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.

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)

AC

B

a

b

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

22

22

22

BAC

ACB

BCA

A = 3C

B = 4

a

b

525

43

4,3

22

22

C

C

BAC

BAif

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 180º in a triangle and we are already using up 90º

For example: if a = 30º b = 90º – 30º b = 60º

AC

B

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(A)

k(C)

k(B)

a

b

AC

B a

b

How does that help us? Take a triangle where angle b is 60º and

angle a is 30º 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!

A = 1

C = 2

B

30º

60 º

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 sinecos…short for cosinetan…short or tangent and the ratios are already calculated, you just

need to use them

So what are the formulas?

hyp

oppsin

hyp

adjcos SOH

adj

opptan

CAHTOA

Sin is Opposite over HypotenuseCos is Adjacent over HypotenuseTan is Opposite over Adjacent

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

AC

B

b

adjhyp

opp

b C is always the hypotenuse

A is the adjacent (near the angle)

B is the opposite (across from the angle)

LongestNear

Across

But if we switch angles…

looking at the triangle in terms of angle a

AC

B

a

opphyp

adja

C is always the hypotenuse

A is the opposite (across from the angle)

B is the adjacent (near the angle)

LongestAcross

Near

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

adj

opptan

A = 3C

B = 4

a

b

Lets solve it

adj

opptan

A = 3C

B = 4

a

b

75.04

3tan a

scalculatorour check

36.87º a

Another tangent example…

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

adj

opptan

A = 3C

B = 4

a

b

13.53

33.1tan3

4tan

b

b

b

Calculating a side if you know the angle you know a side (adj) and an angle (25°) we want to know the opposite side

adj

opptan

A C

B = 6

25°

b

80.2

647.0

625tan6

25tan

A

A

A

A

Another tangent example

If you know a side and an angle, you can find the other side.

adj

opptan

CA = 6

25°

b

B87.1247.0

625tan

6

625tan

B

B

B

B

An application

65°

10m

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

4.21

14.210

65tan1010

65tan

opp

opp

opp

opp

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 = 10A

25°

b

B

Lets do sin first

we want to find angle a since we have opp and hyp we

use sin

hyp

oppsin

C = 10

a

b

B

A = 5

30

5.0sin10

5sin

a

a

a

And one more sin example

find the length of side A We have the angle and

the hyp, and we need the opp

hyp

oppsin

C = 20

25°

b

B

A 45.8

2042.0

2025sin20

25sin

A

A

A

A

And finally cos

We use cos when we need to work with the hyp and adj

so lets find angle bhyp

adjcos

C = 10

a

b

B

A = 4

42.66

4.0cos10

4cos

b

b

b

23.58 a

66.42 - 90 a

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

C = 5

65°

B

So lets calculate

so Spike will have fallen 4.53m

C = 5

65°

B

53.4

591.0

565sin5

65sin

65sin

opp

opp

opp

opp

hyp

opp

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

2m

88°

B

So lets calculate

Lucretia’s walkman fell 57.27m

2m

88°

B

27.57

264.28

288tan2

88tan

88tan

opp

opp

opp

opp

adj

opp

What are the steps for doing one of these questions?

1. Make a diagram if needed

2. Determine which angle you are working with

3. Label the sides you are working with

4. Decide which formula fits the sides

5. Substitute the values into the formula

6. Solve the equation for the unknown value

7. 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:

The second triangle

note that they share a side 45m long

a and b are heights!

60°

45m

40°

b

a

The First Triangle

We are dealing with an angle, the opposite and the adjacent

this gives us Tan

60°

45m

a

77.94m a

451.73a

4560tan45

60tan

a

a

The second triangle

We are dealing with an angle, the opposite and the adjacent

this gives us Tan

45m

40°

b

37.76mb

450.84b

4540tan45

40tan

b

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