The Three Basic Trigonometric Ratios

Trigonometry is defined as the branch of mathematics that deals with relationsbetween sides and angles of triangles. The word trigonometry come from the Greekword trigonon "triangle" (from tri- "three" + gonia "angle") and word metron "ameasure".

Trigonometry has its foundations in two geometric concepts: The PythagoreanTheorem and Similar Triangles.

The Pythagorean Theorem

If you are given a right angle triangle, the square of the longest side(hypotenuse) is equal to the sum of the squares of the other two sides.

The Pythagorean Theorem is usually written as: 2 2 2c a b

Similar Triangles

If two triangles ABC and DEF whose corresponding angles are equal, then

AB BC ACDE EF DF

or , ,AB DE AB DE AC DFBC EF AC DF BC EF

The ratios of specific sides on the Right Angle Triangle have special names. The samenamed ratio on all other Right Angle Triangles all have the same value.

A

B C

x

y z

D

E F

x

y z

ca

b

Definition of the Three Basic Trigonometric Functions

Sine Cosine Tangent

sinOpposite O

Hypotenuse H cos

Adjacent AHypotenuse H

tanOpposite OAdjacent A

SOH CAH TOA

Note: The three orange words on the bottom row, SOH-CAH-TOA produce aword that can assist you in remembering the definitions of Sine, Cosine,and Tangent.

Note: The side opposite the angle θ is called Opposite, the side joining theangle with the hypotenuse is called the Adjacent. The θ angle cannot be the right angle.

Example

Solve for the length of x and y in the following diagram.

Solution:

Both triangles are similar ( AA

The corresponding sides have t

Thus, 40 14014040

3.5

x

x

HypotenuseOpposite

Adjacent

4

28

)

he following ratio:40 285 x

0

x5

The symbol represents similarity.

Example

Determine the value of x and y to one decimal place.

Solution:

Triangle ABC ADE AA

Let’s find the ratio of ABC to ADE

14 9.5 23.514 14

ADAB

Now,

15 23.514

23.5 2108.9

DE ADBC AB

xxx

, and

21 23.521 14

294 14 493.514 199.5

14.25

AE ADAC AB

y

yyy

Example

Evaluate the three trigonometric ratios of the angle as shown below.

Solution:

5sin 0.38461312cos 0.9231135tan 0.4167

12

OHAHOA

135

12

14 21

15

9.5x

y

A

B

C

D

E

Example

Solve (find the value of all sides and angles) the following triangle

Solution:

The missing angle is 90 28 62

Side “a”: 7.2sin 28

7.20.46947156

15.3

OH a

aa

Side “b”:

22 2

2 2

2

7.2

15.3 51.84

234.09 51.84182.2513.5

a b

b

b

b

Example

From the top of a building 80m high, the angle of depression to the base of a secondbuilding is 32. From the same point, the angle of elevation to the top of the secondbuilding is 10. Calculate the height of the second building.

28

7.2

a

b

Solution:

We need to determine the value of “a”, but to do this we first need to find thedistance RP.

In PST ,

80tan 32

80tan 32

128.0268

OA ST

ST

Now, in PQR ,

tan 10128.0268tan 10 128.0268

22.575

a

a

The height of the building is then 80 22.575 102.575 m

Example

A tree casts a shadow 3 times its height. Find the elevation of the sun

Solution:

Let represent the angle of elevation of the Sun.

Then

1

1tan3 3

1tan 303

hh

The angle of elevation is 30

By Alternate AngleTheorem , 32PST

Value of b, the heightof first building

3h

h

A

B C

The special angle

triangle 1, 2, 3

Example

In ABC , we have 90C , c=6 cm and a=4 cm. Find

a) the length of bb) the values of sin(A), cos(A), and tan(A)c) the values of sin(B), cos(B), and tan(B)

Solution:

By using the Pythagorean Theorem, we can determine b

2 2 2

2

2

6 4

36 16

36 1620

20

2 5

b

b

b

b

b

For Angle A, the adjacent side measures 2 5 and the opposite side measures 4

sin

4623

oppositeAhypotenuse

cos

2 565

3

adjacentAhypotenuse

tan

42 525

2 55

oppositeAadjacent

6 cm

A

C B

b

4 cm

Opposite

4

Adj

ace

nt

Hypotenuse62 5

A

C B

20 4 5

4 5

2 5

For Angle B, the adjacent side measures 4 and the opposite side measures 2 5

sin

2 565

3

oppositeB

hypotenuse

cos

4623

adjacentB

hypotenuse

tan

2 545

2

oppositeB

adjacent

Note: 2sin cos3

A B and 5cos sin3

A B

Example

A satellite is orbiting 172 miles above the earth’s surface. (See diagram) When it isdirectly above the point T, the angle S is found to be 73.5°. Find the radius of theearth (correct to three significant figures).

Solution:

Let the radius of the earth be r miles.

Then:

sin 73.5172

172 sin 73.5

sin 73.5 172 sin 73.5

172 sin 73.5 sin 73.5

1 sin 73.5

172 sin 73.51 sin 73.5

4004.8

rr

r r

r r

r r

r

r

r

Hence the radius of the earth is approximately 4000 miles (correct to threesignificant figures.)

Adjacent

4

Op

posi

te

Hypotenuse62 5

A

C B

Radius, r

r

T

73.5°

172 mi

C

S

(r + 112) mi

Example

An airplane is flying at an altitude of 5325 ft, directly above a straight stretch ofhighway along which a car and a bus are traveling towards each other on oppositesides of the highway. The vehicles are on opposite sides of the airplane, the car atan angle of depression of 41.6° and the bus at an angle of depression of 57.1° fromthe plane. How far apart in miles are the vehicles correct to 3 decimal places?

Solution:

Let the distance the car is to the left of the point directly below the plane be x ft, andthe distance the bus is to the right of the point directly below the plane be y ft

Then the car and the bus are (x + y) ft apart.

We will treat each Right triangle separately and use basic trigonometric ratios tosolve for x and y. Since we have the measure of the angle and its adjacent side, andwe wish to find the opposite side, we use the tangent ratio.

tan 41.653255325 tan 41.6

4727.75623165

x

x

tan 57.153255325 tan 57.1

8231.19677122

y

y

4727.75623165 8231.1967712212958.9530029

x yft

12958.95300295280

2.454347159642.454 mile

Therefore the vehicles are approximately 2.454 miles apart

41.6° 57.1°

5325 ft

yx

There are 5280 ftin a mile

Example

From a point 244.0 meters from the base of a building, the angle of elevation to thetop of the building is 15.8°. The angle of elevation from the same point to the tip of aantenna on top of this building is 19.5°. What is the height of the antenna to 2decimal places?

Solution:

Let the height of the building be h meters an

Then:

tan 15.824469.045040

h

h

and

tan 19.5244

236 tan 19.

86.40493

x h

x h

x h

Solving for x:

86.4049386.40493 69.0450417.35989117.36

x h

Hence the pole is approximately 17.36 m tall

15.8°19.5°

244 m

x

h

d the height of the an

5

0

tenna be x meters.