Right Triangle TrigonometryTrigonometry is based upon ratios of the sides of right triangles.

The six trigonometric functions of a right triangle, with an acute angle , are defined by ratios of two sides of the triangle.

θ

opphyp

adjThe sides of the right triangle are:

the side opposite the acute angle ,

the side adjacent to the acute angle ,

and the hypotenuse of the right triangle.

A

A

The hypotenuse is the longest side and is always opposite the right angle.

The opposite and adjacent sides refer to another angle, other than the 90o.

Right Triangle Trigonometry

S O H C A H T O A

The trigonometric functions are: sine, cosine, tangent, cotangent, secant, and cosecant.

opp

adj

hyp

θ

sin = cos = tan =

csc = sec = cot = opp

hypadjhyp

hypadj

adjopp

oppadj

hypopp

Trigonometric Ratios

Finding the ratios

The simplest form of question is finding the decimal value of the ratio of a given angle.

Find using calculator:

1) sin 30 =

sin 30 =

2) cos 23 =

3) tan 78 =

4) tan 27 =

5) sin 68 =

Using ratios to find anglesIt can also be used in reverse, finding an angle from a ratio.To do this we use the sin-1, cos-1 and tan-1 function keys.Example:1. sin x = 0.1115 find angle x.

x = sin-1 (0.1115)x = 6.4o

2. cos x = 0.8988 find angle x

x = cos-1 (0.8988)x = 26o

sin-1 0.1115 =

2nd sin( )

cos-1 0.8988 =

2nd cos( )

Calculate the trigonometric functions for .

The six trig ratios are 4

3

5

sin =54

tan =34

sec =35

cos =53

cot =43

csc =45

cos α =54

sin α =53

cot α =34

tan α =43

csc α =35

sec α =45

What is the relationship of

α and θ?

They are complementary (α = 90 – θ)

Calculate the trigonometric functions for .

Finding an angle from a triangle

To find a missing angle from a right-angled triangle we need to know two of the sides of the triangle.

We can then choose the appropriate ratio, sin, cos or tan and use the calculator to identify the angle from the decimal value of the ratio.

Find angle C

a) Identify/label the names of the sides.

b) Choose the ratio that contains BOTH of the letters.

14 cm

6 cmC

1.

C = cos-1 (0.4286) C = 64.6o

14 cm

6 cmC

1.h

a

We have been given the adjacent and hypotenuse so we use COSINE:

Cos A = hypotenuseadjacent

Cos A = ha

Cos C = 146

Cos C = 0.4286

Find angle x2.

8 cm

3 cmx

a

o

Given adj and oppneed to use tan:

Tan A = adjacentopposite

x = tan-1 (2.6667) x = 69.4o

Tan A = ao

Tan x = 38

Tan x = 2.6667

3.

12 cm10 cmy

Given opp and hypneed to use sin:

Sin A = hypotenuseopposite

x = sin-1 (0.8333) x = 56.4o

sin A = ho

sin x = 1210

sin x = 0.8333

Cos 30 x 7 = k 6.1 cm = k

7 cm

k30o

4. We have been given the adj and hyp so we use COSINE:

Cos A = hypotenuseadjacent

Cos A = ha

Cos 30 = 7k

Finding a side from a triangle

Tan 50 x 4 = r

4.8 cm = r

4 cm

r

50o

5.

Tan A = ao

Tan 50 = 4r

We have been given the opp and adj so we use TAN:

Tan A =

Sin 25 x 12 = k

5.1 cm = k

12 cm k

25o

6.

sin A = ho

sin 25 = 12k

We have been given the opp and hyp so we use SINE:

Sin A =

x =

x

5 cm30o

1. Cos A = ha

Cos 30 =x5

30 cos5

x = 5.8 cm

4 cm

r

50o

2.

Tan 50 x 4 = r

4.8 cm = r

Tan A = aoTan 50 = 4

r

3.

12 cm10 cm

y

y = sin-1 (0.8333) y = 56.4o

sin A = ho

sin y = 1210

sin y = 0.8333

Example: Given sec = 4, find the values of the other five trigonometric functions of .

Solution:

Use the Pythagorean Theorem to solve for the third side of the triangle.

tan = = cot =115

151

15

sin = csc = =415

154

sin1

cos = sec = = 4 41

cos1

15

θ

4

1

Draw a right triangle with an angle such that 4 = sec = = .

adjhyp

14

A surveyor is standing 115 feet from the base of the Washington Monument. The surveyor measures the angle of elevation to the top of the monument as 78.3. How tall is the Washington Monument?

Applications Involving Right Triangles

Solution:

where x = 115 and y is the height of the monument. So, the height of the Washington Monument is

y = x tan 78.3

115(4.82882) 555 feet.

Fundamental Trigonometric Identities

Co function Identitiessin = cos(90 ) cos = sin(90 )sin = cos (π/2 ) cos = sin (π/2 )tan = cot(90 ) cot = tan(90 )tan = cot (π/2 ) cot = tan (π/2 )sec = csc(90 ) csc = sec(90 ) sec = csc (π/2 ) csc = sec (π/2 )

Reciprocal Identities

sin = 1/csc cos = 1/sec tan = 1/cot cot = 1/tan sec = 1/cos csc = 1/sin

Quotient Identities tan = sin /cos cot = cos /sin

Pythagorean Identities

sin2 + cos2 = 1 tan2 + 1 = sec2 cot2 + 1 = csc2