When you have a right triangle there are 5 things you can know about it..

 the lengths of the sides (A, B, and C)  the measures of the acute angles (a and b)  (The third angle is always 90 degrees)

a C B b A

If you know two of the sides, you can use the Pythagorean theorem to find the other side

A



C

2 

B

2

B



C



C

2 

A

2

A

2 

B

2

if A

 3 ,

B

 4

C



C



C



A

3 2 25 2 

B

2  4 2  5

a C B = 4 b A = 3

And if you know either angle, a or b, you can subtract it from 90 to get the other one: a + b = 90

 This works because there are 180 º in a triangle and we are already using up 90º     For example: if a = 30 º b = 90 º – 30º b = 60 º

C b A a B

But what if you want to know the angles?

 Well, here is the central insight of trigonometry:  If you multiply all the sides of a right triangle by the same number (k), you get a triangle that is a different size, but which has the same angles:

k(C) b C b A k(A) a a B k(B)

How does that help us?

 Take a triangle where angle b is 60 º and angle a is 30

º

 If side B is 1unit long, then side C must be 2 units long, so that we know that for a triangle of this shape the ratio of side B to C is 1:2  There are ratios for every shape of triangle!

C = 2 60 º A = 1 30 º B

But there are three pairs of sides possible!

 Yes, so there are three sets of ratios for any triangle  They are mysteriously named:  sin …short for sine  cos …short for cosine  tan …short or tangent  and the ratios are already calculated, you just need to use them

So what are the formulas?

sin tan  cos   

opp hyp

 

adj hyp opp adj

SOH CAHTOA

Some terminology:

 Before we can use the ratios we need to get a few terms straight  The hypotenuse (

hyp

) is the longest side of the triangle – it never changes  The opposite (

opp

) is the side directly across from the angle you are considering  The adjacent (

adj

) is the side right beside the angle you are considering

A picture always helps…

 looking at the triangle in terms of angle b

b

 A is the adjacent (near the angle)  B is the opposite (across from the angle)  C is always the hypotenuse Longest

hyp C B A b

Near

adj opp

Across

But if we switch angles…

 looking at the triangle in terms of angle a  A is the opposite (across from the angle)  B is the adjacent (near the angle)  C is always the hypotenuse Longest

a hyp C B a adj

Near

A

Across

opp

Lets try an example

       Suppose we want to find angle a what is side A?

the opposite what is side B?

the adjacent with opposite and adjacent we use the… tan formula

a

tan  

opp adj

C b B = 4 A = 3

Lets solve it

tan  tan

a



opp



adj

3  4 0 .

75 check our calculator s a  36.87º

C a B = 4 b A = 3

Another tangent example…

 we want to find angle b  B is the opposite  A is the adjacent  so we use tan tan

b

tan

b

 4 3  1 .

33

b

 53 .

13 

a C

tan  

opp adj

B = 4 b A = 3

Calculating a side if you know the angle

 you know a side (adj) and an angle (25

°

)  we want to know the opposite side tan

A

 25  tan 

A

25 6   6

A

 0 .

47  6

A

 2 .

80

25 ° C

tan  

opp adj

b A B = 6

Another tangent example

 If you know a side and an angle, you can find the other side. tan 25   6

B

tan  

opp adj B

 6 tan 25 

b

B B

 6 0 .

47  12 .

87

25 ° C B A = 6

An application

 You look up at an angle of 65 ° at the top of a tree that is 10m away  the distance to the tree is the adjacent side  the height of the tree is the opposite side tan

opp

65   10 

opp

 10 tan 65 

opp

 10  2 .

14

opp

 21 .

4

65 ° 10m

Why do we need the sin & cos?

 We use sin and cos when we need to work with the hypotenuse  if you noticed, the tan formula does not have the hypotenuse in it.  so we need different formulas to do this work  sin and cos are the ones!

C = 10 b A 25 ° B

Lets do sin first

 we want to find angle a  since we have opp and hyp we use sin sin  sin

a

 5 10 sin

a

 0 .

5

C = 10

a

 30 

a B



opp hyp

b A = 5

And one more sin example

 find the length of side A  We have the angle and the hyp, and we need the opp sin

A

 25  sin 

A

25 20   20

A

 0 .

42  20

A

 8 .

45

25 °

sin

C = 20

 

opp hyp

B b A

And finally cos

  We use cos when we need to work with the hyp and adj so lets find angle b cos  

adj hyp

cos

b

 4 10

C = 10 b

cos

b



b

 0 .

4 66 .

42  a  90  -

a

66.42



B

a  23.58



A = 4

Here is an example

 Spike wants to ride down a steel beam  The beam is 5m long and is leaning against a tree at an angle of 65 ° to the ground  His friends want to find out how high up in the air he is when he starts so they can put add it to the doctors report at the hospital  How high up is he?

How do we know which formula to use???

 Well, what are we working with?

 We have an angle  We have hyp  We need opp  With these things we will use the sin formula

B C = 5 65 °

So lets calculate

sin 65  

opp hyp

sin 65 

opp

 

opp

sin 5 65   5

opp

 0 .

91  5

opp

 4 .

53  so Spike will have fallen 4.53m

B C = 5 65 °

One last example…

 Lucretia drops her walkman off the Leaning Tower of Pisa when she visits Italy  It falls to the ground 2 meters from the base of the tower  If the tower is at an angle of 88 ° to the ground, how far did it fall?

First draw a triangle

 What parts do we have?

 We have an angle  We have the Adjacent  We need the opposite  Since we are working with the adj and opp, we will use the tan formula

88 ° 2m B

So lets calculate

tan 88  

opp adj

tan 88 

opp

 

opp

tan 2 88   2

opp

 28 .

64  2

opp

 57 .

27  Lucretia’s walkman fell 57.27m

88 ° 2m B

What are the steps for doing one of these questions?

3.

4.

5.

6.

7.

1.

2.

Make a diagram if needed Determine which angle you are working with Label the sides you are working with Decide which formula fits the sides Substitute the values into the formula Solve the equation for the unknown value Does the answer make sense?

Two Triangle Problems

 Although there are two triangles, you only need to solve one at a time  The big thing is to analyze the system to understand what you are being given  Consider the following problem:  You are standing on the roof of one building looking at another building, and need to find the height of both buildings.

Draw a diagram

 You can measure the angle

40 °

down to the base of other building and up

60 °

to the top as well. You know the distance between the two buildings is 45m

60 ° 40 ° 45m

Break the problem into two triangles.

 The first triangle:

a

 The second triangle

60 ° 45m 40 ° b

 note that they share a side 45m long  a and b are heights!

The First Triangle

 We are dealing with an angle, the opposite and the adjacent  this gives us Tan tan

a

 60  tan 

a

60 45   45 a  1.73

 a  77.94m

45

60 ° 45m a

The second triangle

 We are dealing with an angle, the opposite and the adjacent  this gives us Tan tan

b

 40  tan 

b

40  45  45 b  0.84

 45 b  37.76m

45m 40 ° b

What does it mean?

 Look at the diagram now:  the short building is 37.76m tall  the tall building is 77.94m plus 37.76m tall, which equals 115.70m tall

60 ° 40 ° 45m 77.94m

37.76m