio.

cos θ = \frac{length of adjacent side to ∠ θ}{hypotenuse} ⇓ cos θ = \frac{7}{2 5}

To solve this equation, the inverse of the cosine function is needed.

cos θ = \frac{7}{2 5} ⇕ m ∠ θ = cos^{- 1} \frac{7}{2 5}

To find the value of

cos^{- 1} \frac{7}{2 5},

a calculator should be used. Just like before, the calculator must be set in degree mode. This is done by pushing

M O D E

and selecting Degree in the third row.

Next the value of $cos^{- 1} \frac{7}{2 5},$ is calculated by pushing $2 N D,$ followed by $C O S,$ and $7 / 25 .$

It was found that

m ∠ θ \approx 7 4^{\circ} .

c In the diagram, it can be seen that the hypotenuse of the right triangle is

29

and that the length of the opposite side to

∠ θ

20 .

The trigonometric ratio that relates these two sides is the sine ratio.

sin θ = \frac{length of opposite side to ∠ θ}{hypotenuse} ⇓ sin θ = \frac{2 0}{2 9}

To solve this equation, the inverse of the sine function can be used.

sin θ = \frac{2 0}{2 9} ⇕ m ∠ θ = sin^{- 1} \frac{2 0}{2 9}

To find the value of

sin^{- 1} \frac{2 0}{2 9},

a calculator should be used. Just like in Parts A and B, the calculator must be set in degree mode by pushing

M O D E

and selecting Degree in the third row.

Next the value of $sin^{- 1} \frac{2 0}{2 9},$ can be calculated by pushing $2 N D,$ followed by $S I N,$ and $20 / 29 .$

Thereofre,

m ∠ θ \approx 4 4^{\circ} .

Pop Quiz

Practice Finding Angles Using Trigonometric Ratios

In the following right triangles, two side lengths are given. By using the corresponding trigonometric ratio, find $m ∠ θ .$ Round the answer to nearest degree.

Discussion

Reciprocal Trigonometric Ratios

Apart from the sine, cosine, and tangent ratios, there are three other trigonometric ratios that are worth mentioning.

Consider the right triangle $△ A B C .$

The so called reciprocal ratios are written in the next table.

Name	Definition	Notation
Cosecant of $∠ θ$	$\frac{Hypotenuse}{Length of opposite side to ∠ θ}$	$csc θ = \frac{hyp}{opp}$
Secant of $∠ θ$	$\frac{Hypotenuse}{Length of adjacent side to ∠ θ}$	$sec θ = \frac{hyp}{adj}$
Cotangent of $∠ θ$	$\frac{Length of adjacent side to ∠ θ}{Length of opposite side to ∠ θ}$	$cot θ = \frac{adj}{opp}$

These ratios can be defined in terms of sine, cosine, and tangent.

Rule

Reciprocal Identities

The trigonometric ratios cosecant, secant, and cotangent are reciprocals of sine, cosine, and tangent, respectively.

$csc θ = \frac{1}{sin θ}$

$sec θ = \frac{1}{cos θ}$

$cot θ = \frac{1}{tan θ}$

Proof

Consider a right triangle with the three sides labeled with respect to an acute angle $θ .$

Next, the sine, cosine, tangent, cosecant, secant, and cotangent ratios are written.

sin θ = \frac{opp}{hyp} cos θ = \frac{adj}{hyp} tan θ = \frac{opp}{adj} csc θ = \frac{hyp}{opp} sec θ = \frac{hyp}{adj} cot θ = \frac{adj}{opp}

The reciprocal of the sine ratio will now be calculated.

sin θ = \frac{opp}{hyp}

Solve for

\frac{1}{sin θ}

MultEqn

$LHS \cdot \frac{1}{sin θ} = RHS \cdot \frac{1}{sin θ}$

1 = \frac{opp}{hyp} (\frac{1}{sin θ})

MultEqn

$LHS \cdot \frac{hyp}{opp} = RHS \cdot \frac{hyp}{opp}$

\frac{hyp}{opp} = \frac{1}{sin θ}

RearrangeEqn

Rearrange equation

\frac{1}{sin θ} = \frac{hyp}{opp}

It has been found that

\frac{1}{s i n θ},

which is the reciprocal of

sin θ,

is equal to

\frac{hyp}{opp} .

By the definition, the cosecant of

θ

is also the ratio of the lengths of the hypotenuse and the opposite side to

∠ θ .

Therefore, by the Transitive Property of Equality,

\frac{1}{s i n θ}

is equal to

csc θ .

⎩ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎧ \frac{1}{sin θ} = \frac{hyp}{opp} . csc θ = \frac{hyp}{opp} ⇓ csc θ = \frac{1}{sin θ}

By following a similar procedure, the other two identities for secant and cotangent can be proven.

Explore

Finding Reciprocal Identities

If the sine, cosine, and tangent ratios are known, then their reciprocals cosecant, secant, and cotangent can be calculated without too much effort.

LaShay is really good at her favorite subject, Geometry. She has been appointed by Jefferson High's principal to do some tutoring for some of her classmates after school. To do so, she drew a right triangle. She then asked her peers to find all six trigonometric ratios with respect to the marked angle $θ .$

Help LaShay's classmates find the trigonometric ratios!

Hint

Identify the hypotenuse of the right triangle and the opposite and adjacent sides to $∠ θ .$

Solution

The $hypotenuse$ of the right triangle and the $opposite$ and $adjacent$ sides to $∠ θ$ will be identified.

It can be seen that the hypotenuse is

101

and the lengths of the opposite and adjacent sides to

∠ θ

are

20

and

99,

respectively. With this information, the sine, cosine, and tangent ratios can be found.

sin θ cos θ tan θ = \frac{2 0}{1 0 1} = \frac{9 9}{1 0 1} = \frac{2 0}{9 9}

The reciprocals of the above ratios are the cosecant, secant, and cotangent of

∠ θ .

sin θ = \frac{2 0}{1 0 1} cos θ = \frac{9 9}{1 0 1} tan θ = \frac{2 0}{9 9} \Rightarrow csc θ = \frac{1 0 1}{2 0} \Rightarrow sec θ = \frac{1 0 1}{9 9} \Rightarrow cot θ = \frac{9 9}{2 0}

Closure

Calculating a Side of a Right Triangle

With the topics learned in this lesson, the challenge presented at the onset can now be solved. Previously, it was learned that the Leaning Tower of Pisa has a tilt of

4

degrees. The hammer dropped by the worker landed

3

meters away from the base of the tower.

With the given information, the vertical distance traveled by the hammer can be calculated. Write the answer to the nearest tenth.

Hint

Draw a right triangle and identify the given information.

Solution

If the tower has a tilt of

4^{\circ},

then the acute angle formed by the ground and the tower itself is the difference between

9 0^{\circ}

and

4^{\circ} .

9 0^{\circ} - 4^{\circ} = 8 6^{\circ}

A right triangle can be drawn with an acute angle whose measure is

8 6^{\circ} .

Furthermore, the length of the

a d j a c e n t

side to this angle is

3

meters, and the length of its

opposite

side is unknown.

The trigonometric ratio that relates an angle of a right triangle with its

opposite

and

a d j a c e n t

sides is the tangent ratio.

tan θ = \frac{length of opposite side to ∠ θ}{length of adjacent side to ∠ θ} ⇓ tan 8 6^{\circ} = \frac{x}{3}

This equation can be solved for

x,

which is the vertical distance traveled by the hammer.

tan 8 6^{\circ} = \frac{x}{3}

Solve for

x

MultEqn

$LHS \cdot 3 = RHS \cdot 3$

tan 8 6^{\circ} \cdot 3 = x

RearrangeEqn

Rearrange equation

x = tan 8 6^{\circ} \cdot 3

UseCalc

Use a calculator

x = 42.901998 \dots

RoundDec

Round to $1$ decimal place(s)

x \approx 42.9

The distance traveled by the hammer is about

42.9

meters.

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Catch-Up and Review

Hint

Solution

Hint

Solution

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Solution

Proof

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Solution

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Proof

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Solution

Hint

Solution